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 Title:
 Problem on Pellian equations
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 Greiner, John William, 1945
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 1958
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 English
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 vi, 145 leaves : illus. ; 28 cm.
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Integers ( jstor ) Mathematical procedures ( jstor ) Mathematical tables ( jstor ) Mathematical theorems ( jstor ) Necessary conditions ( jstor ) Number theory ( jstor ) Prime numbers ( jstor ) Sufficient conditions ( jstor ) Symmetry ( jstor ) Diophantine analysis ( lcsh ) Dissertations, Academic  Mathematics  UF Mathematics thesis Ph. D
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 Thesis  University of Florida.
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 Bibliography: leaf 143.
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 Manuscript copy.
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 Vita.
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PROBLEM ON PELLIAN EQUATIONS
By
JOHN WILLIAM GREINER
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
February, 1958
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ACKNOWLEDGMENT
The writer wishes to express his sincere appreciation to Professor Edwin H. Hadlock, Chairman of his Supervisory Committee, for a generous contribution of time, energy, and helpful criticism throughout the preparation of this work and to Professors W. R. Hutcherson, F. W. Kokomoor, D. E. South, and L. E. Henderson, all of whom served as members of his Supervisory Com:nittee. To Professor Hadlock is due a large measure of gratitude for sympathetic encouragement and guidance during the writer's entire doctoral program. Finally the writer wishes to acknowledge that without the help of his wife, Lillian Greiner, his graduate studies could not have been begun and that without her continued confidence in him his graduate studies could never have been completeC.
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TABLE OF CONTENTS
ACKNOWLEDGMENT . . . . . . . . . . . . . . . . . . .
LIST OF TABLES . . . . . . . . . . . . . . . . . ..
Chapter 2 2
I. THE EQUATION nx dny =1, WHERE 8n'dn = D,
A PRODUCT OF DISTINCT ODD PRIMES . . . . . .
II. CONDITIONS FOR ONE OF 8 1  dy2 = 1 TO HAVE
A SOLUTION WHERE 5 'd = D, A PRODUCT OF
TWO DISTINCT ODD PRIMES . . . . . . . . . .
III. CONDITIONS FOR ONE OF 8 x2 _ d l2 = 1 TO HAVE
A SOLUTION WHERE 8 'd = D, A PRODUCT OF
THREE DISTINCT ODD PRIMES . . . . . . . . .
IV. CONDITIONS FOR ONE OF 6 x2  d1y = 1 TO HAVE
A SOLUTION WHERE 8 1di D, A PRODUCT OF
FOUR DISTINCT ODD PRICES . . . . . . . . .
BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . .
BIOGRAPHICAL SKETCH . . . . . . . . . . . . . . . . .
Page ii IV
1.
76
143 144
I
V.
V t
III
A
LIST OF TABLES
Table Page
1. Data for Example Illustrating Theorem la . . . 25 2. Data for Example Illustrating Theorem la . . . 26 3. Excludents of Equations (72)  (78) . . . . . 43
4. Conditions to Exclude Equations (74)  (78)
when X * 2 = (pp3) = +1,
P1 ; 1 (mod 4) . . . . . . . . . . . . . . . 49
5. Conditions to Exclude Equations (74)  (78)
when A y1, (pljp2) = (p 3) +1,
P2 =3 3 (mod 4) . . . . . . . . . . . . . . 53
6. Conditions to Exclude Equations (73) and
(75)  (79) when X y , (pip2) =
(pl1p3) = +1, and p1 3 (mod 4) . . . . . 62
7. Conditions to Exclude Equations (73) and
(75)  (78) when X, ( 2
y10 (p11p2)
(pljp3) = +1, and p2 =3 (mod 4) . . . . . 65
8. Conditions for Equation (74), p2p 3 X2  1 2 =
1, to Have a Solution when X y . . . . 67
9. Conditions for Equation (76), p1p3x2 p2y2
1, to Have a Solution when y . . . . . 69
10. Conditions for Equation (78), p1p2x2 P p y2 =
2 3
1, to Have a Solution when 1 y1 . . . . . 72 11. Excludents of Equations (119)  (134) . . . . 77 12. Conditions to Exclude Equations (122)  (134) when NJ y1, (pIjp2) (p11p3) (IP 4) =
+l, p 1 1 (mod 4) . . . . . . . . . . . . . 92
iv
LIST OF TABLES  Continued
Table Page
13. Conditions to Exclude Equations (122) (134), when X y2 Ip3 ( 4
+1, p2 P3 ! P4 m 3 (mod 4) . . . . . . . . 97 14. Conditions to Exclude Equations (122)  (134), when X y1, (plIP2) = (pllp3) = (pIp14) =
+1, p2 p 3 m 3P4 1 (mod 4) . . . . . . . 101 15. Conditions to 2Exclude Equations (122)  (134), when XJ y1, (pal p2) = (PlIP3) = (P1P4) =
+1, P2 3p3 r P4 * 1 (mod 4) . . . . . . . 102 16. Conditions to Exclude Equations (122)  (134), when N yl, (PlIp2) (p11p3) ( pp4)
+1, 3P2 P3 m P4 1 (mod 4) . . . . . . . 103 17. Conditions by Which Eqation (121) Has a
Solution when X, yl, and (p1lp2) =
(p1lp3) (PlIP4) +1 . . . . . . . . . . 105 18. Conditions to Exclude Equations (121) and (123)  (134) when X Y y2 1P) P
(p3Ipl) = (P4P) = +1, p1 3 (mod 4) . . . l0
19. Conditions to Exclude Equations (121) and
2
(123)  (134) when X, y , (p1lp2)
(pllp3) (PlIP4) = +1, p2 p3 P P4 1
(mod 4) . . . . . . . . . . . . . . . . . . 111
20. Conditions to Exclude Equat ons (121) and (123)  (134) when X, # yl, (plIp2) =
(Pdp ) = +(PlIp4) = +1, 3P2 w 3P3 M P4 E 1
(mod 4 . . . . . . . . . . . . . . . . . . 115
21. Conditions to Exclude Equations (121) and
(123)  (134) when X y , (PIP2)
+(pp3) = (p11P4) =+1, 3p2 pa 3P4 1
(mod 4? . . . . . . . . . . . . . . 116
LIST OF TABLES  Concluded
Page
Conditions to Exclude Equations (121) and
(123)  (134) when 2 , (PlP2)
(pl1p3) = (plIp4) = +1, 2 3P3 = 3p a 1 (mod 4) . . . . . . . . . . . . . . . . . . . 117
22. 23.
24. 25. 26.
11
124 133
,~.
~
9 p
 o F '
vi
Table
Conditions under Which Equation (122) Has a Solution when NJ Yl, (PlIP2) =" l (p 3Ip) (p I 4) = +1 . . . . . . . . . . . . . . . .
Conditions to Exclude Equations 121)  (128) and (130)  (134), when )J / yl, and (p p21p 3 =(pp21) (P3p41p) = +1 .
Headings for Substitution in Tables 17, 23,
and 24 . . . . . . . . . . . . . . . . . . .
Congruences Showing Insolvability, Equations
(121)  (131), (133), and (134) . . . . . . .
CHAPTER I
THE EQUATION 8nX2  dny2 = 1, WHERE Bn.dn= D,
A PRODUCT OF DISTINCT ODD PRIMES
If the integers t and u satisfy the Pellian equation
(1) t2  Du2 = ,
where D is a natural number not a perfect square, the number t + uz/D
is called a solution of the equation (1). Among the infinite number of solutions of (1) there is a least positive solution t1+u1,/, in which t1 and ul have their least positive values.1 This number ti+u /D is called the fundamental solution of equation (1).
All solutions of (1) with positive t = tn and u = un are given by the formula2
(2) tn + ugD = (t1 + u/D)n , n = 1,2,t 
where
1Trygve Nagell, Introduction to Number Theory (New York: John Wiley and Sons, Inc., 1951), p. 197.
2 lbid.
2
(3) tn= + k 21 t(k)2kulkD
and
(4i) un k (kl)t2k+uklDkl.
Let . Let t1+ul%,D be the fundamental solution of (1) with tj odd. Then tn of (3) is odd and un of (4) is even, where in the expansion of (2), tn is found by equating all terms free of\, and un is found by equating all terms containing\/ f. Further tn > un'
Proof. Since tj is odd, tj is odd. Because ul is even, then by relations (3) and (4), it follows that tn is odd and un is even. Since tn+un4/5 is a solution of (1), n = 1, 2, 3,  2 t Du = 1. It follows that tn = Du + 1,
and hence t > Du2. Since D > 1, then tn > unTheorem 1. Let equation (1) have the fundamental solution ti+ui,/D in integers tj and ul, where
(5) t1 = 2X + 1 and D = G2H,
G2 being the largest odd square dividing D, and H is the product of distinct odd primes, plp2...p.1 Then there are at most 28 possible equations of the form
lLeonard Eugene Dickson, Studies in the Theory of Numbers (Chicago: The University of Chicago Press, 1930), p. 35.
IM1
(6) nX2 dny2 1
in which equation (6) has a solution in integers x =xn and y = yn, where
(7) t = 2X + 1
and un are given by (3) and (4) respectively,
(g) y2 (X,G2 2/4), 27yn = Gun, d2 = H,
and n, 8n, and dn are all integers, n = 1, 2, 3, .., with (xn, dn) = 1.
Proof. Since tj is odd, it follows from (1) that u is even. By Lemma 1, tn is odd and un is even. Since relation (2) gives all solutions of (1) with positive tj and ul, it follows from equation (1) that t 1 = 0 = G2Hu (mod 9), and un is a multiple of 4. Thus
(9) un = 4Mn
defines an integer Mn. It follows from (1), (5)2, (7), and
(9) that 4 + 4x + 1 = G2H(16 2) + 1, so that
(10) ( + l)= 4G2M2H.
Now let y be the g. c. d. of X and 4G2M . Then dn and xn
I.
4
are defined by
2
(11) = dnyi and 2Mn = xnyn
respectively, where dn and xn are relatively prime. Hence by (10) and (11), dnyi( + 1) = x y2H, or
(12) dn(? + 1) = 2H.
Since (dnxn) = 1, dn divides H in (12), so that (8)4 defines an integer Sn' From (8)3, which is (11)1, (8)4, and (12), it follows that xn and Yn are solutions of (6). Using (9) and the fact that y2 (,G2 2), (9), results. Also (8)2 follows from (11)2 and (9). Finally, since there are 28 divisors of H, there are 29 values of 8n, and hence at most 28 equations (6).
Henceforth let G = 1 so that H = D, an odd natural number which is squarefree. Corollary 1. Let Pn be a positive divisor of D. Let
(13) tn = 2(D/P2)M + 1 and un = 2BnMn
be given by (3) and (4) respectively where (Bn,D/Pn) = 1, Bn and Mn both > 0. Then
(14) xn = Bn and yn = Mn
are solutions of
5
(15) Pnxn  (D/Pn)yn* 1,
if and only if tn and un of (13) are solutions of (1).
Proof. Let tn and, un of (13) be solutions of (1). From (7), (S)1 and (13),
yfi w (fr,4BjMn/4) =no
n
since (BnD/Pn) = 1. Thus by (7), (5)3, and (13), (D/Pn)M = dn 2 and hence dn = D/Pn* Then by (8)4, Bn = Pn so that (6) takes the form (15) with solutions xn, yn. Therefore, by
(8)2 and (13)2, it follows that 2xnMn = 2Bann; hence xn = Bn*
Conversely, let xn and yn of (14) be solutions of
(15). Then from (13)  (15),
2j Du2 + D2 4D4 + 1  D(4B MS)
n Pn
4DM (DM2 + pn  P ) + P 4Dy Pn +n  P n ]
p2
n
n4Dy yP n nx n)+P P
= _ _ _ _ _ _ _ _ __ _=_ _.
P2 p2
n n
j ,
6
Hence tn and un of (13) are solutions of (1). Example 1. Let the given equation be
t2  209u2 = 1
where 209 = 1119 = D and having the fundamental solution tj + uj\/b~  46551 + 3220\/209. Hence t1 = 46551  2Xi + 1 = 2(23275) + 1 = 2(19352) + 1, which is of the form (13) where D/P1 = 209/P1 = 19, so that Pi = 11 and Ml = 35. Also, 3220 = ul = 2B,(35), and hence B1  46. By (14) and (15), x, = 46 and yi = 35 is a solution of 11x2  19y2 = 1.
Checking: 11(462)  19(352) = 23276  23275  1. Example 2. Let the given equation be t2  65u2 . 1
with the fundamental solution 129 + 16\/;, where 2X1 + 1 = 2(64) + 1 = 2(1.2) + 1; hence by (13), and D/P1  65/P1 = 1, so that P, = 65 and M1 = ,. ul = 2B1(8), and hence B1 = 1. By (14) and (15), y = 8 is a solution of
ti = 129 = M, . g2, Also 16 = x, = 1 and
65x2 2. 1.
Checking: 65(1)2  (g)2 = 1. In addition t3 + u3 \/D = 9596369 + l065009\S7. Hence
7
t3 9586369 = 2X3 + 1 2(4293194) + 1 = 2(120722) + 1, so that by (13), M = 20722, D/P3 = 65/P3 = 1; thus P3  65 and M3 = 2072. Also 1065008 = U3 = 2B3(2072); hence B = 257. By (14) and (15), x3 = 257 and y3 = 2072 is a solution of 65x2  y2 = 1.
Checking: 65(2572)  (20722) = 4293185  4293184 = 1. Corollary 2. If D contains at least one odd prime p = 3 (mod 4), then equation
(16) y2  Dx2 W l
of (6) cannot arise with solutions xn and yn'
Proof. Equation (16) is a special case of (6) where Sn = D and dn = 1, and by Theorem 1 has solutions xn, yn' But y2 = 1 (mod p) has no solution. Hence (16) cannot arise. Corollary 3. Let D contain no prime factor p = 3 (mod 4), and tj of (13) be given by tj = 2MJ + 1. Then equation (16) of
(6) arises having the fundamental solution Yl + x1\/, where yl = M, and x1 = B1 are given by (13)2'
Proof. By hypothesis and (13)1, D/P1 = 1, so that P1 = D; hence, by (14) and (15), x, = B1 and Y, = M1 is a solution of (16). To show that y1 + xi\/6 Is the fundamental solution of (16), let its fundamental solution be Y + X\/U. Then
(y + X V5)2 = y2 + DX2 + 2YX V
is the fundamental solution of (1) where
t1 MY2 + DX2
and u, = 2YX.1
It follows by hypothesis and (17) that
2M + 1 = 2yl + 1 ti = Y + DX
Since Y + X\/5 is the fundamental solution of (16), then from
(19) it follows that 2y2 + 1 = DX2  1 + DX2 or 2y2 + 2= 2DX2, so that
y + 1 = DX2.
Since y, + xi\/ is a solution of (16) it follows from (16) and (19) that Dx2 = DX2; thus x, = X. Hence from (13)2, (14), and (17)2, it follows that yj = Y.
An example of Corollary 3 is Example 2, page 6, in
which D = 65 contains no prime factor p 3 (mod 4) and ti = 2(52) + 1. The solution x= 1, y1 = S is the fundamental solution of
y2  65x2 = .
Example. The equation t2  5u2 _ 1 in which D 2(22) + 1, is another example of Corollary 3. Also by (13), 4 = ul = 2B1Ml = 2B1(2), so that by (14) and Corollary 3, y1 = 2, x1 = 1 is the solution of
= 5, ti = Hence Ml = 2. B1 = 1. Hence fundamental
I
1Nagell, op.Olat., p. 201.
(17)
(18)
(19)
9
y2 5x2 .
Corollary 4. Let tj + uliV/ be the fundamental solution of
(1) where tj = 2N, + 1, yj = (xl,ul/4), and 2xlyl = Ul, then
(20) X2 Dy2 = 1
of (6) cannot arise with the solution xl, y1.
Proof. Equation (20), which is of the form (1), is a special case of (6) with n = 1 where 81 = 1 and di = D, and by Theorem 1 has the solution xi, yl. But by hypothesis u1>1yll and ui>x1l, and by Lemma 1, t1> ul. Hence ti>xi. Therefore jxlI+lyll\/D < ti+u1\/D, which is not possible, since ti+uiV5 is the fundamental solution of (1) where ti and u, have their least positive values. Hence (20) cannot arise with the solution xi, y1. Corollary 5. If in (1), D = p is a prime = 1 (mod 4), then equation (16), y2  Dx2 = 1, of (6) arises and has a solution, X11 y1, given by (9).
Proof. Let tl+uV\/D ti+ui\/p be the fundamental solution of (1) in which ti is odd, and where D = p is a prime =1 (mod 4). Thus relation (8)4 in Theorem 1 gives 81di p. Therefore, since p is a prime, one of 81 and dl is unity. Hence (6) becomes either
(21) x2  py =
10
or
(22) Y  px2 = 1.
By Theorem 1 one of (21) and (22) arises and has a solution. But by Corollary 4 equation (21) is not possible. Therefore equation (22) arises and has a solution, xi, y1, given by (S).
We note that Corollaries 1 and 4 imply that if Pn is a positive divisor of D different from 1, then
PnX2  cD/Pn)Y2 = 1
of (6) must arise. The following corollaries place definite restrictions on the form that (6) may take according as the subscript n in Theorem 1 is odd or even, and according as 8n 1, D or # 1, D, The following Corollary 6 considers the relations of Theorem 1 when n is even, and places no restrictions on the nature of D other than that in its original definition, where it is a squarefree odd natural number. Corollary 6. Let ti+u1%/i be the fundamental solution of (1) where t1 = 2N + 1. Let t2k = 22k + 1, 22
1 1 2k 2X~k + 11 2k = kuk/)
and 2x2ky2k " u2k, where t2k and u2k are defined by (2), n = 2k k =1, 2, 3, *. Then
(23) k k = Dy2
and the equation
X2  Dy2 = 1
( 24)
iv
of (6) always arises and has solutions x x2k' Y = Y2k, where
x2k = tIk and Y2k = uk. Further x2+y2\/ is the fundamental solution of (24) which is of the form (1).
Proof. In (2), since n = 2k, k = 1, 2, 3, , t2k and u2k are given by
(25) (ti + u\/D) 2k  t2k + U2kVD'
Now by (2)
(t +u\/U)2k = [(t1+u\,/)k )2. (tk+ukV 2 = tk+Dk + 2tkuk\/ and it follows that, from (25),
(26) t2k = t2 + Du2 and U2k = 2tkuk*
Hence by hypothesis, (26)1, and the fact that tk and uk are solutions of (1),l
t2k 2c +Du  ( +1) + Du1 1 2Du
t2k"2k1 t+Du (Du
2 2 2 2'
so that
(27) A2k 2 *
From the hypothesis and (26)2, and (27),
~2I = (Xk U2/) (2 42 2 U2
Y2k 2k, u2k/4) = (Dui, 4t kuk/4) = Uk,
since (D, t2) = 1. Thus
1Nagell, op. cit., p. 201.
t  
j
12
(28) 7ik ud
and hence from (27) and (28) relation (23) follows. Consequently in (S)3, d2k = A2k/y7k  Du/uk = D. Hence from (8)4, 82kD = D, so that 82k =*  Thus (6) reduces to (24) which is of form (1). Moreover, from (26)2, (29), and the hypothesis where 2x2ky2k = u2k, it follows that
(29) X2k m k '
Hence from (2S) and (29), x2k, y2k are solutions of (1). Thus when k = 1 in (29) and (29), x2 + y2\/I is the fundamental solution of (24) which is of the form (1). Example. Let the given equation be t2  39u2 = 1
with the fundamental solution tl+ul\/ = 25 + 4\/39. By (25 ), k = 1, t2+u VD = 1249 + 200\/59. By (7), t2 = 1249 = 2X2+ 1 = 2(624) + 1 = 2(42.39) + 1. Hence, X = Du, as asserted in relation (23). Thus by (8)3 and (8)4, d2 = 39 = D, and 82= 1, so that equation (24) of (6) arises. By (28) and (29), for k = 1, Y2 = ul = 4 and x2 = t1 = 25 is the fundamental solution of x2  39y2 = 1, which is of form (1), the given equation. Corollary 7. Let ti+ui/DZ be the fundamental solution of (1) where ti = 2X1 + 1, and D contains no prime factor p =_ 3 (mod 4). Let t2k+1  2X2k+l + 1, y2k+1 " (X2k+l, uk+1/ ), and
A
13
2y2k+lx2k+l =u2k+l, where t2k+1 and u2k+1 are given by (2) for n = 2k+l, k = 0, 1, 2, ... Let equation (16), y2  Dx2
 1, of (6) arise with the fundamental solution y, + xi\/D. Then
(30) X2k+1 = Y7k+l and y2  Dx2 =l
of (6) has the solutions y2k+l X2k+lProof. By hypothesis equation (16) of (6) arises with the fundamental solution yi + xlx/D. Then (y1+x1\/)2 = tl+ul\/ is the fundamental solution of (1). Moreover, for k = 0, 1, 2,
(yl + xlV k+l Y2k+l + X2k+lVD are solutions of (16).2 Since
(y2k+l + 2k+1\5D2  Uyl + X 2k+12 = +
= (t1 + u /5)2k1l (t2k+l + U2k+1 I s and thus
2 = 2 + x2 +2
2k+l + X2k+l\/ 2 k+1 + xk+l + 2k+lx2k+l
= t2k+l + u2k+l\'D,
it follows that
1Nagell, op. cit. 2Nagell, op. cit.
14
2 2
(31) t2k+l = Y2k+1 + Dx2k+l and u2k+l =
From (31), the hypothesis X2k+l = (t2k+l  1)/2, and the fact
2 2
that y2k+1  2k+= 1, it follows that
(y2 + x2 D)l y2 + (y2 + 1)i
X2k+l 2k+l  2k+1 2k+l , or
2k+1' 2 2
(30) From (9)3, and n = 2k+l, d 2 2
3Pd2k+ly2kl ' X2k+l Y2k+l'
so that d2k+1 = 1. By (8)4, with H = D, 82k+1 = D. Hence from (6), equation (30)2 arises with solutions y2k+1s x2k+1' Corollary 8. Let ti+uVD be the fundamental solution of (1) where t1 = 2X, + 1, and D contains no prime factor p Si 3 (mod 4). Let t2k1 22k+1+ , y2k+ = N2k+l, u2k1/)
u =2y X ,and =2 whretd u
2k+l 2k+1 2k+1 2k+l 2k+l where t2k+1 an 2k+l
are given by (2), n = 2k + 1, k = 0, 1, 2, * . Then equation (30)2, y2  Dx2 = 1, of (6) arises with solutions Y2k+lt X2k+l*
Proof. By hypothesis and (8) , k = 2
d 2 so that d2k+ 1. Hence by (9)4, 22k+l =
and (6) takes the form (30)2 with solutions y2k+1' x2k+10 This completes the proof of the corollary.
If the integers xi and y1 satisfy the equation
(32) six_ d1y 1
where x and y1 have their least positive values and 8 1d = D
b~A.. ' 6"
15
of (1), 81 D, 1, then we define the number
X, + y
to be the fundamental solution of (32). Also we define xn and yn by
(33) x + yn xV + y
where n = 2k+l, k = 0, 1, 2, ', and xn is found by equating all terms containing \/j, and y is found by equating all terms containing \/dj. In (33) n cannot be even since the right side of (33) for n even is of the form U + V\/D, where U and V are integers and D = 81d, 81 9 D, 1, which relation Nagell has shown to be impossible.1 In this connection we quote two lemmata by Nagell without proof.
Lemma 1. Let x, y, x , yl, a, b, and a1 be rational
numbers # 0, such that Va, Vb and Va are irrational.
Then we can never have a relation of the form (15) X\/K + y/F = x1VW + y1.
Lemma 2. Let x, y, x1, y1, a, b, a1 and b1 be
rational numbers 9 0, such that \/a, \/b, .a, V
\/a and Vab1 are irrational. Then the relation
(16) x\/a + y /b = x1Vi/ + yg/b
lTrygve Nagell, "On a special class of Diophantine equations of the second degree," Arkiv For Matematik, Vol. 3, No. 2 (1954), p. 53.
U
2p
16
holds only in the following cases: It is either x\/a =
xJ\/IT or X\/a =yl\/bl.
Corollary 9. Let D = 81d, where 61 1,D and tj of (13)
be given by t= 2d1M2 + 1. Then equation (32), 81x2 dly2 = 1, of (6) arises having the fundamental solution xl, yl where x, = B1 and y1 M1 are given by (13)2.
Proof. Since 61 / D,1 then d 1,D in the relation
D = 8 1d1. By hypothesis and (13)1, D/P1 = d1 so that Pi = D/di = 61. Hence by (14) and (1i), x, = B1, y1 = M, is a solution of (32). To show that x1, y is the fundamental solution of (32), let its fundamental solution be X, Y.
Then, since the number
(xV/F + Y )2 = 8X2 + dY + 2XY/D is the fundamental solution of (1), where
(34) ti = 81X2 + d y2 and u1 = 2XY,
it follows by hypothesis, (14) and (34) that
(35) 2d M + 1 1 2d y + 1 = ti = 8 + d .
Since X, Y is the fundamental solution of (32), then from
(35), it follows that 2d y2 + 1 = (1 + d Y2) + d12 or 2dy 2 = 2d Y2. Hence y1 = Y. Thus from (13)2, (14) and
libid., p. 54. 2Ibid., p. 53.
17
(34)2, it follows that xl  X. This completes the proof. Corollary 10. Let t, + ul\/D be the fundamental solution of
(1) where t, = 2X1 + 1 and D = 61d1, 81 # 1,D. Let t2k+l =
2 2
2X2k+1 + 1, Y2k+ . (X2k+l, u2k+1/4), 2x2k+ly2k+l =U2k+, where t2k+1 and u2k+l are given by (2), n = 2k+1,k 0, 1, 2, **. Then all solutions x2k+lp Y2k+1 of (32), six2  2
= 1 are obtainable from
 2k+l
(36) x2k+l\/61 + Y2k+l \/Vi = (x1V~ + y ,k/ l
k = 0, 1, 2, .. , where x1V/&j + yi\/ is the fundamental solution of (32), and
(37) x2k+1 = tkxl + ukyldl and Y2k+l tkyl + ukx1i8, k = 0, 1, 2, , where to = 1 and uo = 0 when k = 0.
Proof. Since 81 # 1, D, then by Corollary 9, equation
(32) of (6) arises with the fundamental solution xi\ + yl\/. Then
(38) (xlVX + yi\ =) t1 + u1\/D
is the fundamental solution of (1).1 From (33), n = 2k+l, k = 0, 1, 2, . , and relations (38) and (2), and the fact that 81d1 = D, it follows that
1 Ibid.
1s
(xl V + yVy)2k+1 . (Xi\8 YiVdY)2k(xl/8 + y
(39) = (tk + uk\/Th (xVlY + y 1V )
= (tkxl+ukyldl) \/8 + (tkyl+ukXl~l) VdT, where we define to  1 and uo = 0 when k = 0. Also from (33), n = 2k+1, k = 0, 1, 2, ,
(40) (x1 + yiVd)2k+l = x2k+ 1j + y2k+lVHence from (39) and (40), relations (37) are established. We next show that (37) are solutions of (32). For, from (37), the hypothesis 81d= D, and the fact that x1Vy7 + y is
the fundamental solution of (32), 2lx k+ldly~k+l = 1(tkxl + ukyldi)  dl(tkyl + ukxl8l)
=81t 2x 2+ l2y2d2
k x1 + uyd 1+ 28ldltkukxlyl
 ditkyl  d1uix(8S  2Sldltkukxlyl
= t2(x81X  dly2) + u2(5 d 2y  2d x )
= t2.(j) + u2(Ddly2  D81x2)
= t2 + Du4(djy 2  six2) = t Du2 = 1. Hence (37) give solutions of (32).
To show that relation (36) gives all solutions of (32) with positive x2k+l and y2k+l, where x 6 + y1 j is the
'~
19
fundamental solution of (32), assume that u\/8 + v\/F is
another solution of (32) with positive u and v and not obtainable from (36).1 Then it follows that for a positive number n = 2k+1, k = 0, 1, 2, ..,
(41) (xi\/j + y1V\/)2k1< u T+ vVd< (\xi/+ y1Vd 2k3.
Hence by (36),
(x2k+lV/l + Y2k+l1) < u\/ + y\/d
< (x2k+l + Y2k+lVdl (xlV \ +y lV )2 Now, x2k+1  Y2k+l\/dl is a positive number, for since
(x2 7  k+l6 + Y2k+l\/) =
2 2 2
X2k+l 1 y2k+ldl = 1,
and (xpk+1V j + Y2k+lVdl) > 0, x2k+lV  2k+\/dl > 0.
Multiplying through by x2k+lV81  Y2k+lV we get
2k+11  y2k+1d1 < (u\/j + v\/)(x2k+lV1  Y2k+1\/) 2k+11  y2+1di)(x1vg + y V72 or, by (39) and the fact that x2k+l, Y2k+l, k = 0, 1, 2, *.., are solutions of (32) it follows that
(42) 1 < (u\/ + v\/') )(x2k+lV  Y2k+lV1) < tj + ul\/D Let
lNagell, Introduction to Number Theory, p. 198.
20
(43) (u\/V + v\/1,)(x2k+V1  2k+lVdl)  w + Z\/$ where, since 81di = D, w = ux2k+181  vy2k+ldi and z = Vx2k+l  uy2k+1' Then it follows that
(44) (u\/8  vV)(x2k+lVl + y2k+l\/dl) =
ux2k+11  vy2k+ldl + (uy2k+1  vx2k+l)V/
= w  Z\/.
Hence multiplying (43) by (44), we get
(u281 _ v2dl)(k+l01  Y2k+ldl) = W2 Dz2,
or, from the assumption that uVI + v\/d is a solution of
(32),
(45) 1 * 1 = i w2  Dz2
Thus w + z\/5 is a solution of (1). Hence, using (42) and
(43),
(46) 1 < w + zV < t 1+ u1 ,
which, if w and z are both positive, is not possible, since ti + u1\/5 is the fundamental solution of (1). Therefore, we must show that w and z are both positive, if u, v is not to be a solution of (32). To that end, we note from (45) that
0 < w2  Dz2 = (w + z\D)(w  zVD) = 1;
w + z VD) (w z VJ
R AMM
hence, since w + z\/ is a positive number > 1 from (46),
(47) 0 < W  z\/3 = 1/(w + z\/5 < 1.
Now, from (47),
(48) 0 < W  z\/5 < 1;
hence,
(49) w < 1 + z\/5.
Replacing w in (46) by the larger number 1 + z VD of (49), we find that 1 < (1 + zV) + z\/5, or 0 < 2z\/5. Thus, since \/D > 0, it follows that z > 0; hence, from (49), w > 0. Therefore, the inequality (46) does not hold. It follows then that the inequalities (41) do not hold, and hence that uV + vVd1 is not a solution of (32). This completes the proof of the corollary. Corollary 11. Let tj + ulV be the fundamental solution of
(1) where tj = 2X, + 1, and D = 81*dl, 81 # 1,D. Let t
2 2
2X2k+l + 1, y2k+l " (X2k+lu2k+1/4)9 2x2k+ly2k+l u2k+l where t2k+l and u2k+l are given by (2), n = 2k+l, k = 0, 1, 2, 0  * . Then
(50) x2k+l " d1y2+1 = d2k+ly2k+l,
k = 0, 1, 2, . , and the equation
(51) 82k+lx  d2k+ly 1
22
of (6) which arises with solutions x = X2k+lo' Y Y2k+1' takes the form (32), 81X2 _dy2=.
Proof. Since the hypothesis of this corollary is
that of Corollary 10, we may use (38), (40), and (2) to get
2k1 + ~+Vd) xl d,2k+132
(x2k+lV/67 + Y2k+l 1) = x1/81 + l
(52) = [(x2\/ +2k
= (t1 + u \/D)k+
= t 2k+ + U2k+lVD
and also
2 2 +2
x2k+l 78y2k+l \d) 2 "X2 kls+ Y2+11 + 2x2k+2k+ Thus, it follows that
(53) tak+l = 8 2k+ + dy2k+ and 2aki 2x2k+ly2k+lt
8lx2k+l + ly2k+l ndU2k1 =
which are solutions of (1). For from (53), the hypothesis 1di = D, and the fact that (36) are solutions of (32),
2 2 2 2 2 2 2
t2k+l  12k+l (8lx2k+l y21)  D(x2k+ly2k+i)
2 4 24 2 2
= 8lx2k+l + d y2k+1 + 2x2k+y12k+1D
2 2
2 2 2 2
= (81x2k+l  dly~k+1) = (1) = 1.
Hence by (53)1, the hypothesis X2k+l (t2k+l  1)12, and
IX159M B16 '; "w_ '1 V5 M
23
th at ht82 2
the fact that lx2k+l  dly2k+1 = 1, it follows that
Xkl (8 x2 + d y 2 )1 (1 + d y 2 )+ 2  1
1 2k+1 + 2k+l ( 1 + 2lk+1) +1 2k+1
2k+ 2 2
or
2
(54) X l2k+1 " dy2k+l2 2
Thus from (9)3 and (54), x2k+l'= d2k+1y2+1 = d1y2k+1, establishing (50). It follows from ()4 that 82k+l = 811 and hence by (6), when n is odd, equation (51) takes the form (32).
Finally, in order for an equation of (6) to arise
which is different from the original equation (1), it is seen by Corollaries 7  11 that n in Corollary 1 must be odd. In fact Corollaries 6  11 establish the following summarizing Theorem la. It is seen by these corollaries that the different forms of equations (6) to arise are equation (24),
x2  Dy2 =,
2
where t2k = 2Dy2k + 1, with solutions given by 2x2ky2k = U2k, and either equation (30)2,
y2  Dx2 = 1,
if and only if D contains no prime factors 3 (mod 4), and
2
= 2yk+ + 1, with solutions 2y2k+lx2k+l = u2k+1, 2r one of the 2s  2 possible equations (32),
24
2 2
81x  d.y = 1,
2
where t2k+l = 2dly2k+1 + 1, 8d1 = D, 8i 1,D, and where y2k+l is the largest square contained in (t2k+1  1)/2, and with solutions given by 2x2k+1y2k+l = u2k+1 Moreover, k = 1, 2, 3, *, or k = 0, 1, 2, a*, according as the subscript 2k or 2k+1 appears.
By (8)4 with G = 1, namely 8n'dn = D, it is seen that dn and in particular dl contains no square factor, since D
2
is a product of distinct odd primes. Hence from dly2k+l =
2
(t2k+l  1)/2 it is seen that y2k+1 is the largest square contained in (t2k+l  1)/2.
An example of Theorem la wherein equations (24) and
(30)2 of (6) arise is example 2 on page 6 in which D = 65 513 contains no prime factor p = 3 (mod 4), and Al (t,  1)/2 = (129  1)/2 = 64= 82 = y2 = dl1 by Corollary 8 and relation (8) 3. Hence y, = S and dl = 1. By (8)4, 81 = 65. Therefore (30)2 of (6) arises with the fundamental solution x, = 1, yl = 8. Also X2 = (t2  1)/2 = (33,281  1)/2
16,640 = 513.256  2 2 = d2y2 by Corollary 6 and
relation (8)3. Hence by Theorem la, y2 = 16 and d2 = 65. By (W)4, 82 = 1. Therefore (24) of (6) arises with the fundamental solution x2 = t1 = 129, y2 = ul = 16, as in Corollary 6. Thus let the two equations of (6) which arise with solutions be denoted by (A) and (B) where
1**' ?~ / **~
25
y2  65x = 1 x2  65Y2 = 1
equation (30)2, equation (24).
The following Table 1 is selfexplanatory.
TABLE 1
DATA FOR EXAMPLE ILLUSTRATING THEOREM la
n ta 2% + un = 2xnn = d y2
1 129 16 64 = g2
2 33,281 4,129 16,64o = 651
g,56,369 1, o65,00 4, 29,14 = 2072
2,215,249,921 274,767,936 1,107,624,960  65412g2
2 2
n xn yn 8n dn 8nx  dny = 1
1 65 1 (A)
2 129 16 1 65 (B)
257 2,072 65 1 (A)
4 33,281 4,129 1 65 (B)
Notes:
a) xj, yl is not a solution of (B) as shown by
Corollary 4, but by Corollary 7 is the fundamental solution of (A).
b) x = t2, Y4 = u, as in (28) and (29) of
Corollary (6) and are obtainable by (2), (29) and (29) for n = 2k, k = 2.
To illustrate that part of Theorem la wherein
equations (24) and (32) of (6) arise consider the equation
2 2 1
t 55u = 1 with t, + uV\/D = 89 + 125. Since X= (ti 1)/2 = (89  1)/2 = 44 = 4.11 = 11.22 d , k = O,
by (8) 3, y = 2, d, = 1. By (8)4, 8i = 5. Therefore (32) of (6) arises with the fundamental solution xi = 3, y1 = 2.
(A)
(B)
26
Also by Corollary 6 and relation (9)3, X2 = (t2  1)/2 = (15,941  1)/2 = 7,920 511122 55122 = Dy2 = 22 Hence by Theorem la, y2 = 12, d2 = D = 55. By (8)4, 82
1. Therefore (24) of (6) arises with the fundamental solution x2 = tl = 89, y2 = u1 = 12, as in Corollary 6. Thus let the two equations of (6) which arise with solutions be denoted by (A) and (B) where
(A) 5x2  11y2 = 1 equation (32),
(B) X2  55y2 = 1 equation (24).
The following Table 2 is selfexplanatory.
TABLE 2
DATA FOR EXAMPLE ILLUSTRATING THEOREM la
2
n tn = 2Xn + 1 un = 2xnyn N= dnyn
1 89 12 44 = 1122
2 15,941 2,136 7,920 = 55122
2,81?,609 380,196 1,409,g04 = 1135g2
501,74 ,561 67, 72,752 250,937,280 = 5521362
5 99,330,852,249 12,045,369,660 44,665,426,124 11.637222
n y n 8n dn Snx  dny 2
1 3 2 5 11 (A)
2 89 12 1 55 (B)
531 359 5 11 (A)
4 1 2,136 1 55 (B)
5 94,515 63,722 5 11 (A)
Notes:
a) x1, y1 is not a solution of (B) as shown by
Corollary 4, but by Corollary 9, is the fundamental solution of (A).
27
b) x4 = t2, Y4 = u2 as in (28) and (29) of Corollary
(6) are obtainable by (2), (29) and (29) for n = 2k, k = 2.
c) By relations (36) anj (37) of Corollary 10, k = 1, x3\/81 + y3V = (3\/5 + 2V1) = (89 + 12V7)(3V5 + 2V1T) = 531V5 + 359V\/ll.
d) By relations (52) and (53), k = 1, (x3V VY+ /Vi)2 = (531 V5 + 35 \/iT)2 = 2,919,609 + 390,196V5 = t3 + u3\ D' This result for t3 + D is verified by using relations (3) and (4).
In the following chapters use will be made of the preceding results to help establish the conditions under which a particular form of (6) will be solvable when D is a product of two, three, and four distinct odd primes. In Chapter 2, D pl*P2 where p1 < p2 are distinct odd primes. In Chapter 3 D = pl*p2'P3 where p1 < p2 < P3 are distinct odd primes, and in Chapter 4, D = pl*P2'*P34 where p1 < p2 < p3 < P4 are distinct odd primes.
Also the word "excludent" will be used in the following discussions, and it will refer to a condition which is sufficient for a particular one of the 28 possible forms of (6) not to arise.
The symbol (mip) is that of Legendre in which p is an odd prime and m is an integer prime to p and is defined to be the value +1 or 1 according as m is a quadratic residue or nonquadratic residue of p.
I
CHAPTER II
CONDITIONS FOR ONE OF 6 2 2 = 1 TO HAVE
A SOLUTION WHERE =16di D, A PRODUCT
OF TWO DISTINCT ODD PRIMES
Lemma 2. Let D = p1.p2 in equation (1), t2 _ 2
1, where pi < p2 are distinct odd primes. Let (1) have the fundamental solution ti + uVD where ti = 2N,+ 1. If one of the excludents listed to the right of the equations (55)
 (57) of (6), 1x2 _ dly = 1, holds, then that equation has no solution in integers x, y.
Equation Excludents
2 2 2
(55) Pip2x y NJ, l l A l
(56) plx2  2 =1, (PlP2)= 1 or (P21p1) =  1,
(57) P22  P1y " 1, (P21pl) =  1 or (p1lp2) =  J.
Proof. Since D is a product of s = 2 distinct odd primes, p1 and p2' there are 28 = 22 = 4 positive divisors of D. These are 1, p1, p2, and plp2. For tj = 2N, + 1 and ulthe fundamental solution of (1), the equation (6) of Theorem 1, where 81di = D = Pl*P2, has four possible forms, namely, (24), x2  P1P22 = 1, and the three equations (55)
 (57) above. By Theorem la equation (24) and one of (55)
(57) always arise from (6) and have integral solutions x, y.
29
We note the following about equations (55)  (57) of (6).
2
Let X, y1. Then equation (55) cannot arise. For
2
if it did, then by Corollary 7 and relation (30)1, X1 = yl,
2
a contradiction to the assumption that X, yi.
Let (56) arise from (6) with 81 = p1 and dl = p2.
Thus by Theorem 1 or Corollary 9, there exist integers x, y satisfying (56). It follows then from (56) that
(58) pix2 = 1 (mod p2) and py2 M  1 (mod pl).
Now define the integers p' and p" by
(59) p'p1 a 1 (mod p2) and p"p2 1 (mod pl).
Then (59) may be written as
ppx 2= p' (mod P2) and p"p2y 2 ;  p" (mod pl),
which in view of (59) reduce to
(60) x2 L p' (mod P2) and y =  pw (mod p1).
By (59)1, we may write
(p'Ip2) (p'p'p1lp2) = (pip2)Hence, if (pjp2) = (p'1P2) =  1, then (59), is not solvable, and therefore neither is (56). Similarly (56) is not solvable if (p2jlP) =  1. Thus the two excludents listed opposite
(56) are established. By similar reasoning the two excludents
V
1
30
listed opposite (57) are established, and lance equation (57) has no solution if either one holds.
We will make use of the following Lemma 3. The conditions
(61) (pilp) = (pgpi) = + 1,
where pi and pj are distinct odd primes, i, j = 1, 2, i # j, follow if and only if either
(62) (pilpj) = + 1 and pi a 1 (mod 4),
or
(63) (pipj) + 1 and pj3 3 (mod 4).
Proof. If (61) holds, (pj'pi) + 1 may be written as
(64) ( ) ) Pil)4P jKL) = (1)( 1)( i)IPij+l) = + .
Since, by (61), (pi Ipj) = + 1, (64) may be written as }(pji)  iypj+1)
(1) = + 1
which holds if either p1 = 1 (mod 4) or p3 z 3 (mod 4).
Conversely, if (62) holds, then we may write (p3p,) = (1p1)(p31p1) = (+l)(plpj) = (+l)(+l) = + 1. If (63) holds, we may write (p3jpi) = (pi1pj) = + 1, and the proof
31
is complete.
Theorem 2. Let D = p1.p2 where p1 < p2 are distinct odd primes. Let t + u V be the fundamental solution of (1),
2
where ti = 2Xi + 1. If NJ yl, and
(65) (pilpj) = + 1 and Pj = 3 (mod 4),
with 1,j = 1,2, I j, then the equation (56),
2 2
P)x  P2Y
or the equation (57)
p2x  Pl
has a solution in integers x, y according as i = 1 or 2, and where the solutions are obtainable by
(66) y2 =(Xi,/4) and 2x y1 = ul.
Proof. By hypothesis and Theorem la, one of the
equations (55)  (57) of (6) has a solution in integers x, y.
2
By Lemma 2 if X y2, equation (55) of (6) has no solution.
If i = 1 and j = 2 in (62) of Lemma 3, then
(67) (pilp2) = + 1 and p, = 1 (mod 4).
Hence by Lemma 3, (p2 p ) = + 1, and by Lemma 2 the conditions necessary for (56) to have a solution are satisfied. Examining the two excludents of (57) in Lemma 2 under
32
conditions (67), we find that excludent (57)l or (p21pl) (PlIP2) = + 1, and therefore it cannot hold. But excludent
(57)2 or (p1lp2) may be written as (ljp2)(p1lp2) = (11p2)(+l)  (lIp2). Thus if p2 m 3 (mod 4), excludent
(57)2 holds, and equation (57) of (6) has no solution. Hence,
2
if X, # yl and
(68) (plIp2) = + 1 and pi m 3p2 1 (mod 4), equation (56) of (6) arises and has a solution.
Also, if i = 1 and j = 2 in (63) of Lemma 3, then
(69) (p1lp2) = + 1 and p2 = 3 (mod 4).
Hence by (61) of Lemma 3, (p21pl) z + 1, and by Lemma 2 the conditions necessary for (56) to have a solution are satisfied. Under these conditions (69), excludent (57)2 or (p1Ip2) = (11p2)(P11P2) = (l)(+1) =  1. Thus (57) has no solutions
2
and hence does not arise. Therefore if NJ y1 and (69) holds, equation (56) arises and has a solution. Conditions (69) imply that either
(p1lp2) = + 1 and p1 i 3p2 F 1 (mod 4)), or (PlIP2) = + 1 and pl p2 3 (mod 4).
The former are conditions (6>). Thus both the former and the latter are implied in (6) for i = 1.
41'
33
By similar reasoning if i = 2 and j = 1 in the above proof, (65) is established whereby (57) of (6) has a solution, if NJ yl. By Theorem 1 solutions of (56) or (57) are obtainable from ()1 and ()2, n = 1  G, thus establishing (66).
Legendre proved that one of Mx2 Ny2 = 1 1 is
solvable if M, N are primes of the form 4n+3. "One can always deduce these theorems and others like them from a consideration of the VA in a continued fraction form where A = MN."l
The following examples illustrate Theorem 2. Example 1. Which of the following equations,
5x2 _ 11y2 = 1 and llx2  5y2 1, has a solution in integers x, y?
Let p, = 5 and p2 = 11, primes of the form 4n+l and 4n+3 respectively. Then the given equations have the forms of equations (56) and (57). Hence the equation (1) is
t2  55u2 = 1
whose fundamental solution is ti + u1VD =9 + 12\/55. Since ti = 89 = 2X + 1, N1 = 44 = 1122 a square. Also since (p P2) = (5111) = +1 and p2 = 3 (mod 4), the conditions X y2 and (65) of Theorem 2 are satisfied. Hence
1AdrienMarie Legendre, Theorie des Nombres (3rd ed., Paris: Chez Firmin Didot Fr6res, 1830), I, pp. 64 71.
34
by Theorem 2, equation (56), 5x2 11Y2 1t
has a solution in integers x, y. By (66), y1 = 2, xi = 3 is a solution. Checking: 5(3)2  11(2)2 = 45  44 = 1. Note: (p21p1) = (1115) = +1, but that (p1Ip2) = (5111) = 1. Hence one of the excludents of (57) in Lemma 2 holds, and therefore (57) is not solvable in integers. Example 2. Does the equation 11x2  19Y2 = 1,
have a solution in integers x, y?
Let p1 = 11 and p2 = 19, both primes of the form 4n+3. Then the given equation has the form of (56). The equation (1) is then
t2  209u2 = 1
where t1 + ui3 = 46551 + 3220V/T09.1 since t1 = 46551= 2X1 + 1, X1 = 23275 = 19.52.72 a square. Also since (PilP2) = (11119) = +1 and p2 = 19 = 3 (mod 4), the conditions (6r) of Theorem 2 for i = 1 are satisfied. Hence by Theorem 2, equation (56), 11x2 _ 19y2 = 1
has a solution in integers. By (66) y1 = 35, xi = 46 is a
A~IIIEMEAEEI
t1 + u\/ was calculated by the writer.
35
solution. Checking: 11(46)2  19(35)2 23276  23275 = 1. Note: (p21pl) = (19111) = 1, and hence by Lemma 2, (57) has no solution and hence does not arise. Example 3. Does the equation 19x2 _ 71y2 1
have a solution in integers x, y?
Let p1 = 19 and p2 = 71, both primes of the form 4n+3, then the given equation has the form of (56). The equation (1) is then
2 _ 1349u2 = 1
since D = 1349 = 1971 is a product of two distinct odd primes, and ti + u1\/ID = 203 918o6 65951 + 5 55200 34940 Vi349,' where ti = 2X, + 1. Hence X, = 101 95903 32975 = 71'(119835)2
a square. Also since (p1lp2) = (19171) = +1, and p2 = 71
3 (mod 4), the conditions (65) of Theorem 2 for i = 1 are satisfied. Hence by Theorem 2, equation (56),
19x2  71y2 = 1,
has a solution in integers x, y. By (66), y = 119835, x, 231652 is a solution. Checking:
19(231652)2  71(119835)2= 101 95903 32976  101 95903 32975 = 1.
Arthur Cayley, Collected Mathemptical Papers (13 vols.0 Cambridge: The University Press, 1997), XIII, pp. 430  467.
36
Note: (p2p91) = (71119) = 1; hence (57) has no solution by Lemma 2 and therefore does not arise. Examrole 4. Does the equation
193x2 _ 7y2
have a solution in integers x, y?
Let p1 = 7 and p2 = 193, primes of the form 4n+3 and 4n+l respectively, then the given equation has the form of
(57). The equation (1) is then t2  1351u2 = 1
with ti + u1\I = 6175 + 169 V1351, where tj = 271 + 1.1 Hence X1 = 307 7.212 a square. Also since (p2 1P) (19317) = +1 and p1 = 7 r 3 (mod 4), the conditions of Theorem 2 for I = 2 are satisfied. Hence by Theorem 2, the given equation (57),
193x2 _ 7y2 = 1,
has a solution in integers x, y. By (66), y1 = 21, x1 = 4 is a solution. Checking: 193(4)2  7(21)2 = 3088  3087 =
1. Note: (p1Jp2) = (71193) = +1, but (p21Pl) = (19317) = 1; hence by Lemma 2, (56) has no solution. Example 5. Find, if any, a solution of the equation
7x2  3y2 = 1.
Ibid.
sT.L~~ ~% ~ 
37
Let p1 = 3 and P2  7, both primes of the form 4n+3. Then the given equation has the form of (57). The equation
(1) is then
t2  21u2 = 1,
with ti + u1 VD = 55 + 12\/21, where ti = 2X, + 1. Hence X1 = 27 = 3*32 # a square. Also since (p21pl) = (713) = +1 and P1 = 3 E3(mod 4), the conditions of Theorem 2 are satisfied for i = 2. Thus the given equation (57) has a solution in integers, x, y. By (66), y1 = 3, x1 = 2 is a solution. Checking: 7(2)2  3(3)2 = 2  27 = 1. Note: (plIP2) = (317) = 1; hence by Lemma 2, (56) has no solution.
2
Theorem 3. If in Lemma 2, p1 p2 m 1 (mod 4) and A = y then equation (55),
2 2
plp2x  1,
of (6) has a solution in integers x, y where
(70) yl = %T and 2x yl = ul.
Proof. By Lemma 2 if (p2l1P) = (P11P2) 1,
equation (56) and (57) of (6) have no solutions in integers x, y. Now, (plp2) =(p21P) may be written as
(1p2 P 1j2 12Pll='(P21)
38
(PlIP2 ( Pll)' (P2+1) or
( (p2 =2+l) hence,
(1) = (1)
which holds if and only if p1 P2 s 1 (mod 4). Also by Lemma 2 and Corollary 9 of Theorem 1, if X, = y where D contains no prime factors = 3 (mod 4), equation (55) has a solution. By Theorem 1 solutions of (55) may be obtained by ( and ( ,) n = 1 = G, where in (8), y2 = X1, since di = 1. This establishes (70).
Theorem 3 is illustrated by the following Example. The equation (1), t2  95u2 = 1,
has D = 95 = 517, where 5 and 17 are both primes = 1 (mod 4), and t1 + u\/lD = 285,769 + 30,996Vil, t1 = 2N + 1. Hence x = (375)2. If we let p1 = 5 and p2 = 17, the conditions of Theorem 3 are satisfied and equation (55),
85x2  y2 = 1,1
has a solution in integers x, y. By (70), y1 = 378, x = 41 is a solution. Checking: 85(41)2  (37S)2 = 142,885 142,884 = 1.
~ A
39
Note: (p21p1) = (1715) = 1, and (plIp2) = (5)(17) = 1; hence, by Lemma 2 equations (56) and (57) have no solutions.
By Lemma 2 if
(71) (plIp2) = (P2lPl) (p2 1P1) (PP2) = +1, the conditions necessary for both (56) and (57) of (6) to have a solution are satisfied. As in Theorem 3 relation (71) holds if and only if p1 p2.= 1 (mod 4). This last set of conditions is not covered in Theorem 2. Moreover, Theorem 3 =2
asserts that if A1 = y1 and p1 m p2 in 1 (mod 4) equation (55),
2 2
PlP2x  y = 1, of (6) has a solution. Hence, if (71) holds
2
and A1 N y1, equation (55) has no solution by Lemma 2. It follows by Theorems 1 and la that one of (56) and (57) of
(6) has a solution. Thus in (S) d = p1 or p2' andy is the largest square in A1 = (t1  1)/2. This discussion proves the following
Theorem 4. If in Lemma 2, A1 # y2 and p1 p2 1 (mod 4), then either equation (56),
pJX2 _ p2 2
or equation (57),
2 2
Px Pl 1,
has a solution in integers x, y according as d = p2 or p1 in relation (8)X, A = dy , where y2 is the largest square in (t  1)/2 = A1.
40
Legendre proved from an analysis of the VA in a continued fraction form where A = MN that if M and N are two primes of the form 4n+l, one of the two equations, x2  MNy2 =  1 or Mx2  Ny2 = t 1, is possible.
The following example illustrates Theorem 4. Example. Show that the first one of
29x2  53y2 = 1, 53x2  29y2 = 1, has a solution.
Let p1 = 29 and p2 = 53, both primes of the form 4n+l. Then the given equations have the forms of (56) and (57). Hence the equation (1) is
t2 1537u2 = 1
since D = 1537 = 2953 is a product of two distinct odd primes and t + u \D = 75,00,329,577 + 1,915,090,212 V1537, tj = 2Xi + 1.2 Hence X1 = 53(26614)2, and X, a square. Thus by Lemma 2 equation (55) has no solution. Since p1 p2
1 (mod 4), and since relation (71) holds, which is easy to show, then by Lemma 2 the necessary conditions for (56) and
(57) to be solvable are satisfied. Now by Theorem 4 and the
2
relation (9)3 we see X, = dly = 53(26614)2, so that dl = 53 and Y = 26614. Hence by Theorem 4 or Theorem 1, (6) takes the form (56),
1Legendre, op. cit. 2Cayley, op, cit.
I
41
29x2  53Y2= 1,
with the solution y = 26614, xi = 35979.
Checking:
29(35979)2  53(26614)2 = 37,540,164,799  37,540,164,798 = 1.
4
 '
fn
W
CHAPTER III
CONDITIONS FOR ONE OF x2 dy = 1 TO HAVE
A SOLUTION WHERE 81.dl = D, A PRODUCT
OF THREE DISTINCT ODD PRIMES
Lemma 4. Let D = plp2p3 in equation (1), t2 Du = 1, where p1 < p2 < p3 are distinct odd primes. Let (1) have the fundamental solution ti + ul\/5 where ti = 2X1 + 1. If one of the excludents listed in Table 3 to the right of the equations (72)  (78) of (6), 81X  dly2 = holds, then that equation has no solution in integers x, y.
Proof. Since D is a product of s = 3 distinct odd primes, p1, p2, and p3, there are 28 = 2 = 9 positive divisors of D. These are 1, p1, p21 P3 P1P2, p2p3, and plp2p3 For t1 = 2X1 + 1 and ul, the fundamental solution of
(1), the equation (6) of Theorem 1, where 81.dl = D = plp2p3,
2 2
has eight possible forms, namely, equation (24), x  PlP2p3y = 1, and the seven equations (72)  (79). By Theorem la, equation (24) and one of (72)  (78) always arise from (6) and have integral solutions x, y. We note the following about (72)  (79) of (6).
Let Y2 y Then equation (72) cannot arise. For
if it did, then by Corollary 7 and relation (30)1, x1 = 2
42
43
TABLE 3
EXCLUDENTS OF EQUATIONS (72)  (78)
Equations Excludentsa
(72) pp2p3 2 Y2 = 2
(73) plx2  p23 2 (11p2) (p11p3) (p2p3l1p)
(74) p2p3x2 ply2 ' 1 (plP2) (p1lp3) (p2p3lpl)
(75) p2x2 _ l3 2 1 (p21p1) (p2jp3) (plp31P2)
(76) p l2y2 21 2(P21P) (P21p3) (p1p3IP2)
(77) pp 2  = (p3jpl) (p~1P2) (P1p2lp3)
(78) PlP2x 2 p3y 2=1 (p3 1i1) (p3 1P2) (PlP21p3)
aA1l excludents 1.
2
a contradiction to the assumption that N
Let (73) arise from (6) with 6 = p andd = p2p3. By Theorem 1 or Corollary 9 there exist integers x, y satisfying (73). It follows from (73) that
(79) pix2 a 1 (mod p2), lx 2= 1 (mod p3), and
2
~p3y 2 1 (mod pp).
Define the integers p', p", and p"' by
(8O) p'p, 1 (mod p2 1 1 (mod p3), and
P"' P2i 3 1 (mod p1).
I4~F~ ~ ~ 2
44
Then (79) may be written as
pIp12 p' (mod p2), p"p1x 2 a p" (mod p3), and
P"p332 = p"' (mod pl), which in view of (S0) reduces to
(51) x2 M p' (mod p2), X 2 = p" (mod p3), and
y2 M p"' (mod pl).
By (0)1 we may then write (p'Ip2) = (p2'p 1 p2 (*1I29.
Hence if (pIjp2) = (P'1P2) = 1, then (79) is not solvable and, therefore, neither is (73). Similarly (73) is not solvable if either (p 'p3) = 1, or (p2p3 pl) = 1. Thus the three excludents listed opposite (73) in Table 3 are established. By similar reasoning the three excludents listed opposite the equations (74)  (78) in Table 3 are established.
We will make use of the following lemmata.
Lemma 5. The conditions
(82) (PiP) = (PilPk) = jPk i9) = +1,
where pi, p,, and Pk are distinct odd primes and I, j, k = 1, 2, 3, follow if and only if either
(93) (Pi1Pj) = (PiiPk) = +1 and p = 1 (mod 4),
45
or
(84) (p p lp = (p =pk +1 and p j 3pk (mod 4).
Proof. If (82) holds, consider (P Pk i) = +1, which may be written as
(85) ( (  k  1) i(P)+3(p i (Pi1)'  }k')=+.
Pj Pk)
Using (82) and combining, relation (85) becomes
(+i) (+i) (i) i(P).Jp+Pk ) = +1,
which holds if and only if either p = 1 (mod 4) or P M 3pk (mod 4). Hence (83) and (84) are established.
Conversely, if (83) holds, we may write ( =
(1Pi)(pJ1Pi)(Pk1Pi) = (+l)(PiPJ)(PiPk) = (+l)(+l)(+l) = +1. Hence (82). If (84) holds, then PjPk = 3 (mod 4), and we may write (p =Pki k i k 1
This completes the proof. Lemma 6. The conditions
(86) (PiIPJ) = (PiPk) = jPkIPi = +1,
where P., Pj, and Pk are distinct odd primes and i, j, k = 1, 2, 3, follow if and only if either
(87) (p ) = (PiPk) = +1 and Pi M 3 (mod 4),
~i
46
or
(8S) (pjpJ) = (PiIPk) = +1 and p3 r pk(mod 4).
Proof. If (86) holds, then the relation (P pi) = +1, may be written as
(99) (p 1p )(p ) = +1.
Also since (piipj) = +1, it may be written as
(p3 lp)(1) p+1).p)= +1, or, multiplying both sides by (p p),
(90) ( ) i+1) = (ppp)).
Similarly, since (p ip ) = +1, i k
(91) +(Pi})(Pk . PklPi
Substituting (90) and (91) in (89), and combining, we get
k +) .l(P)+Pk2) = +1
which holds if and only if either p. a 3 (mod 4) or p k
(mod 4). Hence (97) and (98) are established.
Conversely, if (87) holds, then (p3 1pi) = (pk =
+1. Hence (pPPi) P i k1) = (+')(+') = +1. If
(88) holds, then since pj r Pk (mod 4), (P pk 91I =
47
(PI1P )(Pi1Pk) = +1. This completes the proof of the lemma.
We seek the necessary and sufficient conditions for each of the equations (72)  (78) to have a solution. These will now be determined in Theorems 5 and 6 which follow. Theorem 5. Let D = pIp2p3 where p1 < p2 < p3 are distinct odd primes. Let ti + ui\/5 be the fundamental solution of
(1), t_ Du2 = 1, where t= 2?% + 1. If Xl 2 y1,
(92) (Pi PJ) = (PilPk) =k) = +1,
and either
(93) pi 1 (mod 4) and Pj a 3Pk (mod 4),
or
(94) Pi 3 (mod 4) and Pi = 3pk (mod 4),
then either equation (73), (75), or (77) of (6) has a solution in integers x, y according as i = 1, 2, or 3 with J, k = 1, 2, 3, and j < k, and where the solutions are given by
(9) 2 2
(95) y = (lu1/4) and 2xlyl = ul.
Proof. By hypothesis and Theorem la, one of the
equations (72)  (78) of (6) has a solution in integers x, y. Since NJ # y2, then by Lemma 4, equation (72) of (6) has no solution.
Let i = 1, j = 2, k = 3 in conditions (83) of Lemma
5, then
(96) (p1lp2) ' (p11p3) = +1 and p1 = 1 (mod 4).
Hence by Lemma 5, (p2p3 1p) = +1, and thus by Lemma 4 the conditions necessary for (73) to have a solution are satis2
fled. Now all excludents in Table 3, except NJ yl, can be expressed in terms of one or more of the Legendre symbols, (p11P2), (p 11p 3), and (p2 3), Hence when conditions (96) hold the values of (p2jp3) and the least positive residues modulo 4 of p2 and p3 can be arranged in 23 = 8 different arrangements or cases, as shown in the three columns headed "Conditions" in Table 4. From taking each of the eight cases in turn we determine those excludents of Table 3 which are consistent with conditions (96). The results are recorded in Table 4 in the column headed "Excludents of Table
3 which hold." Hence when an excludent of Table 3 holds the associated equation has no solution by Lemma 4. This fact is recorded in Table 4 in the last column which has the heading "Equations excluded." For example when the conditions (96) and those of case (1),
p2 p3 z 1 (mod 4) and (p21 p3) = +1
hold, we find that none of the excludents in Table 3 with respect to the equations (74)  (79) can hold, that is each one = +1. Hence when the conditions (96) and those of case
TABLE 4
CONDITIONS TO EXCLUDE EQUATIONS (74)(78) WHEN 2
(p11p2) = (pp3) = +1, p1 m 1 (mod 4) Conditions
p p 2 (p ) Equations
Case ( 2 3 Excludents of Table 3 which hold excludeda
(mod 4) ____1 1 1 +1 none none
2 1 1 1 (75) (76)2,3' ( , (78)2 (75)(79)
3 1 3 +1 (74)2, (76)2' (77) B
4 1 3 1 (74)2 (75)23 (76)39 (7)23 A
5 3 1 +1 (74)1, (75)39 (78)2 C
6 3 1 1 (74)1, (75) 21, (76) , (77)2,3 3
7 3 3 +1 (74) (76)2, (77) B
S3 3  (74)1,2 , (75)2,3' 2,3
equations equations equations
(74)(18).
(74) , 76) and (77).
(74), (75), and (79).
p.
a A
B
C
denotes denotes denotes
I
50
(1) hold, none of the equations (74)  (7,5) is excluded. This fact is recorded in Table 4 in the column headed "Excludents of Table 3 which hold." Also, when conditions
(96) and those of case (2),
p2 E p3 =1 (mod 4) and (p2 =hold, excludent (74), cannot hold since (1p2)(P1P2) =
(+l)(+l) = +1; excludent (74)2 cannot hold since (l1p3)(p1p3) = (+l)(+l) = +1; excludent (74)3 cannot hold since (P2ip9(p3Ipl) = (P1p2)(p1p3) = (+')(+') = +1. Hence equation (74) is not excluded. Also: excludent (75)l cannot hold; excludent (75)2 does hold since (p21p3) = 1; excludent
(75)3 holds since (%I(PilP2(p3iP2) = (+1)(+l)(p2lp3) =
(+l)(+l)(l) = 1. Hence by Lemma 4 equation (75) has no solution. Proceeding in this manner with the remaining excludents, it is found that excludents (76)2,3, (,, and
(78)2,3 hold, and hence equations (76)  (78) have no solutions. We record in Table 4 in the column headed "Excludents of Table 3 which hold," those excludents which are consistent with conditions (96) and those of case (2). Continuing in this manner with the remaining cases (3)  (8), results are found as indicated in Table 4. We see that when the conditions (96) and those of either case (4) or case (6) hold, namely,
(97) p1 r p2 = 3p3 = 1 (mod 4) and (p(p1p3)
(p2 p3) = +1,
I&
51
or
(99) PM 3P2 P3  1 (mod 4) and (pjlp2) = (pl1p3) =
(p2lp3) =
equations (74)  (78) have no solutions. By Theorem la, since X y2, one of (73)  (79) has a solution. Hence when conditions (97) or (98) hold and X y2, equation (73) of (6) has a solution. Conditions (97) and (98) are restated in (92) and (93) for I = 1, j = 2, k = 3. By symmetry interchanging the subscripts 1 and 2 in (97) and (99) conditions are obtained by which (75) of (6) has a solution. This establishes (92) and (93) for I = 2, j = 1, k = 3. Also by symmetry interchanging the subscripts 2 and 3 in the conditions obtained for (75) to have a solution produces those by which (77) of (6) has a solution. This establishes (92) and (93) for i = 3, j = 1, k = 2. Thus (92) and (93) are established completely.
The results (92) and (94) will now be derived. Let I = 1, j = 2, k = 3 in (84) of Lemma 5, then
(99) (p1lp2) (pl1p3) = +1 and p2 3p3(mod 4).
Hence by Lemma 5, (P293 = +1, and thus by Lemma 4 the conditions necessary for (73) to have a solution are satisfied. Now all excludents of Table 3, except NJ # y2, can be expressed in terms of one or more of the Legendre symbols,
52
(PlIP2), (Pl1P3), and (p2jp3). Hence when conditions (99) hold the values of (p21p3) and the least positive residues modulo 4 of p1, P2, and p3 can be arranged in eight different arrangements or cases, since in view of (99)3 P2 and p3 cannot be in the same class modulo 4. These eight cases are listed in the four columns of Table 5 under the heading "Conditions." From taking each of the eight cases in turn we determine those excludents of Table 3 which are consistent with conditions (99). The procedure is the same as that used in developing Table 4. The results are recorded in the last two columns of Table 5. Hence it is seen from Table 5 that equations (74)  (79) have no solutions by the conditions
(99) and those of cases (2), (4), (6), and (8). But we observe that the conditions (99) and those in cases (2) and (6) of Table 5 are the same as conditions (96) and those of cases (4) and (6) of Table 4. Thus, from the new conditions of cases (4) and (8) of Table 5, equations (74)  (79) have no solutions if either
(100) 3p, p2 3p3 1 (mod 4) and (p1Ip2) = (pl1p3) =
(p21p3) +1,
or
(101) 3p, E 3p2 p3 1 (mod 4) and (pIjp2) = (p11p3) =
(D 2p ) = +1.
?i:, .7":;, !;V .. "i;
TABLE 5
CONDITIONS TO EXCLUDE EQUATIONS (74)(78) WHEN X (p 2) ( ) = +1, p 3p (modt 4)
1 2 1 3 2 3
Conditions
p 2 = 2p p)= Equations
Case 1 )2 3 3 Excludents of Tble 3 which hold excludeda
(mod 4)EcldetofTbe3wihod
1 1 1 3 +1 (74)2' (76)29 (77)3 B
2 1 1 3 1 (74)2, (75)2,3, (76)39 2' (7)23 A
3 3 1 3 +1 (74)23 (76)23 12,3 B
4 3 1 3 1 (74)2,3 ,, (75)2,3 , (76)1 3 ' ( 7)1 2 ' ( A2,3
.5 1 3 1 +1 (74)1, (75)30 2 C
6 1 3 1 1 (74)1, (75)2, (76)23 (77)2,3 (7)3 A
7 3 3 1 +1 (74) 3, (75) , (78)12 C
8 3 3 1 1 (74)1, (75)l2 (76)23, (77)2, (7) A
m  ,3
'A denotes B denotes
C denotes
equations (74)(7g). equations (74), (76), and (77). equations (74), (75), and (79).
U,
I
*i~ ,~.., A. &'~
54
2
By Theorem la, since NJ yl, one of (73)  (78) has a solution. Hence when conditions (100) or (101) hold, and
2
NJ y1, equation (73) has a solution. Conditions (100) and (101) are written as (92) and (94) for I = 1, j = 2, k = 3. By the same argument as above using the property of symmetry with regard to the subscripts, (92) and (94) are established completely. By Theorem 1 solutions of (73), (75 or (77) of
(6) are obtainable by (95). This completes the proof of the theorem.
The following six examples illustrate Theorem 5 wherein tj + uj\DV was calculated by the writer unless otherwise specified.
Example 1. Show that
13x2  1,247y2 = 1,
has a solution in integers x, y and that its fundamental solution is given by x1 = 10,176, y1 = 1,039.
Since 1,247  2943, the numbers 13, 29, and 43 are three distinct odd primes where if p1 = 13, p2 = 29, and p3 = 43, the given equation has the form (73). Equation (1) is then
t2  16,211u2 .
since D = 1329.43 = 16,211 and its fundamental solution is tj = 2N, + 1 = 2,692,325,375, u1 = 2,145,729. Hence NJ = 2943(l,039)2 # a square. Also since (pl1p2) = (13129) = +1,
55
(pljp3) = (13143) = +1, and (p21p3) = (29143) = 1, and in addition p1 = 13  1 and p2 = 29 a 3'43 = 3p3 in class modulo 4, the conditions (92) and (93) of Theorem 5 are satisfied for i = 1, j = 2, k = 3. Hence by Theorem S the given equation has a solution in integers x, y. By (95) Yr = 1,039, x = 10,176 is a solution. Checking: 13(10,176)2  1,247(1,039)2 = 1,346,162,609  1,346,162,687
1. Moreover by Corollary 9 the solution xl, y1 is the fundamental solution; for by (34), t1 = 81xi + d y1 and Ul = 2xiyi. Checking: t1 = 13(10,176)2 + 1,247(1,039)2 = 1,346,162,688 + 1,346,162,687 = 2,692,325,375; ul = 2(10,176)(1,039) = 2,145,728. Remark: Since by Corollary 9, on page 16, the solution xl, yl is the fundamental solution of the given equation, we can obtain all positive solutions by relation (36) of Corollary 10, on page 17.
Example 2. Show that the fundamental solution of
3x2 143y2 = i,
is given by x = 504, y = 73.
Since 143 = 11'13, the numbers 3, 11, and 13 are three distinct odd primes where if p = 3, p2 = 11, and p = 13, the given equation hRs the form (73). Equation (1) is then
t2  429u2 = 1,
56
since D = 3'11'13 = 429 and its fundamental solution is ti = 2N1 + 1 = 1,524,095, u1 = 73,584. Hence X1 = 762,047 = 11'13.(73)2 a square. Also, since (pilp2) = (3111) = +1, (pIjp3) = (3113) = +1, (p21P3) = (11113) 1, and in addition 3 = p1 = 3 (mod 4) and 11 = p2 3p3 313 (mod 4), the conditions (92) and (94) of Theorem 5 are satisfied for i = 1, j = 2, k = 3. Hence by Theorem 5 the given equation has a solution in integers x, y. By (95), y1 = 73, xi = 504 is a solution. Checking: 3(504)2  143(73)2 = 762,o4g 762,047 = 1. By Corollary 9, x1, y1 is the fundamental solution.
Example 3. Show that
13x2  51y2 = 1,
has a solution in integers x, y and that its fundamental solution is given by x, = 2, y1 ='l.
Since 51 = 3'17, the numbers 3, 13, and 17 are three distinct odd primes where if pi = 3, p2 = 13, and p3 = 17, the given equation has the form of (75). Equation (1) is then
t2 663u2 = 1,
since D = 31317 = 663 and its fundamental solution is ti 2N1 + 1 = 103, u1 = 4. Hence X1 = 51 a square. Also (p21Pl) = (1313) = +1, (p21p3) = (13117) = +1, (p1p3) = (3117) = 1, and P2 = 13 m 1 (mod 4), p1 = 3 G 3p3 = 317
Lp
57
(mod 4). Hence conditions (92) and (93) are satisfied for I = 2, j = 1, k = 3. Hence by Theorem 5 the given equation has a solution in integers x, y. By (95), y1 1, x, = 2 is a solution. Checking: 13(2)2  51(1)2 = 52  51 = 1. By Corollary 9, xl, y1 is the fundamental solution. Example 4. Show that
7x2  37y2 1,
has a solution in integers x, y and that its fundamental solution is given by x, = 208, y1 = 59.
Since 97 = 329, the numbers 3, 7, and 29 are three
distinct odd primes where if pi = 3, p2 = 7, and p3 = 29, the given equation has the form (75). Equation (1) is then t2  609u2 . 1,
since D = 37'29 = 609 and its fundamental solution is tj = 2X1 + 1 = 605,695, u1 = 24,544.1 Hence Xl = 329'(59)2 g a square. Also (p21p1) = (713) = +1, (p21p3) = (7129) = +1, (PlIP3) = (3129) = 1, and 7 = p2 = 3 (mod 4), 3 = pi M 3P3329 (mod 4). Hence the conditions (92) and (94) are satisfied for I = 2, j = 1, k = 3. Thus by Theorem 5, the given equation has a solution in integers x, y. By (95), yl = 59, xi = 208, is a solution. Checking4 7(209)2  97(59)2 = 302,949  302,547 = 1. By Corollary 9, x 1, yl is the fundamental solution.
Example 5. Show that
Legendre, op. cit., Table X.
29x2  35y2 = 1,
has a solution in integers x, y and that its fundamental solution is given by x, = 78, y1 = 71.
Since 35 = 5'7, the numbers 5, 7, and 29 are three
distinct odd primes where if p1 = 5, P2 = 7, and p3 = 29, the given equation has the form (77). Equation (1) is then t2  1,015u2 = 1,
since D = 5*7'29 = 1,015 and its fundamental solution is ti = 2X1 + 1 = 352,971, u1 = 11,076.1 Hence X= 5'7'(71)2 a square. Also (p31p1) = (2915) = +1, (p31P2) = (2917) = +1, (p1Ip2) = (517) = 1, and p3 = 29 a 1 (mod 4), pi = 5 G
3 7 = 3P2 (mod 4). Hence the conditions (92) and (93) are satisfied for i = 3, j = 1, k = 2. Thus by Theorem 5 the given equation has a solution in integers x, y. By (95), yl = 71, x1 = 78 is a solution. Checking: 29(79)2 3F(71)2 = 176,436  176,435 1. By Corollary 9, xi, y1 is the fundamental solution.
Example 6. Show that
31x2  15y2 = 1,
has a solution in integers x, y and that its fundamental solution is given by x, = 16, y1 = 23.
lCayley, op. cit., p. 430.
4'~. i~5~JI
59
Since 15 = 3'5, the numbers 3, 5, and 31 are three distinct odd primes where if p = 3, p2 = 5, and p3 = 31, the given equation has the form (77). Equation (1) is then t2  465u2 = 1,
since D = 3531 = 465 and its fundamental solution is ti = 2X1 + 1 = 15,971, ul = 736.1 Hence NJ = 3'5'(23)2 9 a square. Also (p31pl) = (3113) = +1, (p31p2) = (3115) = +1, (plIp2) = (315) = 1, and p3 = 31 = 3 (mod 4), p1 = 3 a 3.5 3P (mod 4). Hence the conditions (92) and (94) are satisfied for I = 3, j = 1, k = 2. Thus by Theorem 5 the given equation has a solution in integers x, y. By (95), y1 = 23, x, = 16 is a solution. Checking: 31(16)2  15(23)2 7,936  7,935 =
1. By Corollary 9, x1, y1 is the fundamental solution. Theorem 6. Let D = Plp2p3 where p1 < p2 < p3 are distinct odd primes. Let t1 + u\/f be the fundamental solution of (1),
2 _ 2 2
t  Du2 = 1, where t1 = 2X1 + 1. If NJ 9 yl, and either (102) (pi1P ) = (Pi1Nk k) = +1 and p1 = 3 (mod 4),
or
(103) (piIp ) = (P Pk =k) = +1
and p1 a p = Pk a 3 (mod 4),
1Legendre, oP. cit., Table X.
6o
or
(104) (pl1p3) = (PilPk) = +1 and pi 3ru 3Pk 1 (mod 4), then either (74), (76), or (78) of (6) has a solution in integers x, y according as i = 1, 2, or 3, where j, k = 1, 2,
3 with j < k, and where solutions are given by (105) y2 (l,ul/4) and 2x=y  u1.
Proof. By hypothesis and Theorem la, one of the
equations (72)  (79) of (6) has a solution in integers x, y.
2
Since NJ y , then by Lemma 4, equation (72) of (6) has no solution.
Let i = 1, j 2, k = 3 in conditions (97) of Lemma 6, then
(106) (plIp2) (plp3)  +1 and p1 a 3 (mod 4).
Hence by Lemma 6, (p2p3Il) = +1; thus, by Lemma 4 the conditions necessary for (74) to have a solution are satisfied. Now all excludents of Table 3, except A, y , can be expressed in terms of one or more of the three Legendre symbols: (107) (p lp2), (p11p3), and (P2 3 *
Hence when conditions (106) hold, the values of the three symbols in (107) and the least positive residues modulo 4 of P2and p3 can be arranged in eight different arrangements or
61
cases as shovn in the six columns headed "Conditions" in Table 6, where the third, fourth and fifth columns are easily derived from conditions (106) and those of the first two columns in the table. From taking each of the eight cases in turn we determine those excludents of Table 3 which are consistent with each case. The procedure is the same as that used in developing Table 4. The results are recorded in the last two columns of Table 6. Hence from Table 6, equations (73) and (75)  (78) have no solutions by the conditions of cases (2), (4), (6)  (9). By Theorem la, since
2
NJ # yl, one of (73)  (79) of (6) has a solution. Hence equation (74) has a solution when either
(109) 3PL a P G P3 i 1 (mod 4) and
+(pl1p2) = +(pllp3) = (p21p3) +1, or
(109) 3pE p2 ~ 33 1 (mod 4), and
+(PP2 (pp3) =(P2p3) = +1,
or
(110) 3p1 e 3p2 p3 1 (mod 4), and
(plp2) = +(pljp3) = (p21p3) = +1, or
(111) p p2 z p3 3 (mod 4), and
(pilP2) = ( 1ll3) = +(p2lp3) = +1,
TABLE 6
CONDITIONS TO EXCLUDE EQUATIONS (73) AND (75)(78) WHEN X, yj,
(pl1p2) = (pl1p3) = +1, AND p, = 3 (mod 4)
Conditions
Case P2MI 3 Excludents of Table 3 which hold e *
1 1 1 +1 +1 +1 (73) 3 (76), (78), B
2 1 1 +1 +1 1 (73)3, (75)2,3, (76)all, (77)2,3' Aall
3 1 3 +1 1 +1 (73)23 (76)12 1,2 B
4 1 3 +1 1 1 (73)2,3 (75)23, (76) 1,3 (77)23 (7)1,2 A
5 3 1 1 +1 +1 (73)1,3, (76)1,3, (781,2 B
6 3 1 1 +1 1 (73) , (75)23, (76)12, (77)23' A
7 3 3 1 1 +1 (73)all' (75)3, (76)12, 20 A
8 3 3 1 l a 77)al' 2, (76)1,30 (77)3, (78)l2 A
aA denotes B denotes
equations (73), (75)(75). equations (73), (76), and (78).
Iu
63
or
(112) P1 P2 P3 a 3 (mod 4), and
(pjP2)  (plp3) = (P21P3) = +1
In view of (106), conditions (108)  (110) and (112) establish (102), while condition (111) establishes (103) for i = 1, j = 2, k = 3. By symmetry interchanging the subscripts 1 and
2 in (102) and (103), we obtain the conditions by which equation (76) has a solution. This establishes (102) and (103) for i = 2, j = 1, k = 3. Also by symmetry interchanging the subscripts 2 and 3 in the conditions obtained for (76) to have a solution, we get those by which equation (78) has a solution. This establishes (102) and (103) for i = 3, J = 1, k = 2, completing the proof of (102) and (103).
The result (104) will now be derived. Let i = 1,
3 = 2, k = 3 in conditions (88) of Lemma 6, then
(113) (PlIP2) = (pl9p3) = +1 and p2 p3 (mod 4). Hence by Lemma 6, (p2p3 p1) = +1; and thus by Lemma 4 the necessary conditions for (74) of (6) to have a solution are satisfied. Now all excludents in Table 3, except NJ y, can be expressed in terms of one or more of the Legendre symbols,
(114) (p jp2) (p 11p 3' and (p2 3
64
Hence when conditions (112) hold the values of the three symbols in (114) and the least positive residues modulo 4 Of pl, p2, and p3 can be arranged in eight different arrangements or cases as shown in the six columns headed "Conditions" in Table 7, where the fourth, fifth and sixth columns are easily derived from (113) and the conditions of the first three columns. From taking each of the eight cases in turn we determine those excludents of Table 3 which are consistent with each of the eight cases. The procedure is the same as that used in developing Table 4. The results are recorded in the last two columns of Table 7. Hence from Table 7, equations (73) and (75)  (78) have no solutions by the conditions of cases (3), (4), (6), (7), and (8). But the conditions of cases (6) and (8) are implied in (102) for i = 1, j = 2, k = 3, and similarly the conditions of case (7) are implied in (103). Hence from the new conditions of cases
(3) and (4), equations (73) and (75)  (78) have no solutions if either
(115) p1 3P2 3p3 a 1 (mod 4) and (p192) =  (pl1p3)
+(p21p3) +1,
or
(116) p1 3p2 3p3 m 1 (mod 4) and (p1p2) = (p1p3)
(p2lp3) = +1.
A 1 j
TABLE 7
CONDITIONS TO EXCLUDE EQUATIONS (73) AND (75)(78) WHEN A
(PI (1 3)  +1, AND p2 3 (mod 4)
Conditions
Case P1j 3 (= ( (A ) Excludents of Table 3 which hold
1 1 1 1 +1 +1 +1 none one
2 1 1 1 +1 +1 1 (75)2, 7 23 ' , , B
3 1 3 3 1 1 +1 (73)1,29 (75),v3, (76) 7)1,2 ,3 A
4 1 3 3 1 1 l (73)129 (75)12, (76), 39 ( 9 A
5 3 1 1 +1 +1 +1 (73)3, (76)1, (7) C
6 3 1 1 +1 +1 1 (73) 3 (75)23, (76)all ( (7 )Aal
7 3 3 3 1 1 +1 (73)all, (75)3, (76)12, ( ( A
8 3 3 3 1 1 1 (73)all' 2 ( ,3 3 1,2 A
aA denotes equations (73), (75)(79).
B denotes equations (75)( 7).
C denotes equations (73), (76), and (78).
,\
66
2
By Theorem la, since NJ y1, one of (73)  (78) has a solution. Hence (74) has a solution when either (115) or (116) holds. Conditions (115) and (116), in view of (113), establish (104) for i = 1, j = 2, k = 3. By the same argument as above using the property of symmetry with regard to the subscripts, (104) is established completely. By Theorem
1 solutions of (74), (76), or (78) are obtainable from (9)l and (9)2, n = 1 = G, thus establishing (105). This completes the proof of the theorem.
Conditions (102), (103), and (104) present seven
different sets of conditions by which each of the equations
(74), (76), and (75) has a solution. The Tables 5, 9, and 10 indicate these for (74), (76), and (75) respectively. Seven examples follow each table. Since the illustration of Theorem 6 by examples would be similar to that used with regard to Theorem 5, only pertinent data are given in outline form, wherein t1 + u\/D was calculated by the writer unless otherwise specified.
Example 1. [case (1) of Table 5]
P, = 3, P2 = 13, p3 = 73
(pIjp2) = (3)113) = +1, (pljp3) = (3173) +1, (p21p3) (13173) = 1. D = 3'13*73 = 2,547, ti = 2X1 + 1 = 372,007 and u, = 6,972, whence J= 3.2492 # a square. By (105), Y, = 249, x, = 14 is a solution of 13'73x 2 3y = 1. Checking: 949(14)2  3(249)2 = 186,004  156,003 = 1.
IM'
67
TABLE 9
2 2
CONDITIONS FOR EQUATION (74), p2P3x  1ly 2
2
TO HAVE A SOLUTION WHEN N yJ
Conditions Conditions of
) (..p Theorem
Case p1= p2. p3=  where
(mod 4) 2 ___ P3 I = 1, j = 2, k = 3
(1) 3 1 1 +1 +1 1 (102)
(2) 3 3 3 +1 +1 1 (102)
( ) 3 1 3 +1 +1 1 (102)
3 3 1 +1 +1 1 (102)
(5) 3 3 3 +1 +1 +1 (12)
(6) 1 3 3 +1 +1 +1 (104)
(7) 1 3 3 +1 +1 1 (14)
Example 2. [case (2) of Table $]
p1= 7, P2 = 11 p3 = 23,
(PlIP2) = (7111) = +1, (p11p3) = (7123) = +1, (p21p3) (11123) 1. D = 7'11l23 = 1,771, t1 = 2X, + 1 = 505 and
ul = 12, whence X1 = 7'(6)2 a square. By (105), Yj = 6,
x = 1 is a solution of 11,23x2 _ 7y2 = . 253(1)2  7(6)2 = 253  252 = 1.
Checking:
Example 3. [case (3) of Table 8]
p1 = 3, p2 = 13, p3 = 19'
(plp2) (3113) = +1, (Plp p3) = (3119) = +1, (p21p3) (13119) = 1. D = 31319 = 741, t1 = 2N1 + 1 = 7,352,695 and u= 270,108, whence Ni = 3'11072 a square. By (105),
1Legendre, on, cit., Table X.
68
y = 1,107, x1 = 122 is a solution of 13'19x 2 23y 1 Checking: 247(122)2  3(1107)2 = 3,676,348  3,676,347 = 1. Example 4. [case (4) of Table 8]
p1 3, p2 = 7, p3 = 13,
(P1IP2) = (317) = +1, (p1 p3) = (3113) = +1, (p2p3) (7113) = 1. D = 3713 = 273, t1 = 2N1 + 1 = 727 and u = 44,1 whence X1 = 3'(11)2 ? a square. By (105), y1 = 11, x, = 2 is a solution of 7'13x2 3 3y2 = 1. Checking: 91(2)2  3(11)2 = 364  363 = 1. Example 5. [case (5) of Table 8]
P= 3, p2 = 7, p3 = 31,
(.plP2) = (317) = +1, (p jp3) = (3131) = +1, (P2 3 (7131) = +1. D = 3731 = 651, ti = 2Xi + 1 = 1,735 and ul = 68,2 whence = 3 2(17) a square. By (105), y, = 17, xi = 2 is a solution of 7'31x2 2 = 1. Checking: 217(2) 2 3(17) 86  867 = 1. Example 6. [case (6) of Table 8]
Pi = 5, P2 = 7, P3 = 479
(p1Ip2) = (517) = +1, (pl p3) = (5147) = +1, (p2lP3) (7147) = +]. D = 5'7'47 = 1,645, ti = 2X + 1= 9,460,519,131,041 and u1 = 233,255,566,728,3 whence X= 5(972,652)2 a square. By (105), Yi = 972,652, xi = 119,907
libid. 2 t.p.
3Whitford, o'.ci. p. 201.
69
is a solution of 7#47x2  5y2 = 1. Checking: 329(119,907)2  5(972,652)2 = 4,730,259,565,521 4,730,259,565,520 = l.
Example 7. [case (7) of Table 8]
P= 5, P2 = 7, P3 = 23,
(PlIP2) = (517) = +1, (pilp3) = (5123) = +1, (p21p3) = (7123) = 1. D = 5'7'23 = 805, ti = 2N + 1 = 1,514,968,641 and u, = 53,392,104,1 whence X = 5.(12,308)2 / a square. By (105), Y1 = 12,30, x, = 2,169 is a solution of 723x2 5y2 = 1. Checking: 161(2,169)2  5(12,3O=)2 757,434,321
 757,434,320 =
1 .
TABLE 9
CONDITIONS FOR EQUATION (76), pp3x2
TO HAVE A SOLUTION WHEN X y
p2 2
Conditions Conditions of
 (2 tP2N=Theorem 6
Case pl= 2 3 P2 P where
(mod P3,1 i  2, j =1, k =3
(1) 1 3 1 +1 +1 1 (102)
(2) 3 3 3 +1 +1 1 (102)
) 3 3 +1 +1 1 (102)
3 3 1 +1 +1 1 (102)
(5) 3 3 3 +1 +1 +1 (10)
(6) 3 1 3 +1 +1 +1 (1c4)
(7) 3 1 3 +1 +1 1 (104)
1Legendre, op. cit., Table X.
Z~A~kC 
70
Example 1. [case (1) of Table 9]
P1 = 5, P2 = 11 P3 = 979
(p21Pl) = (1115) = +1, (p2p3) =(11197) = +1, (p,1p3) (5197) = 1. D = 5'197 = 5,335, ti = 2X1 + 1 = 951,138,997,479 and u1 = lu,652,890,092, whence Ni 11'(196,693)2 # a square. By (105), y1 = 196,693, x1 = 29,622 is a solution of 5'97x2 _ 11y2 = 1. Checking: 485(29,622)2  11(196,693)2 = 425,569,499,740 425,569,498,739 = 1.
Example [case (2) of Table 9]
Pi = 3, P2 1, P3 103,
(P2=Pl) (1113) = +1, (P2Ip3) = (111103) = +1, (p11p3) = (31103) = 1. D = 3'11'103 = 3,399, ti = 2X1 + 1 = 61,799 and u1 = 1,060, whence X1 = 11(53)2 a square. By (10P), y = 53, x, = 10 is a solution of 3'103x2 _ 11y2  1. Checking: 309(10)2  11(53)2 = 30,900  30,899 = 1. Example 3. [case (3) of Table 9]
P1 = 5, P2 = 11, p3 = 23,
(p21Pl) = (1115) = +1, (p2lp3) = (11123) = +1, (pilp3) (5123) = 1. D = 51123 = 1,265, ti = 2X1 + 1 = 206,999 and u, = 5,920 whence X1 11(97) a square. By (105), y = 97, x1 = 30 is a solution of 5'23x2 11Y2 = 1. Checking: 115(30)2 _ 11(97)2 = 103,500  103,499 = 1.
1Cayley, op. cit., XIII, p. 430.
71
Example 4. [case (4) of Table 9]
P = 3, P2 = "I p3 53,
(p2 1P) = (1113) = +1, (p21p3) = (11153) = +1, (plip3) (3153) = 1. D = 3'1153 = 1,749, tj = 2XI + 1 31,356,358,199 and ul = 749,774,590, whence i= 1'(37,753)2 a square. By (105), yl = 37,793, x, = 9,930 is a solution of 3'53x2 _ 11Y2 = 1. Checking: 159(9,930)2 11(37,753)2 = 15,679,179,100  15,678,179,099 = 1. Example 5. [case (5) of Table 9]
p, = 3, P2 = 1, p3 = 23,
(P2l91) = (1113)  +1, (p21p3) = (11123) = +1, (pljp3) (3123) = +1. D = 3'11'23 = 759, ti = 2X1 + 1 = 551 and ul 20,1 whence l '(5)2 $ a square. By (105), y1 = V, X
2 is a solution of 3.23X2 _ 11Y2 = 1. Checking: 69(2)2 11(5)2 = 276  275 = 1.
Example 6. [case (6) of Table 9]
p= 3, p2 = 5, p3 = 23,
(P2ll) = (513) = +1, (p21P3) = (5123) = +1, (pI'p3) = (3123) = +1. D = 3'5'23 = 345, ti = 2X1 + 1 = 6,761 and u = 364,2 whence X= 5(26)2 $ a square. By (105), yr = 26, xi = 7 is a solution of 3.23x 2 5y 2 = 1. Checking: 69(7)2
 5(26)2 = 3,391  3,380 = 1.
1Legendre, op. cit., Table X. 2Ibid.
Examole 7. [case (7) for
72
Table 9]
p= 3, p2 =5, p3 = 103,
(p21lp) = (513) = +1, (p2lp3) = (51103) = +1, (p 'p3) (31103) = 1. D = 35103 = 1,545, ti = 2X1 + 1 = 123,491,961 and u1 = 3,141,516,1 whence NJ = 5(3,514)2 a square. By (105), y1 = 3,514, x, = 447 is a solution of 3'103x2  5y2 = 1. Checking: 309(447)2  5(3,514)2 =
61,74o,91  61,74o,98o =
1.
TABLE 10
CONDITIONS FOR EQUATION (79), plP22 2
TO HAVE A SOLUTION WHEN 2 y2
y1
Conditions Conditions of
Theorem 6
Case pl= P2 P3M (Y = where
(mod 4) _ 7P2) \P2J 1 3, j = 1, k 2
(1) 1 1 3 +1 +1 1 (102)
(2) 3 3 3 +1 +1 1 (102)
(4) 1 3 3 +1 +1 1 (102)
3 1 3 +1 +1 1 (102)
(5) 3 3 3 +1 +1 +1 (104)
(6) 3 3 1 +1 +1 +1 (104)
(7) 3 3 1 +1 +1 1 (104)
Example 1. [case (1) of Table 10]
p1 = 5, p2 = 17, p3 1 9'
(p3lpl) = (1915) +1, (p3 92) = (19117) = +1, (pilp2) "
lWhitford, op. cit., p. 201.
I., ?  . 49! 4FF e1  .
73
(5117) = 1. D = 5'1719 = 1,61r, t = 2N + 1 = 3,2S4,569 and u1 = 81,732, whence N = 19.(294)2 a square. By (105), yl = 294, xi = 139 is a solution of 5'17x2 _ 19Y2 = 1. Checking: 95(139)2 _ 19(294)2 . 1,642,285  1,642,284 = 1. Example 2. [case (2) of Table 10]
P1 = 7, P2 11, p3 = 19'
(p31p1) = (1917) = +1, (p3 1p2) = (1911) = +1, (p1LP2) (7111) = 1. D = 7'll19 = 1,463, ti = 2X1 + 1 = 153 and u = 4,2 whence X1 = 19(2) 2 a square. By (105), y1 = 2, xi
1 is a solution of 7'llx2 _ 19Y2 = 1. Checking: 77(1)2 19(2)2 = 77  76 =
Example 3. [case (3) of Table 10]
p1 = 5, P2 = 7, P3 = 19,
(P3 Il) = (1915) = +1, (p31p2) (1917) = +1, (p1lp2) (517) = 1. D = 5'7'19 = 665, t1 = 2Nl + 1 = 13,719 and u = 532,3 whence X = 19(19) 2 a square. By (105), y1 = 19, x = 14 is a solution of 5#7x2 _ 19Y2 = 1. Checking: 35(14)2  19(19)2 = 6,86o  6,859 = 1. Example 4. [case (4) of Table 10]
P, = 3, p2 = 5, p3
(p3 jpi) = (1113) = +1, (p31p2) = (1115) = +1, (p11P2) (315) = 1. D = 3'511 = 165, t1 = 2N1 + 1 = 1,079 and u, =
lIbid. 2Cayley, oi, cit., XIII, p. 430.
3Legendre, op, cit., Table X.
74
84, whence i= 11'(7)2 $ a square. By (105), y1 = 7, x1 =
6 is a solution of 3'5x2 11y2 = i. Checking: 15(6)2 11(7)2 = 540  539 = 1.
Example 5. [case (5) of Table 10]
pi 3, p2 1 p3 131,
(p 3p1) = (13113) = +1, (p 3p2) = (131111) = +1, (p11p2 (3111) = +1. D = 3*11'131 = 4,323, t = 2X1 + 1 = 263 and Ul = 4, whence Ni = 131'(1)2 # a square. By (105), y1 = 1, xi = 2 is a solution of 3llx2 131Y2 = 1. Checking: 33(2)2  131(1)2 = 1.
Example 6. [case (6) of Table 10]
p1 = 3, p2 = "I p3 = 17,
(p31pj) = (1713) = +1, (p31p2) = (17111) = +1, (p11p2) = (3111) = +1. D = 31117 = 561, ti =2X + 1 = 522,785 and U1 = 22,072,1 whence Ni = 17'(124) 2 a square. By (105), Y = 124, x, = 89 is a solution of 311x2 _ 17y2 1. Checking: 33(89)2  17(124)2 = 261,393  261,392 = 1. Example 7. [case (7) of Table 10]
P = 7, p2 = 11, p3 13,
(p3 pi) = (1317) = +1, (p31p2) = (13111) = +1, (p1Ip2) = (7111) = 1. D = 71113 = 1,001, t1 = 2X1 + 1 = 1,060,905 and u1 = 33,532,2 whence Xi = 131(202)2 # a square. By (105),
1Legendre, op. alt., Table X.
2Cayley, op. cit., XIII, p. 420.
4, 1 A
75
y= 202, xi = 83 is a solution of 7llx 2 13Y 2 1.
Checking: 77(83)2  13(202)2 = 530,453  530,452 = 1.
An illustration of Corollary 8 for equation (72),
2 2
PlP2p3x  1
is the following
Example. Let p1 = 5, P2 = 13, p3 17, all odd primes of the form 4n+l. Then D = 5'13'17 = 1,105 in (1), t2  Du2 = 1t ti = 2X1 + 1 = 1,623,132,289 and u1 = 48,808,432.1 Hence X, = (284,882)2. By Corollary 8 and relations (8)3 and (8)4, yl = 284,822, xi = 857 is a solution of (117) 1105x2 _ Y2 = 1.
Checking: 1105(857)  (284882)2 811,566,145  811,566,144
= 1. We note that in Table 3 the excludents (73)1, (74)1, ..., (78) all hold in this example, thus by Lemma 4 equations
(73)  (78) have no solutions. Since in Theorem 1 one of (72)
 (78) has a solution, equation (72) has a solution, which in this case is (117).
~Ibid.
* *>~
A
CHAPTER IV
CONDITIONS FOR ONE OF 8 x2 2 = 1 TO HAVE
A SOLUTION WHERE 8 i'di D, A PRODUCT
OF FOUR DISTINCT ODD PRIMES
Lemma 7. Let D = p p2p 3p4 in equation (1),
t2 Du2 = 1, where p1 < p2 < p3 < P4 are distinct odd primes. Let (1) have the fundamental solution ti + u1iD where ti = 2X1 + 1. If one of the excludents listed in Table 11 to the right of the equations (119)  (134) of (6), 81x2 dly = 1, holds, then that equation has no solution in integers x, y.
Proof. Since D is a product of s = 4 distinct odd primes, pi, p2' p3 P4' there are 28 = 24 = 16 positive divisors of D. These are
1,
p4, P2p 3'
1 p2 4,
(118)
pip
plp~ p2p4 pi 3 4'
p2p
1~ 3 ~
P3,
pip4, p1p2p 3 1 2 3
4'
For t1 = 2N + 1 and u1, the fundamental solution of (1), the equation (6) of Theorem 1, where 81'di = D = PiP2P3P4, has sixteen possible forms, namely equations (119)  (134). By Theorem la equation (119) and one of (120)  (134) always
76
now"
TABLE 11
EXCLUDENTS OF EQUATIONS (119)  (134)
Equations
(119)
(120) (121) (122) (123)
(124) (125) (126) (127) (128)
=
=
1
1
1
1
1
1
1
1
1
1
Excludents a
2
pip2p 3P4 X2 _y2 px2 _ p2p 3p4y 2 p2 34x 2 2
P 2 2 2
phPPx P2P2
p 3 p2 2 2 2 932~9 242 p 242 2
plp2P4x 2 p2
P 2 _ p 2P3 2 p p2 3 2 2
(p 11p3)
( P21 3 (p2 1p3) (p3
(p3 1P2) (p4 p2) (P41 P2)
aAll excludents = 1.
(Table 11 is concluded on the next page.)
(p 2 12
(pl 1P2) (2 l)
(p21 p) (p3 1 pi)
(P41 )
v
none
42
L~ 1 (p P4)
(p2 1 P4) (p3 1P4) (p3 1P4) (P41 P3) (P4 1 P3)
(p 3P41 P2) (pip 3P14IP2) (plp2p1 P3) (plp2p4'p3) (pp231P4)
Equations
 p 3p4Y2
 p p2y 2
 p2p4y 2
 plp3y 2
 p2p3 y2
 PlPy2
i
1
=1 =1 =1 =1
TABLE 11  Continued
Excludents
4 1
(P1p2 3 ) (Plp21 P3) (p'p31 P2)
(pip3l IP2) (PlP1 P2) (Plp~l 2)
(pip3 PO (p1p 31 4
(plp41 3) 1 l4l P3)
(p3P4lPl)l (p3P41 P9 (P2P41 P')
(p2P41 P (p2p31 p1)
(p2p31 p1)
________________________
(p3P4j'p2) (p3pQl P2 (p2P41 P3)
(2P3 )
(24Ip3) P (p2p3I PO
(129) (130) (131) (132) (133)
(134)
plp2x 2 p3p4.2 plp3x2 p2P4x 2
p2p3 x2
4j
79
arise from (6) and have integral solutions x, y. We note the following about (120)  (134).
2
Let X y1. Then equation (120) cannot arise.
2
For if it did then by Corollary 7 and (30)j, X, yl, a contradiction to the assumption that NJ # y2e
Let (121) arise from (6) with 81= p1 and d, = p2p34* By Theorem 1 or Corollary 9 there exist integers x, y satisfying (121). It follows from (121) that
plx2 = 1 (mod p2), 9, 2 , 1 (mod p3), (135)
pJx2 a 1 (mod p), and p2p3p4y2 = 1 (mod p1). Define the integers p', p", p ' I pR by
p'p 1 (mod p2 p p = 1 (mod p3), (136)
pb 11 1 (mod p1), and p p2(34 . 1 (mod pl). Then (135) may be written as
p'p x2 M p' (mod p2), P"P1x2 p" (mod p3),
f"p 2 = p"' (mod p4), and p (4) ppP4y2 (mod pl),
which in view of (136) reduce to
x2 z p' (mod p2), x 2 P" (mod p3), (137)
x2ap'" (mod P4), and y2 ~ .(4) (mod p9).
80
By (136)1 we may write (p'1p2) (P'*p'11P2) (PlP2)* Hence if (pl1p2) (P'jP2) 1, then (135)l is not solvable and, therefore, neither is (121). Similarly (121) is not solvable if either (pljp3) = 1, or (PlIP4) = 1, or (p2p3P4lP) = 1. Thus the four excludents listed opposite (121) in Table 11 are established. By like reasoning the four excludents listed opposite equations (122)  (134) in Table 11 are derived.
We will make use of the following lemmata. Lemma 5. The conditions
(135) (P ) = Pk = = ) = +1,
where pi, pi, pk' p, are distinct odd primes and i, j, k,
= 1, 2, 3, 4, follow if and only if either
(139) (p ) = p k ii = +1 and p, 1 (mod 4),
or
(140) (p i ) = (p I k (p i ) = +1 and
P G pk & p = 3 (mod 4),
or
(141) (p ) pk= ) = +1 and exactly two of
Pit PkI p, a 1 (mod 4).
Proof. If (138) holds, consider (p jPkPIPi) = +1, which may be written as (pj pj)(pkIPj)(P 1i) = +1, or
(142) A(Pi1)'Pj)+(pil).(pk)+}(pi1).(pIl)
Pj k=PI)+
By (139) and (142),
= +1, or
j(p l)(p +p p
= +1,
which holds if .nd only if either p1 ! 1 (mod 4) or (PJ+PkPI3)/2 = 0 (mod 2). Thus pJ+pk~Pj = 3 (mod 4), which holds if and only if either p k 3 (mod 4) or
exactly two of pji ,k' P1E 1 (mod 4). Hence (139)  (141) are established.
Conversely, if either (139) or (140) or (141) holds we may write
pi [ p [k +P1
Pi pj+PkPi3
= +1.
Hence (138). This completes the proof of the lemma.
Lemma 9. The conditions
(+) (+1) (+1) ( 1)} (p 1) ( p +p k P
82
(143) (PiI P) = (Pil~k) = (PIkP) =) = +1,
where pi, ps ,Pk, and p, are distinct odd primes, and
i, 1,
k, j = 1, 2, 3, 4, follow if and only if either (144) (p11pJ) = (P11Pk) = (pi p0) = +1 and p. 3 (mod 4), or
(p11p1) = (P= (pPkiP) = +1 and
(145)
P a P pi 1 (mod 4),
or
(146)
(piip3) = (P= (Pi1i) = +1 and
exactly two of p1, Pk' PI 3 (mod 4).
Proof. If (143) holds, then the relation (P Pk 1i) = +1 may be written as
(p p 1i)(pk pi ) = +1.
Also, since (p11pj) = +1, it may be written as
(P p )(l) (pi+l)/2'(pl)/2 upon multiplying each side by (p3 jp1),
(148)
(l) (pi+l)/2'(p31)/2
= +1; or
= (pi 1PI)
1
(147)
Similarly since (p.1Pk) = (Pi pi) = +1,
(1)(pi+l)/2 (pk1)/2 k ) and
(149)
(1) P 1+1)/2'(pll)/2 (lp)
Substituting (149) and (149) into (147) and combining, we get
)(~ p +l)/2(p +pk 3)/2
(&V.~ =4  ) = +1,
which holds if and only if either p1 3 (mod 4) or (Pj+Pk +P 3)/2 z 0 (mod 2). Thus PJ+Pk+P = 3 (mod 4), which holds if and only if either pj pk  z1 a 1 (mod 4) or exactly two of pi, pk' P= 3 (mod 4). Hence (144)  (146) are established.
Conversely, if either (144) or (145) or (146) holds, we may write (p1Ipj)(p1IjpXp1Ip1) = +1, or
(p Ip)(p ki Ppi Pi+)/2(p +pk 3)/2 +1,
so that (p jPkP pi)(+l) = +1. Hence (143). This completes the proof of the lemma.
Lemma 10. Let pi, PJ, Pk' pi be four distinct odd primes where 1, j, k, = 1, 2, 3, 4. The conditions (150) (pIp Pk = 1 P k 9= PI 9 ) = +1,
9.
hold if and only if
(151) (PiPlPk)  (p polI) (PkPPi) +1
and one of (152)  (163) hold (mod 4), where
Pj= Pj= Pk= Pj=1,
3Pjs3Pjs3Pk43Pj 1, PIS pj= Pk 1, = 3i=3Pj::3Pk P I, Pj3 PJ 3pk3 1
(158)
(159)
(160) (161) (162) (163)
3Pi j3Pka POPa 1, Pi= Pj= Pi p1i 1, Pi 2pf=3 ke p 1,
But the conditions
(164) (p Ipj 1Pk) PpiO (PkPj ii  PkP IPJ) = +1,
hold if and only if (151) and one of (169)  (169) hold (mod 4), where
Pi73P9 pka p= 1, 3Pi pgO3p3 1
(167) (169)
3pi;; p PO Pis1,
3pj P~ P1 p1= 1.
Proof. Since p., P pk' p, are four distinct odd primes where 1, j, k, I = 1, 2, 3, 4, there are 2 = 16 possible arrangements of the least positive residues (mod 4) of these four primes, namely, (152)  (163) and (165)  (169) above. Let (151) hold and consider (152)  (159). In each P= p (mod 4), hence
(152) (153)
(154) (155)
(156) (157)
(165) (166)
85
(169) Pipj = 1 (mod 4) and (lLp ) = (lIlP). Thus the first two members of (151), or (PPj k
1, may be written as
( p )=(Pk i PkI ) = +1 and
((,p)p ) =P ) = +1,
respectively, so that (170) (Pki = k j) and (p=p p p .
From (151)3 ,
1 = (Pk = k i '
which by (169)2 and (170) may be written as
1 = (lIPJ)(PkP)(PIIPJ) kIJ
Hence when (151) and one of (152)  (159) hold, (PkPIIPJ) +1.
Again let (151) hold and consider the remaining
arrangements (160)  (163) and (165)  (169). In each p3r (mod 4); hence
(171) pipj i 1 (mod 4) and (ljp,) = (llp
Now multiplying (151)i and ( 15)2 by (lIpk) and (lip1) respectively, we obtain
(ppJPk (IPk) and (p p pI) = (l1pI).
These, by (171)1, may be written as (172) (PkLPiPj) = (llpk) and (p1 p p,) = (lip ). Multiplying (172), and (172)2 by kJ) and (plI p) respectively, we get
(173) (pkIPi)(PkIP (lIPk) and (p jp1)(pijpj)(lipj). Now multiplying corresponding terms of (173)i and (173)2 together we obtain
(174) (PkPj Pi) ~ (PkPIIPJ)( lIPkPI)*
Hence multiplying corresponding terms of (171)2 and (174) produces
(1751 (PkPt IPi) = (PkPOIP )('IPkPi). Since by (151)3, (PkPjIPi) = +1, (175) reduces to (176) 1 = (pkPlP (lIPkPOi
Consequently in (176), (PkPIPJ1) = +1, if and only if PkPI =
3 (mod 4); and (pkPilPj) = 1, if and only if PkPI a 1 (mod 4). Hence, remembering that p1 3P3 (mod 4), (PkPJIPj) = +1 when
I.,.: L
97
(151) holds simultaneously with one of (160) (163); and, (PkPjP ) = 1 when (151) holds simultaneously with one of (165)  (168).
Conversely, if (150) holds, consider (150)4, or (.PkP IPJ) = +1, which may be written as
1 +
.p + 1 ,
or
(177)
= +1.
Now from (150)1 and (150)2,
(pilpk) (PPk)
and (p bi ) = (piliP).
Hence (177) may be written as
= +1.
If we let Pk (mod 4) in (179), it may be written
as
= +1,
or upon multiplying both sides by (lpi),
=(lPj).
((Ip )/2
(179)
(179)
(Pg IPkP (' (p l)/2(Pkc+pi)/2
(pj ipkPj(1 (p 1)/2(pki+p )/2
(Pk'p1pj( (pll)/2(2pk) /2
 p,  D9
pkP +
This last reduces to
(180) (l) =(1p)
since by (150)3, (pkpjIpi) = +1. We note that (10) holds if and only if p, = pj (mod 4). Hence when (150) holds, pk = and pi m p, (mod 4), one of (152), (153), (156), and (157) holds.
If we let Pk z 3PI (mod 4), then by (150)3 and (179), (151) 1 P (Pp IkPI)*
Hence pi, pj = 1 or 3 (mod 4). Thus when (150) holds and Pk = 3P (mod 4), one of (154), (155), (155), (159), (160) (163) holds.
Now if (164) holds, consider (164)4, or (Pk
1, which may be written as
(192 (pJPk~j(_l)(pjl)/2(pk+pl)/2=
By (164)1 and (164)2' relations (179) hold, and hence (182) may be written as
(pjl)/2*(Pk+P )/2
(153) (p1I Nk~ ) 1
Now let p (mod 4) in (153), then it may be written as
(p 1)/2
(p k ii )(l
or after multiplying both sides by (1jp,), we get
89
(pkP1 iP 9( l)? (p(l)/2
Hence, since by (164)3, or (Pk 91P)= +1,
(1)(pjl)/2
This last holds if and only if pi = 3p3 (mod 4). Thus when (164) holds, Pk ; p, and P. P 3Pj (mod 4), one of (165) (168) holds. Since all possible arrangements of the least positive residues modulo 4 of the four primes have been exhausted with respect to (179) and (193), the proof of the lemma is complete.
We seek the necessary and sufficient conditions for each of the equations (120)  (134) to have a solution. These will now be determined in thAe theorems which follow. First we will be concerned with equations (121), (123), (125), and (127) and will make use of Lemma 8. Theorem 7. Let D = plp2p3P4 where pi < p2 < p3 < P4 are distinct odd primes. Let t1 + u\/D be the fundamental solution of (1), t2  Du2 = 1, where t1 = 2N, + 1. If , y2,
(14) (Pi1PJ) = (PiiPk) = (pipi) = +1, pi 1 (mod 4), and either
(195) exactly two of' p p, pk ,1 = 1 (mod 4) and
at least two of (PJIPk)' (PjiPO, (PkP =I 1,
90
or
(186) p 3 (mod 4) and
(PJIPk) k ) = +1,
or
(187) P k P 3 (mod 4) and
(p = (PIP) k ) = +1,
then either equation (121), (123), (125), or (127) has a solution in integers x, y according as i = 1, 2, 3, or 4, with j, k, I = 1, 2, 3, 4 where j < k < I. Solutions are given by
( ) 2 2 /4) and 2xjy = u1.
y1 it 1,/4
Proof. By hypothesis and Theorem la one of (120) (134) of (6) has a solution in integers x, y. By Lemma 7, since NJ y , equation (120) has no solution.
Let i = 1, j = 2, k = 3, 1 = 4 in conditions (139) of Lemma 8, then
(189) (PlIP2) = (p11p3) = (P11P4) = +1 and p1 = 1 (mod 4). Hence by Lemma 8, (P2P3P4i1P) = +1; and thus by Lemma 7 the necessary conditions for (121) to have a solution are satis2
fied. Now all excludents of Table 11, except X, y1, can be expressed in terms of one or more of the six Legendre symbols,
91
(P1P9, (p2 p3), (p19p4), (p2jp3), (P21p4), and (p31p4). Hence when conditions (189) hold, the values of (p21p3 ), (P21P4), (p31P4), and the least positive residues modulo 4 Of P21 p3, and P4 can be arranged in 26 = 64 different arrangements or cases as shown in the six columns headed "Conditions" in Table 12. From taking each of the sixtyfour cases in turn we determine those excludents of Table 11 which are consistent with conditions (189). The procedure is the same as that used in developing Table 4. The results are recorded in the last column of Table 12, omitting the column which designates the particular excludent which holds, as was done in Tables a, 5, and 6.
We see from Table 12 that equations (122)  (134) have no solutions under the conditions (199) and those of any one of the fourteen cases:
2: c, e, g, h, 6: c, e, g, h,
which are recorded in Table 17 s Theorem la, since NJ yl, one c has a solution in integers x, y. and those of any one of the case
2
hold, and # yl, equation (121 Conditions (199) and those of th
4: c, e, g, h, 9: d, e,
Ls cases (1)  (14). By )f the equations (121)  (134)
Hence when conditions (199) .s (1)  (14) in Table 17 .) of (6) has a solution. ie first twelve cases of
92
TABLE 12
CONDITIONS TO EXCLUDE EQUATIONS (122)(134) WHEN X y
(p 11p2 = 1 1P3 ) 1Ip4) = +1, p, v 1 (mod 4)
Conditions Equations
Case Ip 3 aP4j (P2Ip3)= (P2P4)= (P3IPO excluded
(mod 4)
1(a)
1(b) 1(c) 1(d)
1(e) 1(f)
1(g)
1(h) 2(a)
2(b)
2(c) 2(d)
2(e) 2(f) 2(g)
2(h)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1 2.
1
+1 +1 +1
+1
1
4.
1
1
+1 +1
+1 +1
1
1
1
1
+1 +1
1
1
+1
+1
1
1
+1
+1
1
1
+1 +1
1
1
+1
1
1
+1
1
+1 +1
1
+1
1
1
+1
1
+1 +1
1
none
B
C
D C
E C
C
F
G
A
H A I A A
aA denotes equations
B denotes equations C denotes equations D denotes equations (133), and (134).
E denotes equations F denotes equations
(130), (132), and (133).
G denotes equations
H denotes equations
(122)(124)
(125)(129), (131)(134). (123)(134
(123)9, (12 ) , (127) (130)9,
(123)(126) (129)(132). (122), (124), (126), (127),
(122), (124)(128) (130)(134)(122) (124) , (126)(130)t, (132
I denotes equations (122)(127), (129)(133).
(Table 12 is continued on the next page.)
'1.* 
93
TABLE 12  Continued
Conditions Equations
Case 2 3 4 2 I 2 4 3 excluded
od 4) _ 2_ _ __3
3(a) 3(b)
3(c)
3(e) 3(f)
3 (g)
3(h)
(a) 4(b)
4(c) 4(d)
4(e) 4(f)
4(g)
4(h) 5(a) 5(b)
5(c) 5(d)
5(e) 5(f)
5()
5(h)
1
1
1
1
3
1
1
1
1
1
1
1
1
1
1
3
3
3
3
3
3
3
3
3
3
3
1
1
1
1
1
1
1
1
3
3
3
3
3
3
3
3
+1 +1 +1 +1
1
1
1
1
+1 +1 +1 +1
1
1
1
1
+1 +1 +1 +1
1
1
1
1
+1 +1
1
1
+1 +1
1
1
+1 +1
1
1
+1
+1
1
1
+1 +1
1
1
+1 +1
1
1
+1
1
1
+1
1
+1 +1
1
+1
1
1
+1
1
+1 +1
1
+1
1
1
+1
1
+1 +1
1
F
J
K
H
L
I
M
N
J
G
A
K
A
L
A
A
F
G
m
N
K
I
L
K
J denotes equations (130), (131), and (134).
K denotes equations (133), and (134).
L denotes equations (134).
M denotes equations (131)(134).
N denotes equations (129), (132), and (134).
(122), (124),
(122)(125), (127)(131),
(122)(126), (129)(132), and (122), (123), (125)(129), (122), (123), (126), (128),
(Table 12 is concluded on the next page.)
4.
(125)v, (128)9,
94
TABLE 12  Concluded
Conditions Equations
Case P24 (p2 1P3 ) 2p4) 93 4 excluded
(mod 4)
6(a) 3 1 1 +1 +1 +1 N
6(b) 3 1 1 +1 +1 1 M
6(c) 3 1 1 +1 1 1 A
6(d) 3 1 1 +1 1 +1 H
6(e) 3 1 1 1 +1 1 A
6(f) 3 1 1 1 +1 +1 L
6(g) 3 1 1 1 1 +1 A
6(h) 3 1 1 1 1 1 A
7(a) 3 3 1 +1 +1 +1 J
7(b) 3 3 1 +1 +1 1 G
7(c) 3 3 1 +1 1 1 I
7(d) 3 3 1 +1 1 +1 K
7(e) 3 3 1 1 +1 1 M
7(f) 3 3 1 1 +1 +1 N
7(g) 3 3 1 1 1 +1 H
7(h) 3 3 1 1 1 1 1
8(a) 3 3 3 +1 +1 +1 I
8(b) 3 3 3 +1 +1 1 K
8(c) 3 3 3 +1 1 1 K
8(d) 3 3 3 +1 1 +1 A
8(e) 3 3 3 1 +1 1 A
S(f) 3 3 3 1 +1 +1 I
8(g) 3 3 3 1 1 +1 M
8(h) 3 3 3 1 1 l N
Table 17 establish (184) and (15), while (19) and the conditions of the thirteenth and fourteenth cases in Table 17 establish (184), (196), and (187) for i = 1, j = 2, k = 3,
1 = 4. By symmetry, interchanging the subscripts 1 and 2 in the cases (1)  (14) of Table 17, we get the necessary and sufficient conditions for equation (123) to have a solution. Also by symmetry, interchanging the subscripts 2 and 3 in the conditions by which (123) has a solution, we obtain those by which (125) has a solution. Again by symmetry,

Full Text 
PAGE 1
PROBLEM ON PELLIAN EQUATIONS By JOHN WILLIAM GREINER A DISSERTATION PRESENTED TO THE GRADUATE COUNQL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA February, 1958
PAGE 2
1 ACKNOWLEDGMENT The writer wishes to express his sincere appreciation to Professor Edirln H. Hadlock, Chairman of his Supervisory Committee, for a generous contribution of time, energy, and helpful criticism throughout the preparation of this work and to Professors W. R. Hutcherson, P. W. Kokomoor, D. E. South, and L. E. Henderson, all of v7hom served as members of his Supervisory Cominlttee. To Professor Hadlock is due a large raeasiire of gratitude for sympathetic encouragement and guidance during the writer's entire doctoral program. Finally the v:riter vjishes to acknowledge that without the help of his vrlfe, Lillian Greiner, his graduate studies could not have been begiin and that without her continued confidence in him his graduate studies could never have been completed. 11 M
PAGE 3
TABLE OP CONTENTS ACKNOWLEDGMENT Page 11 LIST OP TABLES Iv Chapter p p I. THE EQUATION ^.^x d^y = 1, WHERE ^^'6^ = D, A PRODUCT OP DISTINCT ODD PRIMES 1 II. CONDITIONS FOR ONE OP 6^x^ d^y^ Â«= 1 TO HAVE A SOLUTION WHERE S^'d^ = D, A PRODUCT OP TWO DISTINCT ODD PRIMES 2& III. CONDITIONS POR Om OF 6^x^ d^y^ = 1 TO HAVE A SOLUTION WHERE 6^'d^ = D, A PRODUCT OP THREE DISTINCT ODD PRIMES ^12 IV. CONDITIONS POR ONE OP fi^x^ d^y^ Â» 1 TO HAVE A SOLUTION WHERE 6^d^ = D, A PRODUCT OP FOUR DISTINCT ODD PRIIiES 76 BIBLIOORilPHY l'+3 BIOGRAPHICAL SKETCH ikk 111
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LIST OF TABLES Table Page 1. Data for Example Illustrating Theorem la . . , 25 2. Data for Example Illustrating Theorem la . . . 26 3. Excludents of Equations (72) (76) 43 k. Conditions to Exclude Equations (7^) (72) when 7^ y^, (pj^lp^) = (p^lp^) = = 1 (mod ^9 5. Conditions to Exclude Equations (7^) iJS) vfhen y^, (p^ip2) = (PiiP3) = p^ = 3P3 (mod if) 53 6. Conditions to Exclude Equations (73) and (75) (7^) when X^ ^ (PilPg) = (P^Ip^) Â° +1. Pi = 5 ^2 7. Conditions to Exclude Equations (73) and (75) (725) when X^ ft y^, (p^jp^) = (P1IP3) = +1, and P2 = pj (mod k) 65 6, Conditions for Equation (7^), PoPz^^ " ^^"^^ ~ 2 / 1, to Have a Solution V7hen \ Â°7 9. Conditions for Equation (76), P^PzX^ PpJ"^ = 1, to Have a Solution when ^2. Yj^ o9 10. Conditions for Equation (78), P,PoX^ PrJ"^ = X 2 2 3 1, to Have a Solution when ?^ 72 11. Excludents of Eqviations (119) (13^) .... 77 12. Conditions to Exclude Equations (122) il'}^) when Xj_ 7^ y^, (p^lpg) = (P1IP3) = (Pilpi^) = +1, p^ a 1 (mod if) 92
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LIST OF TABLES Continued Table Page 13, Conditions to Exclude Equations (122) (13^), when f ^V^V'^ = (p^iP^) = (p^lpii) = +1, P2 = P3 a PI], a 3 (mod \) 97 ill. Conditions to Exclude Equations (122) (13^), when X^ f y^, (p^^lpg) = (P1IP3) = (Pilpi].) = +1, P2 = P^ s 3Pi,. a 1 (mod ^) 101 15. Conditions to^Exclude Equations (122) (I3I+), when XjL f y^^, (PilPa) = (P1IP3) = (Pilpi].) = +1, pg = 3P3 3 Pl a 1 (mod \) 102 16. Conditions to Exclude Equations (122) (I3I1), when X^ ^ y^, (p^lpg) * (P1IP3) (PilPl].) = +1, 3P2 S P3 a Pli S 1 (mod \) IO3 17. Conditions by Which Equation (121) Has a Solution when X]_ ^ yj_, and (plIp2) = (P1IP3) = (Pilpi^) = +1 105 1^, Conditions to Exclude Equations (121) and (123) (I3if) when Xi yf, (pgipi) = (P3IP1) = (Pl.IPi) = +1, Pi = 3 (mod it) . . . log 19. Conditions to Exclude Equations (121) and (123) (13^) when Xi f yf, (pl1p2) = (Pllpj) = (PilPi].) = +1, P2 S P3 s PI]. 5 1 (mod 4) Ill 20. Conditions to Exclude Equations (121) and (123) (13^) when Xi 7^ yf, (p^lpg) = (Pllp^) = +(Piipi].) = +1, 3P2 = 3P3 Pi], h 1 (mod 115 21. Conditions to Exclude Equations (121) and (123) (13*+) when \^ ^ y\, (p^lpg) = +(PiIpt) = (p^lPi^) = +1, 3P2 a P3 5 3Pi]. 2 1 (mod k) 116
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LIST OF TABLES Concluded Table Page 22. Conditions to Exclude Equations (121) and (123) (13^) when \ f +(Pi1p2) = (PXIP3) = (PilPi) = +1Â» P2 = 3P3 s 3P1j. 3 1 (mod 117 23. Conditions under Which Equation (122) Has a Solution when Xj_ f y^, (P2^p2) = (P1IP3) = (PllPl^) = +1 24. Conditions to Exclude Equations ^121) (126S) and (130) (13*+), >fhen 7^ y^, and (P3_P2lP3) =^PiP2l^i^ " (P3P[.IPi) =+1 . . . 12I; 25. Headings for Substitution In Tables 1?, 23, and 2^ 133 26. Congruences Showing Insolvability , Equations (121) (131). (133), and (134) 1^12 vi
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CHAPTER I THE EQUATION 6^x2 d^y2 = 1, WHERE ^j^.^'^ D, A PRODUCT OF DISTINCT ODD PRIMES If the Integers t and u satisfy the Pelllan equation (1) t2 Du2 Â» 1, where D Is a natural number not a perfect sq^iare, the number t + uv^ Is called a solution of the equation (1). Among the Infinite number of solutions of (1) there is a least positive solution ti+uiy'^ In which t^ and u^ have their least positive values."^ This number t]_+U]y/D is called the fundamental solution of equation (1). All solutions of (1) with positive t = t^ and u = u^ p are given by the formula (2) tn + u^D = (tL + uj/DP, n Â» 1,2.3. where ^rygve Nagell, Introduction to Number Theory (New York: John Wiley and Sons, Inc., 1951), p. 197. ^Ibid. 1
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and Lemma 1 . Let t]_+U3_v'D be the fundamental solution of (1) with ti odd. Then tjj of (3) is odd and of {^) Is even, where In the expansion of (2), Is found by equating all terms free of\/D, and Is found by equating all terms oontalnlngy^. Further tj^ > u^^. Proof . Since t^ Is odd, t^ Is odd. Because u^ Is even, then by relations (3) and (4), It follows that t^ Is odd and Is even. Since tjj+iajj>/5 Is a solution of (1), n = 1, 2, 3, Du^ = 1. It follows that t^ = Du2+ 1, and hence t > Du^. Since D > 1, then t^ > n^. Theorem 1 . Let equation (1) have the fundamental solution t^+ui\/D In Integers t^ and u^, where (5) ti = 2XL + 1 and D = O^H, 0^ being the largest odd square dividing D, and H Is the product of distinct odd primes, P3_P2' Â• 'Pg ."^ Then there are at most 2^ possible equations of the form 'Â•Leonard Eugene Dickson, Studies In the Theory of Numbers (Chicago; The University of Chicago Press, 1930) Â»
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3 (6) 6^x2 dj^y^ = 1 In which equation (6) has a solution in integers x = and y = yjj, where (7) tÂ„ = 2X^+1 and are given "by (3) and (4) respectively, and Xjj, 6jj, and d^ are all integers, n = 1, 2, 3, with Proof . Since t^^ is odd, it follows from (1) that U]_ is even. By Lemma 1, is odd and is even. Since relation (2) gives all solutions of (1) with positive t^ and uj_, it follows from equation (1) that t 1 = 0 = G^Hu^Cmod g), and Ujj is a multiple of ^, Thus (9) % =Â» ^Mn defines an integer Vl^, It follows from (1), (5)2* ^7)i (9) that + iXn + 1 =Â« G%(i6m) + 1, so that (10) XjjOvn + D = 1+02m2h. Now let y^ be the g. c. d. of and 40^11^. Then and
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are defined by (11) = 2QM^ = Vn respectively, where d^ and x^^ are relatively prime. Hence by (10) and (11), dnyn(Xn + 1) x^yH, or (12) + 1) = xh. Since (djj.x^) = 1, d^^ divides H in (12), so that {&)^_ defines an integer 6^^. Prom (g)j, which is (ll)i, i&)^.t and (12), it follows that and y^j are solutions of (6). Using (9) and the fact that y^ " {X^.kQ^VI^) , (g)i results. Also (8)3 follows from (11)2 and (9). Finally, since there are 2^ divisors of H, there are 2^ values of and hence at most 2^ equations (6) , Henceforth let 0 = 1 so that H = D, an odd natural number which is squarefree. Corollary 1 . Let be a positive divisor of D, Let (13) = 2(D/P^)M^ +1 and u^ = 2B^l^ be given by (3) and (4) respectively vihere iB^,D/'P^) = 1, B^^ and both > 0. Then W) = and y^ = are solutions of
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1 (15) Pn^ (D/Pn)yi Â« 1, If and only If and of (13) are solutions of (1). Proof . Let tjj and of (13) be solutions of (1). Prom (7). (S)i and (I3), since (Bq.D/Pjj) = 1. Thus by (7), (2)3, and (I3), and henoe d^ Â« D/P^j. Then by (S)i, 6jj Â» P^^ so that (6) takes the form (I5) with solutions x^, y^. Therefore, by (g)2 and (13)2. follows that 2x^n^ Â» SB^Mn; hence % = B^. Conversely, let and yj^ of (1^) be solutions of (15). Then from (I3) (15), ^n '^n 4Dm2(Dm2 + Pn PgBg) + pg P2 n ^Pyn^^^Pn^ Pn^)3 ^ Pg p2 p2 n n H
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6 Henoe and of (13) are solutions of (1). Example 1 . Let the given equation be t2 209u2 =1 1 where 209 = 11^19 = D and having the fundamental solution t^ + ui\/D = 46551 + 3220\/209. Hence t^ Â« ^6551 = 2?vi + 1 Â» 2(23275) + 1 = 2(19.352) + 1, which Is of the form (13) where D/Pj^ = 209/Pi = 19, so that = 11 and M;j_ = 35. Also, 3220 = ui = 28^(35), and hence B^ = HS. By (1^^) and (I5), xi = H6 and = 35 Is a solution of 11x2 19y2 = 1. Checking: 11(462) 19(35^) Â« 23276 23275 Â» 1. Example 2 . Let the given equation be t2 65u2 1 with the fundamental solution 129 + l6\/^i where tj^ = 129 Â» 2Xi + 1 = 2(64) + 1 = 2(lg2) + ij hence by (I3), Â» g2^ and D/Pi Â» 65/Pi Â» 1, so that Pj, Â« 65 and = g. Also I6 Â» ui = 2B]^(g), and henoe B^j^ = 1. By (l4) and (15), x^ = 1 and y]^ = S is a solution of 65x2y2= 1. Checklngj 65(1)2 (g)2 = In addition tj + U3 = ^536369 + 106500g\/^. Hence
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7 t3 = 35^6369 = 2X3 + 1 Â» 2(1^29313'^) + 1 = 2(120722) + 1, so that by (13), = 20 72^, D/P3 = 65/P3 = 1; thus P3 = 65 and M3 = 2072. Also 106500g = U3 = 283(2072)1 hence B3 Â» 257. By (l4) and (I5), X3 = 257 and y3 = 2072 Is a solution of 65x2 y2 Â» 1. Checking: 65(257^) (2072^) = 12931^5 '+2931Â«^ Â» 1. Corollary 2 . If D contains at least one odd prime p = 3 i^^oCL ^) , then equation (16) y2 Dx2 Â« 1 of (6) cannot arise with solutions Xq and y^^. Proof . Equation (I6) Is a special case of (6) where <= D and djj Â« 1, and by Theorem 1 has solutions x^, Yn* Bi^t y^ = 1 (mod p) has no solution. Hence (I6) cannot arise. Corollary 3 . Let D contain no prime factor p = 3 (nio^i H) , and t]_ of (13) be given by t^ = 2mJ + 1. Then equation (16) of (6) arises having the fundamental solution y^ + x^ v^, where yj^ Â« M]^ and x^ = B^ are given by (13)2* Proof. By hypothesis and (13)i, D/P^ Â«= 1, so that ?2, = D; hence, by (l4) and (15), x^^ = B^ and y^. = % Is a solution of (16). To show that y^^ + Xq_\/D Is the fundamental solution of (16), let Its fundamental solution be Y + XV^. Then (Y + X\/D)2 = Y^ + DX^ + 2YXVD
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8 Is the fundamental solution of (1) where (17) ti Â» + DX^ and u^ Â» 2YX.^ It follows by hypothesis and (I7) that (IS) 2Mf + 1 2yi 1 Â» ti Â« + DX^. Since Y + Xs/d is the fundamental solution of (I6), then from (Ig) it follows that 2y + 1 Â» DX^ 1 + DX^ or 2yf + 2 Â» 2DX^, so that (19) + 1 = DX^. Since y^ + x^s/D is a solution of (I6) it follows from (16) and (19) that Dxf = DX^; thus = X. Hence from (13)2, (1^), and (17)2* follows that y^ = Y, An example of Corollary 3 is Example 2, page 6, in which D Â«Â» 65 contains no prime factor p = 3 (mod U) and t^^ = 2(g^) + 1, The solution x^ = 1, y^^ = ^ is the fundamental solution of y2 65x2 = 1. Example . The equation t^ 5u2 Â» 1 in which D Â« 5Â» " 2(2 ) + 1, is another example of Corollary 3. Hence = 2, Also by (13), I^Â• = ui = 2BiM;l Â° 2Bi(2), so that = 1. Hence by (1^) and Corollary 3, y^ = 2, x^^ = 1 is the fundamental solution of Nagell, OP. cit. p p, 201.
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9 Corollary 4 . Let + be the fundamental solution of (1) where t^ = + 1, = (X^.u^A), and Sxiy^ = u^, then (20) Dy2 a 1 of (6) cannot arise with the solution xj_, y^^. Proof . Equation (20), which is of the form (1), is a special case of (6) with n = 1 where Â« 1 and dj^ Â« D, and by Theorem 1 has the solution x^i^, y^^. But by hypothesis i^l^lyil and U]_>x]_, and by Lemma 1, t]_> u^. Hence ti>x]_. Therefore Uil + jyil \/5 < ti+u^V^i which is not possible, since t^^+UQ^V^ is the fundamental solution of (l) where t]^ and Uj^ have their least positive values. Hence (20) cannot arise with the solution x^, Yj^* Corollary 5 . If in (1), D = p is a prime = 1 (mod k) , then equation (l6), y^ Dx^ Â« 1, of (6) arises and has a solution, Xj^, y^, given by (g) . Proo f . Let t^+u^^x/D = t^^+Uj^v/p be the fundamental solution of (1) in which t^ is odd, and where D Â«= p is a prime =1 (mod k) , Thus relation (5)^ in Theorem 1 gives ^1*^1 " Therefore, since p is a prime, one of 6j_ and dj^ is unity. Hence (6) becomes either (21) x2 py2 = 1
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10 or (22) yf px = 1, By Theorem 1 one of (21) and (22) arises and has a solution. But by Corollary k eqiiatlon (21) Is not possible. Therefore equation (22) arises and has a solution, x^, y^, given by (&) . We note that Corollaries 1 and kimply that If Pjj is a positive divisor of D different from 1, then P^x^ (D/Pn)y^ = 1 of (6) must arise. The following oorollaries place definite restrictions on the form that (6) may take according as the subscript n in Theorem 1 is odd or even, and according as = 1, D or ^ 1, D, The following Corollary 6 considers the relations of Theorem 1 when n is even, and places no restrictions on the nature of D other than that in its original definition, where it is a squarefree odd natural number. Corollary 6 . Let tj^+uj^^/D be the fundamental solution of (1) 2 2 where tj^ = 2\^ + 1. Let t2i^ = ^'^2k **" '^2\s. ~ ^^2kÂ»^2k/^^Â» and Sxgi^ygk = Ugj^, where tg^j and Ugj^ are defined by (2), n = 2k, k = 1, 2, 3, Â• Â• Â• . Then (23) X2k = I>Uk = Dyi,. and the equation (24) x2 Dy2 = 1
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11 of (6) always arises and has solutions x = Xgj^, y = Y2\s.> where ^2k Â° *k ~ ^^i'ther xg+ygv/^ Is the fundamental solution of (2^) which Is of the form (1). Proof . In (2), since n = 2k, k Â» 1, 2, 3, t^j^ and Ugjj are given by (25) (ti + \x^\/0)^^ Â» tjk + U2k\/D. Now by (2) (ti+Ui\/5)^^ = [(ti+Ui\/5)^]^ = (tic+Uk\/5)^ = t^+^k 2tijUi^\/D, and It follows that, from (25), (26) tgij = t + DuÂ§ and Ugj^ Â« ^tj^Uj^. Hence by hypothesis, (26)]_, and the fact that tj^ and Uj^, are solutions of (1) = ^2k " ^ _ ^k ^k Â• ^ _ (Du^ + 1) tDug 1 _ 2Dug ^2k= 2 2 2 2~* so that (27) Xgk " ^k' From the hypothesis and (26)2, and (27), yL = (>^2k> 4kA) ^t^^^A) = u^, since (D, t^) = 1. Thus ^Nagell, op. clt. . p. 201.
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12 (2g) y^ic = ugj and hence from (27) and (28) relation (23) follows. Consequently in (g)^, = ^2k/V2k " ^k/^k = ^' ^^^4' 62k*D = D, so that = 1. Thus (6) reduces to (2^1) vihioh Is of form (1). Moreover, from (26)2, (28), and the hypothesis where 2x2ky2k " ^2kÂ» follows that (29) X2k " *k . Hence from (28) and (29), Xg^^, y2k solutions of (1). Thus when k = 1 in (28) and (29), + y2V^ is the fundamental solution of (2^1) which is of the form (1). Example . Let the given equation be t^ 39u^ = 1 with the fundamental solution t]L+Ui\/D = 25 + By (25 ), k = 1, t2+u^D = 12i^9 + 200 V39. By (7), t2 = 12^9 = 2X2+ 1 = 2(62^) + 1 = 2{k^39) + 1. Hence, ~ ^lÂ» asserted in relation (23). Thus by (8)3 and (8)^, d2 = 39 = D, and 62= 1, so that equation {2^) of (6) arises. By (28) and (29), for k = 1, y2 = U]_ = 4and X2 = t]_ = 25 is the fundamental solution of x2 39y2 Â» 1, which is of form (1), the given equation. Corollary 7 . Let tj+u^\/D be the fimdamental solution of (1) where t^ = 2Xi + 1, and D contains no prime factor p = 3 (mod h). Let t2k+i " 2X2k+i + 1, yfk+l = ^^2k+l' ^Ik+l^^
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1 13 ^y2k+1^2k+l ^2k+l' w^Â®^Â® ^2ic+l ^2k+l given by (2) for n Â» 2k+l, k Â« 0, 1, 2, Let equation (l6), Dx^ " 1, of (6) arise with the fundamental solution y^, Then (30) X2k+1 = yik+1 Dx^ =1 of (6) has the solutions ygk+l* ^2k+l' Proof . By hypothesis equation (l6) of (6) arises with the fundamental solution y]_ + Then = ti+ui\/D Is the fundamental solution of (1)."^ Moreover, for k = 0, 1, 2. (yi + XiV^^^^ " y2k+i + X2k+i\/D 2 are solutions of (l6). Since
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2 2 (31) t2ic+i = J2k+1 ^^2k+l ^2k+l " ^yj^j+^Xg^+i . Prom (31), the hypothesis Xsk+l Â° ^*2k+l ~ lV2, and the faot 2 2 that 72^+1 ^^2k+l follows that (y^ + D) 1 + (y2 + 1) 1 V = ^^2kH ^kfl^ = ^2k+l ^^2kH ' or '^Sk+l 2 2 2 2 (30)^. Prom (S)^, and n = 2k+l, ^2k+i^2\câ€¢l ' ^2k+l " y2k+lÂ» so that d2k+l ' ^' ^^^l^t with H = D, 52k+l Â° ^* Hence from (6), equation (30)2 arises with solutions V2k^1* ^2k+l* Corollary 8 . Let t2_+u^\/D be the fundamental solution of (l) where t^ = 2X^ + 1, and D contains no prime factor p S 3 2 2 (mod Let t^^+i " ^^2^+1^ ^' ^Sk+l = ^^2k+l' ^2k+l/'^^ ^2kfl ' 2y2ic+i=^2k+l' ^2k+l = yL+1' "^^^^ ^2k+l ^2k+l are given by (2) , n Â» 2k + 1, k = 0, 1, 2, Then equation (30)2, Dx^ = 1, of (6) arises with solutions y2k+lÂ» ^2k+l* Proof . By hypothesis and (2)^ ^2k+l " ^2k+l " ^2k+iyk+l' ^Â° '^2k+l Â° ^' *2k+l " ^Â» and (6) takes the form (30)2 *'^lth solutions 721^1* ^2k+l* This completes the proof of the corollary. If the Integers and y^ satisfy the equation (32) B^x^ d^Y^ = 1, where x^^ and y^ have their least positive values and 5^Â»d^ = D
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15 of (1), 6]_ 7^ D, 1, then we define the number to be the fundamental solution of (32). Also we define and y^^ by where n = 2k+l, k=0, 1, 2, and Is found by equating all terms containing V^, and y^^ is found by equating all terms containing \/d^. In (33) n cannot be even since the right side of (33) for n even is of the form U + V\/D, where U and V are Integers and D = 6.j_'d^, 6^ 7^ D, 1, which relation Nagell has shown to be impossible.^ In this connection we quote two lemmata by Nagell without proof. Lemma 1. Let x, y, Xn , y^, a, b, and a^j^ be rational numbers 0, such that s/a> Vb and \/a^ are irrational. Then we can never have a relation of the form (15) xVa + y\/b = x^^N/aJ + yi. Lemma 2. Let x, y, Xtl, yj^, a, b, aj^ and h^ be rational numbers 7^ 0, such that \/a, \/b, \/a^, \/ab and v/a^^b^ are Irrational. Then the relation (16) x\/a + yV^ = iTrygve Nagell, "On a special class of Dlophantine equations of the second degree," Arklv F6r Matematik . Vol. 3, No. 2 (195^), p. 53.
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16 holds only In the following cases: It is either xy/a = ^l\/Â®l 0^ ^X/Â®' ^ yiS/i^i."^ Corollary 9 . Let D = Si'd^, where S^. ^ ^1 Â°^ be given by t^ 2dj^M^ + 1. Then equation (32), S^x di^y^ = 1, of (6) arises having the fundamental solution x^, where xj_ = B^^ and y^^ = M^ are given by (13)2Â» Proof . Since 5^ ^
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17 (3^)2Â» follows that xi = X, This completes the proof. Corollary 10 . Let t^ + u^^vyD be the fundamental solution of (1) where ti Â» EX^ + 1 and D Â» Si'di, 5i ji 1,D. Let t2ic+l " 2 2 2^2k+l + 1. y2k+l = <^2k+l. U2k+lA). Sxgk+lYSk+l = ^Sk+l. where t2k+i and Vi2k+1 given by (2), n Â» 2k+l, k = 0, 1, 2 2 2, Then all solutions Xgi^+i, Ygk+l 02), S^.^ ~ ^^1^ a 1 are obtainable from 2k+l (36) ^sk+i^vAI + y2k+iVd][ = (xiN/fii + yiV^i) k = 0, 1, 2, v^here x^V^ yiV^ is the fundamental solution of (32), and (37) X2k+i = tijXi + u^yidi and 72^+1 = tj^y^ + k = 0, 1, 2, vihere tg = 1 and Uq = 0 vihen k = 0 , Proof . Since 6^ ji 1, D, then by Corollary 9, equation (32) of (6) arises v^ith the fundamental solution x^vZ&i + (3g) (xiV/57 + yiV^)^ = ti + uiVD is the fundamental solution of (1).'^ Prom (33), n = 2k+l, k = 0, 1, 2, and relations (30) and (2), and the faot that 81 'd^ * D, it follows that Ibid.
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IS (39) = (tic + ui,\/D) (xiV^I + yi\/dl) = (tijXi+Ukyidi)\/5j + (tijyi+%xi6i)\/dj, where we define Â« 1 and Uq = 0 when k = 0. Also from (33), n Â» 2k+l, k = 0, 1, 2, iko) (xix/67 + yi\/d5;)2^+l = X2k+i\/^ + y2k+lVdI. Hence from (39) and (^10), relations (37) are established. We next show that (37) are solutions of (32). For, from (37) i the hypothesis 6i*^= and the fact that x^^V^ is the fundamental solution of (32), Sl^ik+ldiyik+l = ^l^tijxi + ui,yidi)2 di(ti,yi + xx^^i^i)^ Â« SltkXi + Siuiyidi + 2SiditkUkXiyi d^tgyf diugxfsf 25iditijUi,xiyi Â» t^(63^xf dj^yf) + u^(5idfyf sfd^xf) = tg.(l) + ug(Ddiyf Dfi^xf) + Du2(diy2 S^xf) = Du^ = l. Hence (37) give solutions of (32). To show that relation (36) gives all solutions of (32) with positive y^2k+l y2k+l' w^Â®^Â® ^l\/^ yi^^ is the
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19 fundamental solution of (32), assume that u\/^ + vV^i is another solution of (32) vjlth positive u and v and not obtainable from (36).^ Then It follows that for a positive number n = 2k+l, k=0, 1, 2, Henoe by (36) , (x2k+iV^ + y2lc+lVdr) < uV5l + < <^2k+lVA?y2k+lVdr)(xi\AI+yiVdI) Novj, X2iÂ£+i\/5i y2k+l\/^l is a positive number, for slnoe (x2i^lVfiI y21cflV^^<^2k+l\/^ y2k+iVdI) ^2k+l*l y2k+A Â° ^' and (xpi^i\/67 y2k+l\/^) ^ 0. X2k+l\/5l " y2k+l\/^ ^ 0Multiplying through by X2k+iV^ y2k+lV^ SQt ^2k+l*l yfk+l'^l < (^\/8l + v\/d^)(x2k+iVÂ«I y2k+l\/dl) < (xi,+i6i . y2k+idi)(xiV6l + ylVd^)^Â• or, by (3g) and the fact that ^2k+lÂ» y2k+l' 1Â» 2, are solutions of (32) It follows that (42) 1 < (u\/6j + vV/dJ)(x2k+i\/Sr y2k+l\/^^ < t^ + u^X/D . Let Nagell, Introduction to Number Theory , p. 19g,
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20 (43) (u\/6^ + vV/d^)(x2k+iV^ y2k+lV^^ " ^ ^V^, where, since ^I'd^^ Â« D, w = ^2k+l^l ~ """ysk+l^l ^ ^^2k+l " ^^2k+l* follows that W) (u\/67 v\/5I)(x2k+lV6r + y2k+l\/dl) ^2k+l5l vy2k+l^l + (uy2k+l vx2i^i)v^ = w zVD. Hence multiplying (^3) by (kk) ^ we get or, from the assumption that u\/5^ + v\/d^ is a solution of (32). {^3) 1 Â• 1 = 1 = w^ Dz^. Thus w + z\/d is a solution of (1). Hence, using (42) and (46) 1 < w + zV^ < t^ + u^V^Â» which, if w and z are both positive, is not possible, since t^^ + u^VS is the fundamental solution of (1). Therefore, roust show that w and z are both positive, if u, v is not to be a solution of (32). To that end, we note from (^5) that 0 < Dz^ = (w + zy/D)(w zy/D) = 1;
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21 hence, since w + zV^ Is a positive number > 1 from (46), i^J) 0 < w z\/D = l/(w + z\/D < 1. Now, from (^7) , 0 < w zV/D < 1; hence , (^9) w < 1 + zys. Replacing w in (^+6) by the larger number 1 + z\/D of (^9), we find that 1 < (l Â• z\/D) + z\/D, or 0 < 2z V5. Thus, since \/D > 0, it follows that z > 0; hence, from , w > 0. Therefore, the inequality {^6) does not hold. It follows then that the inequalities (^1) do not hold, and hence that uV^ + Is not a solution of (32). This completes the proof of the corollary. Corollary 11 . Let t^ + u^V^ be the fundamental solution of (1) where t^ = + 1, and D = Si'd^, 6^ 1,D. Let tgi^+i = 2 2 2^2k+l ^' y2k+l " ^^2k+l'^2k+l/^^ ' 2^2k+iy2k+l " ^2k+lÂ» where t^^^^ and ^2k+l given by (2), n = 2k+l, k = 0, 1, 2 , Â• Â• Â• . Then ^50) >^2k+l = ^lyfk+l " <^2k+iy2k+l' k = 0, 1, 2, and the equation (51) 'i2k+iy^ = 1
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22 of (6) which arises with solutions x Â« ^2k+l' ^ ~ ^2k+lÂ» takes the form (32), 6^x^ d^y^ = 1. Proof . Since the hypothesis of this corollary is that of Corollary 10, we may use (3^), (^0), and (2) to get ^X2k+l\AI^ y2k+lVd7)2 = C(xi\/67+ yiVd7)2k+l]2 (52) C(xi\/67 + yiVdI)2]2^^ (ti + UiVD)^^"^^ " ^2k+l + ^2k+lV^Â» and also Â— Â— 2 2 2
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23 2 2 the fact that 81X2^+1 ^lYzic+l " ^Â» follows that ^'l4^1 ^ S4^l^ 1 (1 ^ d^y2^^,) + d^y^^^^ . 1 X2k+1 = ^ ^ or (5^) X2k+i = V2k+1Thus from (g)^ and {3k), X2k+i = dgj^iygk+l = ^iy2k+l. establishing (50). It follows from tteit *2k+l ~ *1Â» and henoe by (6), when n is odd, equation (5I) takes the form (32). Finally, in order for an equation of (6) to arise which is different from the original equation (1), it is seen by Corollaries 711 that n in Corollary 1 must be odd. In fact Corollaries 611 establish the following summarizing Theorem la . It is seen by these corollaries that the different forms of equations (6) to arise are equation (2^), x2 Dy2 Â» 1, 2 where tj^j = ^^^zk ^Â» solutions given by 2x^^72^^ = Ujj^, and either equation (30)2, y2 Dx2 = 1, if and only if D contains no prime factor S 3 (mod 4), and 2 ^2k+l " 2y2k+l 1Â» ^^^^ solutions 2y2k+iX2k+i = ^2^+1, 21. one of the 2^2 possible equations (32),
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2h 2 2 5]_x d^y =1, where t2ij+i = ^d^^yilc+l ^Â» ^1*^1 " ^' ^1 ^ where ^212+1 largest square contained in (tg^j+i l)/2, and with solutions given by Sxgi^fiyak+l Â° ^21c4l* Moreover, k = l, 2, 3, ork = 0, 1, 2, according as the subscript 2k or 2k+l appears . By (g)i^ with 0=1, namely 6^**^ ^Â» ^^^^ dj^ and in particular d]_ contains no square factor, since D 2 is a product of distinct odd primes. Hence from d^ygic+i = 2 ^*2k+l " ^^/^ ^^^^ y2k+l largest square contained in (t2icfl l)/2. An example of Theorem la wherein equations ( 2^) and (30)2 of (6) arise is example 2 on page 6 in which D = 65 Â» 5 Â•13 contains no prime factor p = 3 (mod and X^^ = (ti l)/2 = (129 l)/2 = 6if = g2 = , ^^y^ Corollary 3 and relation (S)^. Hence y^ = 3 and d^^ = 1. By fi^ Â« 65. Therefore (30)2 of (6) arises with the fundamental solution x^ =1, yi = Also X2 = (t2 l)/2 = (33,281 l)/2 ? 2 2 Â» l6,6iJ0 = 5 Â•13. 25 6 = 65.16^ = Dy2 = d2y2 by Corollary 6 and relation (2)^. Hence by Theorem la, y2 = I6 and d2 = 65. By (g)i, Â§2 Therefore (2^) of (6) arises with the fundamental solution X2 = t^^ = 129, y2 = u^ *= I6, as in Corollary 6. Thus let the two equations of (6) which arise with solutions be denoted by (A) and (B) where
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25 (A) 65x^ = 1 equation (30)2 (B) 65y^ = 1 equation (2^^). The following Table 1 is selfexplanatory. TABLE 1 , DATA FOR EXAMPLE ILLUSTRATING THEOREM la n Un Â» 2xnyn 5 = ^yn 1 2 I 129 33,231 3,5^6,369 2,215,2i;9,921 16 if, 12s l,065,00g 27^.767,936 6k l6,6iK) ^,293,13^+ 1,107,624,960 = 65.162 = 20722 = 65.4l2g2 n yn Sn 2 2 1 2 I 1 129 257 33,221 g Â•16 2,072 k,12& 65 1 65 1 1 65 1 65 (A) (B) (A) (B) Notes : ^1, yi is not a solution of (B) as shown by Corollary 4, but by Corollary 7 is the fundamental solution of (A) . b) xli Â« y^. = Up as in (2g) and (29) of Corollary (6) and are obtainable by (2), (28) and (29) for n = 2k, k = 2, To Illustrate that part of Theorem la wherein equations (24) and (32) of (6) arise consider the equation 2 2 Â— t 55u =1 with ti + uiVD Â» g9 + I2V55. Since = (t;^ l)/2 = (Â«9 l)/2 ^ kk = k.ll x: 11.22 . ^^y2^ k = 0, l3y (3)^, yi = 2, di = 1. By (g)2^, 61 = 5. Therefore (32) of (6) arises with the fundamental solution x^^ = 3 , y^^ = 2.
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26 Also by Corollary 6 and relation (3)^, = (tj l)/2 = (15,3^1 l)/2 Â» 7,920 = 511122 = 55*122 = Dy = dgysHence by Theorem la, 73 = 12, d2 = D = 55. By (3)i., = 1. Therefore (2^) of (6) arises with the fundamental solution X2 = t]_ = 39, yg = U]_ = 12, as in Corollary 6. Thus let the two equations of (6) which arise with solutions be denoted by (A) and (B) where (A) 5x2 _ iiy2 = 1 equation (32), (B) x2 55y2 Â» 1 equation (2^^). The following Table 2 is selfexplanatory. TABLE 2 DATA FOR EXAMPLE ILLUSTRATING THEOREM la n tn = 2>.n + 1 Un = 2xnyii Xn = ' dnyn 1 2 I 5 39 15,3^+1 2,3iq,609 501,374,561 39,330,352,2^^9 12 2,136 330,196 67,672,752 12,0^^5,369,660 , 7,920 l,409,30if 250,937,230 i<i,665,Il26,12i+ = 1122 = 55122 = 113522 Â« 55.2136"^ 11637222 n yn Sn S^x d^y^ = 1 1 2 I 5 3 39 531 1^5, 341 9^,515 2 12 355 2,136 63,722 5 1 5 1 5 11 55 11 55 11 (A) (B) (A) (B) (A) Notes: a) X]_, y^^ is not a solution of (B) as shown by Corollary k, but by Corollary 9, is the fundamental solution of (A).
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27 b) x^. tg, yi = U2 as in (2^) and (29) of Corollary (6) are obtainable by (2), (22) and (29) for n = 2k, k = 2. c) By relations (36) and (37) of Corollary 10, k = 1, x3\/67 + y3\/d^ 2_^3\/5 + 2Vli)^ = (69 + 12V55)(3V5 + 2Vir) 531^5 + 35^\/ii. d) By relations (52) and (53), k = l^ix^\/Il + J^Vd^^ = (531 V5 + 35Â«VTr)^ Â•= 2,gi9,609 + 3Â«o,i96\/55 = t3 + U3\/D. This result for t3 + u^D is verified by using relations (3) and (i;). In the following chapters use will be made of the preceding results to help establish the conditions under which a particular form of (6) will be solvable when D is a product of two, three, and four distinct odd primes. In Chapter 2, D = ^1*^2 Pi P2 distinct odd primes. In Chapter 3 D = p2_'P2'?3 where P]_ < P2 P3 Q^g distinct odd primes, and in Chapter k, D = ?]_ Â•P2*P3 'Plj. where < P2 < P3 < PI+ are distinct odd primes. Also the word "excludent" will be used in the following discussions, and it will refer to a condition which la sufficient for a particular one of the 2^ possible forms of (6) not to arise. The symbol (mp) is that of Legendre in which p is an odd prime and m is an integer prime to p and is defined to be the value +1 or 1 according as m is a quadratic residue or nonquadratic residue of p.
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CHAPTER II CONDITIONS FOR ONE OP B^^X^ d^^y^ = 1 TO HAVE A SOLUTION WHERE ^x'^^l ~ ^Â» ^ PSODUCT OF TWO DISTINCT ODD PRIMES 2 2 Lemma 2 . Let D Â« Pi'Pg equation (1), t Du = 1, where < pg are distinct odd primes. Let (1) have the fundamental solution t^ + U]_V^ where t;j_ = 2>rj_+ 1. If one of the excludents listed to the right of the equations (55) 2 2 (57) of (6), S^^x d^^y = 1, holds, then that equation has no solution in integers x, y. Equation Excludents 2 2 2 (55) P1P2X y =1, Xi 7^ Yi, (56) pj_x Pgy = 1, (PilPg) = 1 or (Pgip^) Â«= 1, (57) PgX p^y = 1, (PglPi) = 1 or (p^lpg) = 1.Proof , Since D is a product of s = 2 distinct odd primes, p^ and pg, there are 2Â® = 2^ = 4positive divisors of D. These are 1, p^^, Pj, and p^^'Pj. ^'or t^^ = 2X^ + 1 and uj_, the fundamental solution of (1), the equation (6) of Theorem 1, where S^'d^^ = D = P]_Â»P2, has four possible forms, 2 2 namely, {2k}, x P2P2y ~ three equations (55) (57) above. By Theorem la equation (24) and one of (55) (57) always arise from (6) and have Integral solutions x, y, 2g
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29 We note the following about equations (55) (57) of (6), 2 Let Xj. yiÂ« Then equation (55) cannot arise. For 2 if it did, then by Corollary 7 and relation (30)^^, = y]^, 2 a contradiction to the assumption that X]_ y^^. Let (56) arise from (6) with 6^ = and d^ = Pg* Thus by Theorem 1 or Corollary 9, there exist integers x, y satisfying (56). It follows then from (56) tha.t 2 2 (55) pj_x s 1 (mod pg) and pgy S 1 (mod pj^). Now define the integers p' and p" by (59) p'Pi s 1 (mod P2) and p^pg = 1 (mod p;j_) . Then (5^) may be written as 2 2 p'P]_x = p' (mod P2) and p"P2y = p" (mod p^), which in view of (59) reduce to 2 2 ( 60 ) X s p ' ( mod P2 ) and y = p " ( mod pj_ ) . By (59)]_Â» we may write (p'P2) (p'P'PilP2) = (Pi1p2)Â» Hence, if (p^lpj) = (p'iP2) = 1, then {^^)'^ is not solvable, and therefore neither is (56). Similarly (56) is not solvable If (P2IP1) = 1. Thus the two excludents listed opposite (56) are established. By similar reasoning the two excludents
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30 listed opposite (57) are established, and iBnce equation (57) has no solution if either one holds. We will make use of the follovring Lemma 1 . The conditions (61)
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31 is complete. Theorem 2 . Let D = P]_Â»P2 where p^^ distinct odd primes. Let t^ + uj_ the fundamental solution of (l) , 2 where t]_ = ZX^, !Â• If ?^ Yi, a^id (65) ^PllPji = 1 Pj = ^ with ij = 1,2, 1 ?^ J, then the equation {56), 2 2 P]_x P2y =1, or the equation (57) 2 2 P2X P^y = 1, has a solution in Integers x, y according as 1 = 1 or 2, and where the solutions are obtainable by (66) y^ = (Xj^.u^A) and 2xTLyL Â° ^1* froof . By hypothesis and Theorem la, one of the equations (55) (57) of (6) has a solution in integers x, y. By Lemma 2 if X]^ / y^^, equation (55) of (6) has no solution. If 1 = 1 and J = 2 in (62) of Lemma 3, then Â• (67) (P1IP2) = + 1 ani PiSl(modl^). Hence by Lemma 3, (PglP^^) ~ ''" IÂ» ^y I^mma 2 the conditions necessary for (56) to have a solution are satisfied. Examining the two excludents of (57) Lemma 2 under
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32 conditions (67), we find that exoludent (57)i or (pglp^) (p^jpg) = + 1, and therefore it cannot hold. But excludent {57)2 or (pj^lpg) may be written as (llpg) (P1IP2) = (1P2)(+1) Â» (1IP2). Thus If P2 s 3 (mod k) , excludent (57)2 holds, and equation (57) of (^) has no solution. Hence, 2 if 71 (66) (P1IP2) = + 1 and pj_ Â— 3P2 s 1 (mod 4), equation (56) of (6) arises and has a solution. Also, if i = 1 and J = 2 in (63) of Lemma 3, then (69) (P1IP2) = + 1 and P2 s 3 (mod 1+) . Hence by (61) of Lemma 3, (P2IP1) 1Â» and by Lemma 2 the conditions necessary for (56) to have a solution are satisfied. Under these conditions (69), excludent (57)2 Â°^ ('P1IP2) = (1IP2) (P1IP2) (1)(+1) = 1. Thus (57) has no solutions and hence does not arise. Therefore if ^ y^ and (69) holds, equation (56) arises and has a solution. Conditions (69) imply that either (P1IP2) + 1 and p^ s 3p2 s 1 (mod ^) , or (PilP2) 1 and Pi ~ P2 = 3 (mod 'l) . The former are conditions (66). Thus both the former and the latter are implied in (65) for i = 1.
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33 By similar reasoning if i = 2 and J = 1 in the above proof, {65) is established whereby (57) of (6) has a solution, if ?^ Theorem 1 solutions of (56) or (57) are obtainable from (g)j^ and (2)2, n = 1 Â« G, thus establishing (66), Legendre proved that one of Mx^Ny^ = Â± 1 is solvable if M, N are primes of the form in+3. "One can always deduce these theorems and others like them from a consideration of the V/A in a continued fraction form where A = MN,""'" The following examples illustrate Theorem 2. Example 1 . TVhich of the following equations, 5X2 _ lly2 = I gj^d 11x2 5y2 = 1^ has a solution in integers x, y ? Let Pi = 5 and P2 = 11, primes of the form ^+n+l and ^n+3 respectively. Then the given equations have the forms of equations (56) and (57) Â• Hence the equation (1) is 55u^ Â« 1 whose fundamental solution is t^ + ^iV/^^ &3 12V/55* Since t^ = 29 = 2\^ + 1, = ^ij= 11*22 ^ a gq^a^e. Also since (p^lpj) = (5 1 11) = +1 and p^ 5 3 (mod k) , the conditions X, 7^ y^ and (65) of Theorem 2 are satisfied. Hence ^AdrienMarie Legendre, Theorie des Nombres (3rd ed,, Paris: Chez Firmin Didot Pr^res, 1J?30), I, pp. 647I.
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3^ by Theorem 2, equation (5^), has a solution In Integers x, y. By (66), y^^ = 2, = 3 Is a solution. Checking: 5(3)^ 11(2)2 ^ _ ^ Note: (pplPi) = (ll5) = +1, but that (p^^lpj) = (5ll) = 1, Hence one of the excludents of (57) Lemma 2 holds, and therefore (57) Is not solvable In integers. Example 2 . Does the equation 11x2 , i5y2 Â« have a solution in integers x, y? Let p^ = 11 and P2 = 19 Â» both primes of the form ^n+3. Then the given equation has the form of (56). The equation (1) is then t2 209u2 o 1 where t^^ + u^V^^ = ^^6551 Â• 3220 V209.'Since t^ = i+6551 = 2Xj_ + 1, Xj^ Â» 23275 = 19 '52. 72 ji a square. Also since (P1IP2) = (111 19) Â«= +1 and P2 = 19 = 3 (mod , the conditions (65) of Theorem 2 for 1=1 are satisfied. Hence by Theorem 2, equation (56), 11x2 19y2 = 1 has a solution in integers. By (66) = 33, ^1 ^ ^1 ^iV^ was calculated by the writer.
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35 solution. Checkingj 11(^6)^ 19(35)^ = 23276 23275 = 1. Note: (pglPi) = (I9I11) = If and hence by Lemma 2, (57) has no solution and hence does not arise. Example "5 . Does the equation 19x2 7iy2 1 have a solution in integers x, y? Let pj^ = 19 and P2 = 71 1 "both primes of the form ^n+3, then the given equation has the form of (56). The equation (l) is then t^ 13^3vi^ = 1, since D = 13^9 = 19*71 is a product of two distinct odd primes, and t^ + ul\/D = 203 91^06 6595I + 5 552OO 3^^^0 V!ij?9 , where 2\ + 1. Hence \ = 1^1 95903 32975 = 71 '(119335)^ ^ a square. Also since (P2IP2) ~ (19 1 71) = "*"1Â» and pg = 71 = 3 (mod 4), the conditions (65) of Theorem 2 for i = 1 are satisfied. Hence by Theorem 2, equation (5^), 19x2 71y2 = 1^ has a solution in integers x, y. By (66), y^ = 119^35, x^^ = 231652 is a solution. Checkingt 19(231652)2 71(119835)2 Â» 101 95903 32976 101 95903 32975 = 1. Arthur Cayley, Collected Mathematical Paoers (13 vols. Cambridge: The University Press, 1^97), XIII, pp. 430 467.
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36 Note; (pgiPi) = (71I19) = 1; hence (57) has no solution by Lerama 2 and therefore does not arise. Exanrole 4 . Does the equation 193x^ 7y2 = 1 have a solution in integers x, y? Let P]_ = 7 and P2 = 193 Â• primes of the form 4n+3 and respectively, then the given equation has the form of (57). The equation (1) is then t^ 1351u^ = 1, with + ulVD = 6175 + l6gVl35r, where t^^ 2\^ + 1.^ Kenoe = 30^7 = 7*21^ a square. Also since (pglp^) = {193 1 7) s +1 and pj^ = 7 = 3 (mod k) , the conditions of Theorem 2 for i = 2 are satisfied. Hence by Theorem 2, the given equation (57) 1 193x2 7y2 = 1, has a solution in integers x, y. By (66), y^ = 21, = ^ is a solution. Checking: 193(^)^ 7(21)^ = 30Sg 30^57 = 1. Note: (pilpg) = (7I193) = +1, but (P2IP1) = (193i7) = 1; hence by Lemma 2, (56) has no solution. Example 5 . Find, if any, a solution of the equation 7x2 . 3y2 = 1, Ibid.
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37 Let P]_ = 3 and * 7Â» "both primes of the form ^n+3. Then the given equation has the form of (57). The equation (1) is then t2 21u2 = 1, with t]_ + Uj^ = 55 + 12\/2Tf where t^^ = SX^ + 1. Hence = 27 = 3*3^ ?^ a square. Also since (pglp^^) = (7l3) = +1 and = 3 =3{mod k) , the conditions of Theorem 2 are satisfied for i = 2. Thus the given equation (57) i^as a solution in integers, x, y. By (66), ~ ^Â» = 2 is a solution. Checking: 7(2)^ 3(3)^ = 2g 27 = 1. Note: (pip2) " (3 1 7) " 1; hence by Lemma 2, (56) has no solution. Theorem 3 . If in Lemma 2, p^ = p^ s 1 (mod k) and = y^, then equation (55) 1 2 2 P1P2X y = 1, of (6) has a solution in Integers x, y where (70) y^ = V^]L ^^1^1 ^ hiproof. By Lemma 2 if (PglPi) = (PilPg) ~ equation (56) and (57) of (6) have no solutions in integers X, y. Now, (plIp2) = (P2IP1) ^Â® written as i(Pii)i(P2i) (lp2)(PilP2) = (PllP2)(1)
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3g or = (piIp2)(i) i(Pl.l)i(p2+l) hence, (1IP2) = (1) 
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39 Note: (PglPi) = (17l5) = 1, and (5)1^2) = (5)(17) = 1; henoe, by Lemma 2 equations (56) and (57) have no solutions. By Lemma 2 If (71) (P1IP2) = (P2IP1) = (PplPi) = (P1IP2) " "^l* the conditions necessary for both (56) and (57) of (6) to have a solution are satisfied. As in Theorem 3 relation (71) holds if and only if p^^ = Pj. = 1 (niod K) , This last set of conditions is not covered In Theorem 2. Moreover, Theorem 3 asserts that if = y^ and p^^ s P2 s 1 (mod k^) equation (55) i 2 2 P]_P2X y = 1, of (6) has a solution. Hence, if (71) holds 2 and ft yj_, equation (55) has no solution by Lemma 2. It follows by Theorems 1 and la that one of (56) and (57) of (6) has a solution. Thus in (^)j, d^ = p^ or p^, and y^ is the largest square in X]_ = (tj^ l)/2. This discussion proves the following Theorem If in Lemma 2, ^1 / P^ = Pg Â« ^ {mod k) , then either equation (56), or equation (57) Â» 2 2 , Pix P2y = 1, 2 2 P2X P^y = 1, has a solution in integers x, y according as d^ = p^ or p^ in relation (S)j, = *^"^l' ^^'^^^ y^ the largest square in (t^ l)/2 = Xj^.
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Legendre proved from an analysis of the \/A in a continued fraction form where A = MN that if M and N are two 2 2 primes of the form one of the two equations, x MNy = 1 or Mx^ Ny2 = Â± 1, is possible."^ The following example illustrates Theorem Example . Show that the first one of 29x2 53y2 = 1, 53x2 . 29y2 = 1, has a solution. Let piL = 29 and P2 = 53, ^oth primes of the form l^n+1. Then the given equations have the forms of (56) and (57). Hence the equation (1) is t^ 1537u^ =1, since D = 153? = 29 Â•53 is a product of two distinct odd primes and t^ + u^\/D = 75,0?SO,329,577 + 1,915,090,212 Vl537, ^1 +1.^ Hence X^ = 53*(266ll^)2, and a square. Thus by Lemma 2 equation (55) has no solution. Since pj_ = Pg 3 1 (mod k) , and since relation (71) holds, which is easy to show, then by Lemma 2 the necessary conditions for (56) and (57) to be solvable are satisfied. Now by Theorem k and the relation (g)^ we see \ = dj^y^ = 53. (266ll4)2, so that d^ = 53 and y^ = 2661^4, Hence by Theorem k or Theorem 1, (6) takes the form (56), Legendre, op. clt . '^Cayley, op. clt.
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^1 29x2 53y2= 1, with the solution yj_ = 266l4, x^^ = 35979. Checklngi 29(35979)^ 53(2661^+)^ = 37,5^0,16^,7^9 37,5^.16^,733
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CHAPTER III CONDITIONS FOR ONE OF 5]_x^ d^^y^ = 1 TO HAVE A SOLUTION WHERE 61 'dj^ = D, A PRODUCT OP THREE DISTINCT ODD PRIMES Lemma ^ . Let D = pj_P2P3 equation (1), t Du = 1, where pj^ < < P^ are distinct odd primes. Let (1) have the fundamental solution tj_ + u^V^ where tj^ = 2XL + 1. If one of the exoludents listed in Table 3 to the right of the equations (72) (78) of (6), 6,^x^ d^y^ = 1, holds, then that equation has no solution in Integers x, y. Proof . Since D is a product of s = 3 distinct odd 8 3 primes, j>^, P2, and p^, there are 2 2'^ = S positive divisors of D. These are 1, p^^, P2, V^t ^1^2' ^2^3' P^PgP^. ^1 " + 1 and u^, the fundamental solution of (1), the equation (6) of Theorem 1, where 6iÂ«dx = D = 'pjJ>2P^Â» / 1 V 2 2 has eight possible forms, namely, equation (24), x PiPgP^y * 1, and the seven equations (72) (7^)Â» By Theorem la, equation (2^) and one of (72) (7^) always arise from (6) and have integral solutions x, y. We note the following about (72) (76) of (6). p Let y^. Then equation (72) cannot arise. For If it did, then by Corollary 7 and relation (30)^^, = y^, 12
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^3 TABLE 3 EXCLUDENTS OF EQUATIONS {J 2) (yg) Equations Exoludents^ (72) 2 2 P1P2P3X y Â« 1 (73) 2 2 Pix P2P3y = 1 (P1IP2) (P1IP3) (P2P3IP1) (7^) 2 2 P2P3X p^y = 1 (P1IP2) (P1IP3) (P2P3IP1) (75) 2 2 Pgx PiP3y Â» 1 (PglPi) (P2IP3) (PxP3lP2^ (76) 2 2 P1P3X P2y = 1 (P2IP1) ( P2 1 P3 ) (P1P3IP2) (77) PiP2y = 1 (P3IP1) (P3IP2) (PlP2iP3) (78) 2 2 PLP2X P3y = 1 (P3IP1) (P3IP2) (P1P2IP3) ^All exoludents Â» 1, a contradiction to the assumption that ^1 ^ y^* Let (73) arise from (6) with 5^ = p^ and d^ = P2P By Theorem 1 or Corollary 9 there exist integers x, y satisfying (73). It follovrs from (73) that 2 2 (79) P]_x = 1 (mod p^) , Pq^x = 1 (mod p^), and 2 P2P3y s 1 (f"o^ Pi) Â• Define the integers p', p", and p'" by (go) p'p^^ = 1 (mod pg), p"P]^ = 1 (mod p^), and P'" P2P3 = 1 (mod p^) .
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Then (79) may be written as P'Pt^x^ s p' (mod Pg) , p"p^x^ 5 p" (mod p^), and P**P2P3y = p"' (Â™od Pq_), which in view of (gO) reduces to (gl) m p' (mod Pg), s p" (mod p^), and y m p"* (mod pq^) . By (gO)^ we may then write (p'lPg) (p' Â•P'Pj_IP2^ Â° (Pq_1p2)Â» Hence if (p^lpg) = (p'IPg) = "1. ^^^^ ^79)i is not solvable and, therefore, neither is (73). Similarly (73) is not solvable if either (p^p^) = 1, or (PjP^lp^) = 1. Thus the three exoludents listed opposite (73) in Table 3 are established. By similar reasoning the three exoludents listed opposite the equations (7^) (7^) in Table 3 are established. We will make use of the following lemmata. Lemma 5 . The conditions (g2) ^PilPj) = ^Pi'Pk^ = ^PjPklPi^ ' "^^Â» where p^^, Pj , and Pj^ are distinct odd primes and i, j, k = 1, 2, 3, follow if and only if either (33) (PilPj) = (PilPk) = Pl s 1 (mod k),
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^5 or Proof . If (32) holds, consider (P^P^lPi^ = which may be written as Using (32) and combining, relation (35) becomes ^(p,l)^p.+Pk) (+1)(+1)(1) ^ ^ = +1, which holds if and only if either = 1 (mod k) or p^ s 3p (mod k) , Hence (33) and (3^) are established. Conversely, if (33) holds, we may write (P^P^jiPi) = (llPl)(PjlPl)(PklPi) = (+l)(PilPj)(PiiPk) = (+1)(+1)(+1) = +1. Hence (32). If (34) holds, then p^p^^ = 3 (mod 4), and we may write (P^Pj^lPi) = ^Pi'PjPk^ ^ ^^i ' ^ ^^i ' ^k^ ^ This completes the proof. Lemma 6 . The conditions (36) (p^Ipj) = (PilPij) = ^PjPklPi) = where p^, Pj , and p^^ are distinct odd primes and i, J, k = 1, 2, 3, follow if and only if either (37) (PilPj) = (Pilp^) = +1 and p^^ s 3 (mod ^) ,
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^6 or (gg) (Pilpj) = (PilPk) = +1 =! Pic (mod 4), Proof . If (go) holds, then the relation (PjPjclPi) = +1, may be written as (39) ^Pjl^l^^^k'^l^ ' Also since (Pj^lp^) = +1, It may be written as (pjPi)(l) ^ ^ or, multiplying both sides by (Pjlpj^), ,90, .^_^^i(p,.l).i(Pjl)^,^^l^^^_ Similarly, since (p p ) = +1, 1 k (91, ,.Â„*'^^^^'*'^^^' = (p,,p,,. Substituting (90) and (91) in (29), and combining, we get ^P^+1) *i(Pi+Pi,2) (PjlPi)(pJPl) = (1) ^ ^ =^1 which holds if and only if either p^ 3 3 (mod k) or p^ = p (mod k) , Hence (SJ) and (gg) are established. Conversely, if (g/) holds, then (Pjlp^) = (Pj^lp^) +1. Hence (PjPj^lPi) = (Pj I Pi) (p^l Pi) = (+1)(+1) = +1. If (gg) holds, then since Pj Â« Pj^ (mod k) , (PjP^lPi^ "
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^7 (p^pj)(Pj^p^) = +1, This completes the proof of the lemma. We seek the necessary and sufficient conditions for each of the equations (72) (7^) to have a solution. These will now be determined in Theorems 5 and 6 which follow. Theorem 5 . Let D = P1P2P3 where P]_ < P2 < P3 are distinct odd primes. Let t^ + U]_\/D be the fundamental solution of (1), t^ Du^ Â« 1, where t^ = + 1. If ^ y\, (92) ^Pl'Pj^ ^Pi'Pk^ " ^Pj'Pk^ and either (93) Pi a 1 (mod ^) and Pj Â» 3?^^ (mod i) , or (9^) Pi Â» 3 (mod \) and pj = 3pj^ (mod i^) , then either equation (73) Â» (75), or (77) of (6) has a solution in integers x, y according as 1 = 1, 2, or 3 with J, k = 1, 2, 3Â» and J < k, and where the solutions are given by (95) yi = (X^.u^A) and 2x^y3_ = yx^. Proof . By hypothesis and Theorem la, one of the equations (72) (7g) of (6) has a solution in Integers x, y. Since X^ ^ y^, then by Lemma 4, equation (72) of (6) has no solution. Let 1=1, J=2, k=3in conditions (^3) of Lemma
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5 , then (96) (P1IP2) = (P1IP3) = +1 and p^ = 1 (mod U) , Hence by Lemma 5Â» (P2P3iPi) ~ "*"1> and thus by Lemma kthe conditions necessary for (73) to have a solution are satis2 fied, Nov'j all excludents in Table 3i except ^ j^t can be expressed in terms of one or more of the Legendre symbols, (p^lp^), (p^lpj), and (p^lp^). Hence when conditions (96) hold the values of (pglpj) and the least positive residues modulo Hof P2 and p^ can be arranged in 2^ = g different arrajigeraents or oases, as shovm in the three columns headed "Conditions" in Table Prom taking each of the eight cases in turn we determine those excludents of Table 3 which are consistent with conditions (96). The results are recorded in Table 4in the column headed "Excludents of Table 3 which hold." Hence when an excludent of Table 3 holds the associated equation has no solution by Lemma k. This fact is recorded in Table in the last column which has the heading "Equations excluded," For example when the conditions (96) and those of case (1), P2 s pj = 1 (mod k) and (p2lp5) = +1, hold, v^e find that none of the excludents in Table 3 with respect to the equations (74) (7g) can hold, that is each one = +1. Hence when the conditions (96) and those of case
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^9 m tUH >> Ik H << 5z; M o a H r Â— HI 1 H Pi J* H CO + o H M EH 1 Pi a? w Pi n Pi o H Eh Pi CO o H CQ CQC] O Â© Â•H p s H o o H H ^ OS Eh O W P fl Â•d :3 H O X w Pi Pi ni ^ Pi Â•d o H B Pi to 8 9> o to CQ CQ O g CM OJ tVJ OJ OJ OJ 1^ Co CM Co Â•k OJ CU ik OJ Co OJ Co ^ Â— *. LP, Â•b CM CM Â«k cu CVI OJ H H H H + H I H I (Q CQ OQ P +3 o o o S Â« a> 0) a> 'd "d "d K> r<> OJ CM in VO w
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50 (1) hold, none of the equations (7^) (7^0 is excluded. This fact is recorded in Table kin the column headed "Excludents of Table 3 v;hich hold," Also, vrhen conditions (96) and those of case (2), P2 = P3 5 1 and (p2iP3) = r) hold, excludent (7^)i cannot hold since ( 1 1 Pg) (pj^ I = (+1)(+1) = +1; excludent (7^)3 osJ^^ot hold since (lp^)(p^p^) = (+1)(+1) = +1; excludent (7^)^ cannot hold since (pglp^) (P3IP1) = (p^ipg) (PiiP3) = (+1)(+1) = +1. Hence equation (7^) is not excluded. Also: excludent (75)j_ cannot hold; excludent (75)2 ^oÂ®^ hold since (p2iP3) = 1; excludent (75)3 holds since (li^)(p;j_'P2^ ^^3 '^2^ (+1) (+1) (P2 i P3) = (+1){+1)(1) = 1. Hence by Lemma kequation (75) has no solution. Proceeding in this majmer with the remaining excludents, it is found that excludents (76)0 , (77)o 7, and (72)2 3 hold, and hence equations (Jo) (7g) have no solution We record in Table k in the column headed "Excludents of Table 3 v/hich hold," those excludents vrhichare consistent with conditions (96) and those of case (2), Continuin/j in this manner with the remaining cases (3) (g), results are found as indicated in Table 4. We see that when the conditions (96) and those of either case or case (6) hold, namely, (97) P^ = P2 s 3P3 = 1 (mod k) and (p^lpg) = (P1IP3) = (P2IP3) = +1,
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51 or (9^) Pi S 3P2 = P3 Â» 1 (mod and (pj_p2) = (p^lpj) = (P2IP3) ' +1, equations (7'+) (7^) have no solutions. By Theorem la, Since ^ yj^, one of (73) (7^) has a solution. Hence when conditions (97) or (9^) hold and ^^ ^ equation (73) of (6) has a solution. Conditions (97) and (9^) are restated In (92) and (93) for 1=1, j=2,k=3. By symmetry interchanging the subscripts 1 and 2 in (97) and (9^) conditions are obtained "by which (7?) of (6) has a solution. This establishes (92) and (93) for 1 = 2, J = 1 , k = 3 . Also by symmetry interchanging the subscripts 2 and 3 in the conditions obtained for (75) to have a solution produces those by which (77) of (6) has a solution. This establishes (9?) and (93) for 1 = 3, J = 1, k = 2. Thus (92) and (93) are established completely. The results (92) and (,94) will now be derived. Let 1=1, J = 2, k = 3 in (g4) of Lem^Tia 5, then (99) (Pi1p2^ Â° (P1IP3) = +1 and pg s 3P3(mod 14). Hence by Lemma 5, (PjPjlPj^= +1, and thus by Lemma 4 the conditions necessary for (73) to have a solution are satls2 fled. Now all exoludents of Table 3i except y^, can be expressed in terms of one or more of the Legendre symbols.
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52 (PllP2^Â» ^Pi1p3)Â» snd (P2IP3). Hence when conditions (99) hold the values of (P2IP3) sin^ the least positive residues modulo ^ of Pq^, Pj, and p^ can be arranged in eight different arrangements or cases, since in view of (99)31 ^2 ^3 cannot be in the same class modulo 4, These eight cases are listed in the four columns of Table 5 under the heading "Conditions." From taking each of the eight cases in turn we determine those excludents of Table 3 which are consistent with conditions (99). The procedure is the same as that used in developing Table k. The results are recorded in the last two columns of Table S. Hence it is seen from Table 5 that equations (7^) (7^) have no solutions by the conditions (99) and those of cases (2), (4), (6), and (g) . But we observe that the conditions (99) and those in cases (2) and (6) of Table 5 are the same as conditions (96) and those of cases {^) and (6) of Table ^. Thus, from the new conditions, of cases (^) and (g) of Table 5, equations (7^) (7g) have no solutions if either (100) 3p;j_ = P2 = 3P3 = 1 i^od h) and (p^^lpg) = (P1IP3) = (P2IP3) = +1, or (101) 3p^ = 3p2 = S 1 (mod U) and (p^^jpj) = (P1IP3) =
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53 H *< o s TO Ph 1 III CVJ ft * H o + M EH II i W ft w ft II CVJ o ft Ei ft mei Pi'd O vor^i60 r^i^ Â•d "d tO U 03 ?3 o o o Â•H ^ 4^ o o o ^ Pi O (D S "d "d "d
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5^ By Theorem la, since Xj^ ?^ yj_, one of (73) (7^) has a solution. Hence when conditions (100) or (101) hold, and 2 ^1 ^ yiÂ» e
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55 (P1IP3) = <13l^3) = +1, and (pglp^) = (291^+3) = 1, and In addition pj_ = I3 s 1 and P2 = 29 e 3*^3 = 3?^ in class modulo h, the conditions (92) and (93) of Theorem 5 are satisfied for 1=1, j=2, k=3. Hence by Theorem 5 the given equation has a solution in integers x, y. By (95) y^ = 1,039, = 10,176 is a solution. Checking: 13(10,176)2 1,2^^7(1,039)2 = 1,3^6,162, 6gS 1,346,162,6^7 = 1. Moreover by Corollary 9 the solution x^^, y^ is the fundamental solution; for by (3^), t_ = 5j_xj_ + '^x^l ^1 = 2x3_yL. Checking: t;j_ = 13(10, 176)^ + 1,2^17(1,039)^ = 1,3^6,162,688 + 1,3^+6,162,667 = 2,692,325,375; ui = 2(10, 176)(1, 039) = 2,145,728. Remark; Since by Corollary 9, on page 16, the solution x.j_, y]_ is the fundamental solution of the given equation, we can obtain all positive solutions by relation (36) of Corollary 10, on page 17. Example 2 . Shovr tha.t the fimdamental solution of 3x2 _ ii^3y2 ^ Is given by x^^ = 50^, y]_ = 73. Since 1^3 = 11 '13, the numbers 3, 11, and 13 are three distinct odd primes where if p^ = 3, p^ = 11, and p^ = 13, the given equation has the form (73). Equation (l) is then t2 _ l29u2 = 1,
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56 since D = 3 '11 '13 = ^29 and its fundamental solution Is tj^ = 2^l + 1 = 1,52^,095, = 73,5^^. Hence = 1^2,0^7 = 11*13.(73)^ ^ a square. Also, since (P2^p2) = (3 111) = +1, (P1IP3) = (3I13) = +1, (P2IP3) = (III13) = 1, and in addition 3 = Pi = 3 (mod ^) and 11 = Pg = 3P3 = 3*13 (mod k) , the conditions (92) and (9^) of Theorem 5 are satisfied for j, = 1, J = 2, k = 3. Hence by Theorem 5 the given equation has a solution in integers x, y. By (95), y^ = 73, = 50^ is a solution. Cheokingt 3(50^)^ 1^+3(73)^ = 762,Olg . 762,0^7 = 1. By Corollary 9, x^, y]_ is the fundamental solution. Example 3 . Show that 13x2 51y2 = 1, has a solution In integers x, y and that its fundamental solution is given by x^ = 2, y^ = '1. Since 51 = 3*17, the numbers 3, 13, and I7 are three distinct odd primes where if p^ = 3, P2 = 13 , and pj = 17, the given equation has the form of (75). Equation (1) is then t2 663u2 = 1, since D = 3 '13 '17 = 663 and its fundamental solution is t^ = 2X]_ + 1 = 103, ui = 4. Hence X], = 51 7^ a square. Also (P2IP1) = (13l3) = +1, (P2IP3) = (13I17) = +1, (PiiP3) = (317) = 1, and pg = 13 a 1 (mod 4), p^ = 3 s 3P3 = 3*17
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57 (mod H) . Hence conditions (92) and (93) a^^e satisfied for 1=2, J=l, k=3. Hence by Theorem 5 the given equation has a solution in integers x, y. 3y (95), y^ = 1, xj^ = 2 is a solution. Checking: 13(2)^ 51(1)^ = 52 51 = 1. By Corollary 9i X]_, y^ is the fundamental solution. Example k . Show that 7x2 . ^7y2 = 1^ has a solution in integers x, y and that its fimdamentel solution is given by X2_ = 208, y^ = 59. Since S7 = 3 '29, the numbers 3, 7, and 29 are three distinct odd primes where If P]_ *= 3, P2 = 7, and p^ = 29, the given equation has the form (75). Equation (1) is then t2 609u2 = 1, since D = 3*7'29 = 609 ajid its fundamental solution is t2_ = + 1 = 605,695, uL = 2k,3kk.^ Hence = 329(59)2 ^ a square. Also (P2IP1) = (7l3) = +1, (P2IP3) = (729) = +1, (P1IP3) = (3I29) = 1, and 7 = P2 = 3 (mod k) , 3 = = 3.p^ _ 3 '29 (mod k) . Hence the conditions (92) and (94) are satisfied for 1=2, jÂ«l, k=3. Thus by Theorem 5, the given equation has a solution in integers x, y. By (95), y^ = 59, x^ = 20?!, is a solution. Checking: 7(208)2 _ ^^^^^^2 302, gUg 302,8^7 =1. By Corollary 9, x^, y^ Is the fundamental solution. Example 5 . Show that Legendre, op. cit .. Table X.
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3& 29x2 35y^ = 1, has a solution In Integers x, y and that Its fundamental solution Is given by = 7^, yi = 71. Since 35 = 5*7, the numbers 5, 7, and 29 are three distinct odd primes where if = 5i P2 = 7, and p^ = 29, the given equation has the form (77). Equation (1) is then t^ l,015u2 = 1, since D = 5*7*29 = 1,015 and its fundamental solution is tj^ Â» 2XL + 1 = 352,371, = 11,076.'Hence \^ = 5*7*(71)2 a square. Also (P3IP1) = (295) = +1, (P3IP2) = (29l7) = +1, (PllPg) = (5l7) = 1, and P3 = 29 s 1 (mod 4) , p^ = 5 = 3*7 = 3P2 ^^o"^ ^) Â• Hence the conditions (92) and (93) are satisfied fori=3, J=l, k=2. Thus by Theorem 5 the given equation has a solution in integers x, y. By (95), y_ = 71, x^ = 72 is a solution. Checking: 29(7^)^ 3^(71)2 = 176,1^36 176,^35 = 1. By Corollary 9, x^, y^ Is the fundamental solution. Example 6 . Show that 31x2 _ i5y2 ^ 1^ has a solution in integers x, y and that its fundamental solution is given by X3_ = I6, y^ = 23. Cay ley, op. cit .. p. i^Â•30.
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59 Since 15 = 3*5Â» the numbers 3Â» 5Â» and 3I are three distinct odd primes where if = 3, = 5, and p^ = 31, the given equation has the form (77). Equation (1) is then t2 i;65u2 = 1, since D Â« 3 '5 '31 = ^65 and its fundamental solution is t^^ = 2Xl + 1 = 15,^71, ui = 736.^ Hence = 35(23)^ a square. Also (p^lpi) = (3ll3) = +1, (P3IP2) = (3l5) = +1, (P1IP2) = (3l5) = 1, and p^ = 31 = 3 (mod k) , p^ = 3 5 3*5 = 3l^(mod k) , Hence the conditions (92) and (9^^) are satisfied for i = 3Â» J = lÂ» lc = 2. Thus by Theorem 5 the given equation has a solution in integers x, y. By (95), 71 = 23, X]_ = I6 is a solution. Checking: 31(16)^ 15(23)^ = 7,936 7,935 Â» 1. By Corollary 9, xj_, yj_ is the fundamental solution. Theorem 6 . Let D = p^pgpj where p^ < p^ < p^ are distinct odd primes. Let t^ + uj_\/^ tie the fundamental solution of (l), 2 2 p t Du =1, where t^ = 2X^ + 1. If \^ ^ yj, and either (102) (P^IPj) = ^PilPk^ Â° "^Pj'Pk^ Â» +1 and p^ s 3 (mod , or (103) (Pjpj) = (pJPk) = +(PjlPk) = +1 and p^ s p = Pj^ s 3 (mod i^). Legendre, op. cit .^ Table X.
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6o or (10^) (P^Ipj) = (PiIPij) = +1 and p^ s 3Pj = ^ 1 (mod k) , then either (7^*), (76), or (7^) of (6) has a solution In Integers x, y according as 1 = 1, 2, or 3, where J, k = 1, 2, 3 with J < k, and where solutions are given by (105) 71 " (X^.u^A) and 2x^71 = hiproof. By hypothesis and Theorem la, one of the equations (72) (7^) of (6) has a solution In Integers x, y. Since y]_, then by Lemma equation (72) of (6) has no solution. Let 1=1, J = 2, k = 3 1n conditions (557) of Lemma 6 , then (106) (PilPg) = (P1IP3) Â» +1 and P3_ s 3 (mod i^Â•) . Hence by Lemma 6, (pgPjIPi) = +1; thus, by Lemma ^ the conditions necessary for (7^) to have a solution are satisfied. Now all excludents of Table 3, except 7* yj^, can be expressed In terms of one or more of the three Legendre symbols: (107) (p^lPg)! (P1IP3). and (Pglp^). Hence when conditions (I06) hold, the values of the three symbols In (I07) and the least positive residues modulo \ of Pp and p, can be arranged In eight different arrangements or
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61 oases as shovm in the six columns headed "Conditions" in Table 6, v:here the third, fourth, and fifth columns are easily derived from conditions (106) and those of the first two columns in the table. Prom taking each of the eight cases in turn we determine those excludents of Table 3 which are consistent v:ith each case. The procedure is the same as that used in developing Table The results are recorded in the last two columns of Table 6, Hence from Table 6, equations (73) and (75) (7^) have no solutions by the conditions of cases (2), {k) , (6) (g). By Theorem la, since ^1 ^ ^1Â» Â°^ ^''^^ ~ ^^^^ Â°^ ^ solution. Hence equation (7^) has a solution when either (log) 3Pl s s Pj a 1 (mod k) and +(PllP2) = +(PilP3) = (P2IP3) = +1, or (109) 3Pi S P2 s 3P3 S 1 (mod U) , and +(PllP2) = (P1IP3) = (P2IP3) = or (110) 3pj_ s 3p2 = P3 = 1 (mod k) , and (p^lPg) = +(PilP3) = (P2IP3) = +1, or (111) PL = Pg s pj s 3 (mod k) , and (p^lpg) = (P1IP3) = +(P2lP3) = +1,
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62 O ISO 6 1 III H ft Q Â— ' N H H Â«% to to Â«^ CVJ CVJ "^ CVI hOJ ' Â— ^ Â— Â• r Â— ' Â— ' r Â— Â— OJ CVI OJ H H H H H CVI H ^ Â— . m 0] 150 H H ^ s y Â— H hISO CO H CM H OJ CVJ y Â— Â» H CM LPv in H H Â«Â« H H OJ OJ H H cU 1^ + I + I + I + Â• r\r^r^r^T^^T\r^r^ + + I I + + I I + + + + I Â• I I H H H 1^ 1^ H H H Co CVJ ^ Â«o
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63 or (112) (p^lPg) " (P1IP3) = (P2IP3) = In view of (I06) , conditions (10^) (110) and (112) establish (102), while condition (111) establishes (IO3) for 1=1, J = 2, k = 3. By symmetry interchanging the subscripts 1 and 2 in (102) and (IO3), we obtain the conditions by which equation (76) has a solution. This establishes (102) and (103) for 1=2, J=l, k=3Â» Also by symmetry interchanging the subscripts 2 and 3 in the conditions obtained for {76) to have a solution, we get those by which equation (7^) has a solution. This establishes (102) and (I03) for 1=3, J = 1, k = 2, completing the proof of (102) and (I03). The result (104) will now be derived. Let i = 1, J = 2, k = 3 conditions (38) of Lemma 6, then (113) (P^Ipj) = (P1IP3) = +1 and P3 (n^o^ k) . Hence by Lemma 6, (pgP^lp^) = +1; and thus by Lemma k the necessary conditions for (74) of (6) to have a solution are satisfied. Now all excludents in Table 3, except y^, can be expressed in terms of one or more of the Legendre symbols , (114) (Pj^IPj)* (P1IP3), and (p^lp,).
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Hence when conditions (112) hold the values of the three symbols in (11^) and the least positive residues modulo k of Pi, P2, P3 osn be arranged in eight different arrajigements or cases as shown in the six columns headed "Conditions" in Table 7, where the fourth, fifth and sixth columns are easily derived from (113) and the conditions of the first three columns. From taking each of the eight oases in turn we determine those exoludents of Table 3 which are consistent with each of the eight cases. The procedure is the same as that used in developing Table k. The results are recorded in the last tvjo columns of Table 7. Hence from Table 7, equations (73) and (75) (78) have no solutions by the conditions of cases (3), (^) , (6), (7), and (8). But the conditions of cases (6) and (S) are implied in (102) for lt=l^j=:2, k=3, and similarly the conditions of case (7) are implied in (103). Hence from the new conditions of cases (3) and (4), equations (73) and (75) (7?!) have no solutions If either (115) Pi = 3P2 S 3P3 = 1 (mod k) and (p^lp2) = (P1IP3) = +(P2lP3) = +1. or (116) = 3p2 = 3P3 a 1 (nio
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65 H o o Â•H i Q) H ^ 03 O . CO a> H O >^ pa 11 ft I ft + ft I ft. < O < < < CV] H H H _J I 1 oO C5 cO rH 1^ ""^ CU OJ pÂ— ** "* CM H H to CM CVJ t ^ ai Co H H H CxJ f Â— I C3 H H H VD Â«Â« ^<^ OJ m OJ H H Â— * vo in OJ in rv 1 t VI m H H CJ H H cd C5 hH H H H H H H 1 + J + 1 + 1 H H H H H H H + 1 1 + + 1 1 OJ tl o I ft ni III OJ III H ft CI 0} o + + I I + + I I H H K> H H H 1^ CO CQ to C o o o 4J 4^ P c} Cd (d 7i :i Â•:i a* Â©(DO) CQ CO CQ 0) 0) 0) P 4^5 4J O O O Â«
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66 By Theoreni la, since \i ^ y^, one of (73) (7^) has a solution. Hence (7^) has a solution when either (115) or (116) holds. Conditions (115) and (ll6), in view of (113), establish (10^) for i = 1, J = 2, k = 3. By the same argument as above using the property of symmetry with regard to the subscripts, {10^) is established completely. By Theorem 1 solutions of (7^), (76), or (7^) are obtainable from (g)i and (g)p, n = 1 = G, thus establishing (lOg). This completes the proof of the theorem. Conditions (102), (I03), and (10^4) present seven different sets of conditions by which each of the equations (7^), (76), and (78) has a solution. The Tables S, 9, and 10 indicate these for (7^), (76), and (7g) respectively. Seven examr;les follow each table. Since the illustration of Theorem 6 by examples would be similar to that used viith regard to Theorem 5, only pertinent data are given in outline form, wherein tj^ + ^iS/^ was calculated by the writer unless otherwise specified. Example 1 . [case (1) of Table g] Pi = 3, P2 = 13, = 73 (PllPg) = (3)13) = +1, (P1IP3) = (3I73) = +1. (P2IP3) = (13I73) = 1. D = 31373 = 2,g47, t;i^ = + 1 = 372,007 and U;L 6, 972, whence \ = 3*2^19^ a square. By (IO5), = 2^3, xi = 1I+ is a solution of 13'73x^ 3y^ = 1. Checking: 9^9(1^)^ 3(2^9)^ = 126,00^+ 1^6,003 = 1.
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67 TABLE g 2 2 CONDITIONS FOR EQUATION (7^1), P2P3X Piy TO HAVE A SOLUTION WHEN ^ = 1, Case Conditions Pl= PgS (mod 4) Conditions of Theorem 6 where 1. J 2, k = 3 (1) (2) (5) (6) (7) 3 3 3 3 3 1 1 1 3 1 3 3 3 3 1 3 3 1 3 3 3 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 1 1 1 1 +1 +1 1 (102) (102) (102) (102) (io;5) (io4) (10^^) Example 2 . [case (2) of Table S5] Pi = 7, P2 = il> P3 = 23, (P1IP2) = (7I1I) = +1. (P1IP3) = (7I23) = +1, (P2IP3) = (III23) =1. D = 71123 = 1,771, = 2XL + 1 = 505 and ui = 12, whence X]_ = 7'(6) a. square. By (IO5), y^ = 6, = 1 is a solution of ll*23x^ 7y^ = 1. Checking: 253(1)^ 7(6)^ = 253 252 = 1. Example 3 . [case (3) of Table g] Pi = 3, P2 = 13, P3 = 19, (P1IP2) = (3I13) = +1, (PiiP3) = (3il9) = +1, (P2IP3) = (13I19) = 1. D = 31319 = 7^1, = 2X^ + 1 = 7,352,695 1 p and u^ = 270, lOg, whence = 3*1107 ^ a square. By (I05), Legendre, on . clt . . Table X
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6g 2 2 = 1,107. = 122 Is a solution of 1319x 3y = 1Checking: 21^7(122)^ 3(1107)^ = 3,676,3^3 3,676,3^7 = 1. Example k , [case {^) of Table g] P]^ = 3, 7, P3 = 13Â» (.p^p^) = (37) = fi, (p^^Ip^) = (3I13) = +1. (P2IP3) = (7ll3) = 1. D Â«= 3'713 = 273, ti = 2Xi + 1 = 727 and = 1+1+^^ whenoe = 3(ll)^ 7^ a square. By (I05) , yj_ = 11, = 2 Is a solution of 7*13x^ 3y2 = 1. Checking: 91(2)2 . 3(11)2 ^ 354 . 353 ^ 1. Example 5 . [case (5) of Table g] Pi = 3, P2 = 7, P3 = 31, (P1IP2) = (3l7) = +1, (P1IP3) = (3i31) = +1. (P2iP3) = (7I31) = +1. D = 3731 ' 651, ti = 2X1 + 1 = 1,735 and ? 2 = 6g, whenoe Xi = 3*(17) / a square. By (I05) , yi = 17, Xi Â» 2 is a solution of 7*31x^ 3y^ = 1. Checking: 217(2}^ 3(17)^ = S6g g67 = 1. Example 6 . [case (6) of Table g] Pi = 5, P2 = 7, P3 = ^7, (P1IP2) ' (5i7) = +1, (P1IP3) = (51^7) = +1, (P2IP3) = (71^7) = +1. i> = 57^7 = lM5, ti = 2X1 + 1 = 9, '+60,519, 131, 0^1 and Ui = 233, 255, 566, 72g,^ whenoe Xi = 5(972,652)^ a square. By (I05), yi = 972,652, Xi = 119.907 4bid . %bid . 3whitford, OP. cit .. p. 201.
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69 Is a solution of 7*^7x 5y =1. Checking: 329(119,907)^5(972,652)2 = ^,730,259,565,521 ^,730.259,565,520 = 1. Example 7 . [case (7) of Table g] Pi = 5, Pg = 7, P3 = 23, (PllPg) = (5i7) = +1, (P1IP3) = (5I23) = +1, (P2IP3) = (7I23) <= 1. D = 5*7*23 = 305, = 2X^ + 1 = I,5i^,g6g,6iii and = 53,392,10^,^ whence \^ = 5.(12,30^)^ ^ a square. By (105), = 12,30?., = 2,169 Is a solution of 7*23x2 5y2 = 1. Checking: l6l(2,l69)2 5(12, 308)^ = 757, ^3'+, 321 757,^3^,320 = 1. TABLE 9 CONDITIONS FOR EQUATION (76), p^p^x^ p^y^ TO HAVE A SOLUTION WHEN ^ y^ = 1, Case Conditions Pl= I Pte P3S (mod^) ^ Conditions of Theorem 6 where 1 = 2, J = 1, k = 3 (1) (2) \l\ 11! (7) 1 3 1 3 3 3 3 3 3 3 3 3 1 1 1 3 3 1 3 3 3 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 1 1 1 1 +1 +1 1 (102) (102) (102) (102) (103) (ic4) (lOil) Legendre, op. olt ., Table X.
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70 Example 1 . [case (1) of Table 9] P]_ = 5, P2 = 11. P3 = 97, (P2IP1) = (I1I5) = +1. (P2IP3) = (III97) = +1, (P1IP3) = (5l97) = 1. D = 51197 = 5,335, t^^ = sx^ + 1 = g51,13g,997,i<79 and = ll,652,ggO,092, whence = 11(196,693)^ 7^ a square. By (105), = 196,693, = 29,622 is a solution of 5*97x^ lly^ = 1. Checking: i^g5(29, 622)2 11(196,693)2 = ^25,569,^9^,7^0 ^25,569,^92,739 = 1. Example 2 , [case (2) of Table 9] Pi = 3, P2 = 11. P3 * 103, (PglPi) = (III3) = +1, (P2IP3) = (III103) = +1, (P1IP3) = (3il03) = 1. D = 3'11103 = 3,399, t^ = 2Xi + 1 = 61,799 and u]_ = 1,060, whence X^^ = 11(53)^ a square. By (I05) , y^ = 53, = 10 is a solution of 3 '103x2 lly2 = 1. Checking: 309(10)^ 11(53)^ = 30,900 30,^99 = 1. Example 3 . [case (3) of Table 9] Pi = 5, P2 = 11. P3 = 23, (P2IP1) = (lli5) = +1, (P2IP3) = (ll23) = +1, (P1IP3) = (5I23) = 1. D = 5'ii23 = 1,265, t^ = 2X3^ + 1 = 206,999 and ui = 5,520,^ whence Xi = 11*(97)2 ^ a square. By (I05), yj_ = 97, = 30 is a solution of 5*23x2 iiy2 = 1, Checking: 115(30)^ 11(97)^ = 103,500 103,499 = 1. ^Cayley, op. cit .. XIII, p. ^130.
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71 Example ^ . [case {^) of Table 9] Pi = ^' ^2 " ^3 * (P2IP1) = = +1, (P2IP3) = (I1I53) = +1. (P1IP3) = (3i53) = 1. D = 31153 = l,7'^9. 4 = ^^1 ^ = 31,356,353,199 and = 7'^9,77'+,5f50, whence X^^ = ll(37,753)^ a square. By (105), y;j_ = 37,7'^3, = 9,930 Is a solution of 3'53^^ ' = 1Checking: 159(9, 930)^ 11(37,753)^ = 15, 67?', 179, 100 15,672,179,099 = 1. Example 5 . [case (5) of Table 9] Pi = 3, P2 = 11> P3 = 23, (P2IP1) = (11I3) =+1, (P2iP3) = (11I23) =+1, (P1IP3) = (3I23) = +1. D = 31123 = 759. 4 = 2X^ + 1 = 551 and u^ = 20,^ whence = ll(5)^ a square. By (IO5), = 5 , = 2 Is a solution of 3.23x^ Hy^ = !Â• Checking: 69(2) 11(5)^ = 276 275 = 1. Example 6 . [case (6) of Table 9] Pi = 3, P2 = 5, P3 = 23, (P2IP1) = (5i3) = +1. (P2IP3) = (5I23) +1. (P1IP3) = (3I23) = +1. D = 3 Â•5*23 = 3^5, = 2X^ + 1 = 6,761 and u^ = 36^+,^ whence X^ = 5* (26)^ ^ a square. By (I05), y_ = 26, Xi = 7 Is a solution of 3.23x^ 5y^ = 1. Checking: 69(7)^ 5(26)2 ^ 3^3^! . 3^3go = 1^ . Legendre, op. clt.. Table X. IblA*
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72 Example 7 . [case (7) for Table 9] Pt =3, Po = 5, Pt 103 (PglPi) = (5I3) Â» +1. (P2IP3) = (5I103) =+1. (P1IP3) =Â» (3il03) = 1. D = 35103 = 1,5^5, = 2X3_ + 1 = 123,i^gl,96l and ul = 3,l^l,5l6,''' whence X]_ = 5*(3,5l'^)^ ?^ a square. By (I05) , y^^ = 3,51^, x^j^ = ^+'+7 is a solution of 3103x2 5y2 = 1. Checking: 309(^^7)^ 5(3,51^)^ = 61,7^10,931 6i,7io,9f50 = 1. TABLE 10 CONDITIONS FOR EQUATION (7g), P^P^X^ PjY^ 1, TO HAVE A SOLUTION WHEN fi y^ Case Conditions Pl= P2S I P^s (mod 4) Conditions of Theorem 6 where 1 = 3. J = 1, k (1) (2) ii) (5) (6) (7) 1 3 1 3 3 3 3 1 3 3 1 3 3 3 3 3 3 3 3 1 1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 1 1 1 1 +1 +1 1 (102) (102) (102) (102) (101) (104) (104) Example 1 . [case (l) of Table 10] Pi = 5, P2 = 17, P3 = 19, (P3IP1) = (19I5) = +1, (P3IP2) = (19I17) = +1, (PiiPa) ^Whitford, op. cit .. p. 201
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73 (5li7) = 1. D = 51719 = 1,615, = 2X3_ + 1 = 3,25^,569 and uj_ = 51,7}2,'^ whence X^ = 19.(29^)^ fi a square. By (I05), y^^ = 29^, = 139 is a solution of 5*17x^ 197^ = 1. Checking: ^5(139)^ 19(29^)^ Â» 1,6^42,225 1,642,2^^4= 1. Example 2 . [case (2) of Table 10] Pi = 7, P2 = 11, P3 = 19, (P3IP1) (I9I7) =+1, (P3IP2) = (I9I1I) =+1, (PiiPa) = (7ll) = 1. D = 7*11'19 = 1,^63, tj_ = 2Xi + 1 = 1^3 and = k,^ whence Xj_ = 19*(2)^ a square. By (IO5), y^ = 2, x^ = 1 is a solution of 7*llx^ 197^ = 1. Checking: 77(1)^ 19(2)2 = 77 _ 76 = 1. Example 3 . [case (3) of Table 10] Pi = 5, P2 = 7, Pj = 19, (PjiPl) = (1915) = +1, (P3IP2) = (19I7) = +1, (P1IP2) = (5l7) = 1. D = 5'719 = 665, ti = 2Xi + 1 = 13,719 and ui = 532, v'henoe \^ = 19(19)^ a square. By (I05), yj_ = 19, x^ = 1^4is a solution of 5*7x^ 197^ = 1. Checking: 35(1'J)^ 19(19)^ = 6,g6o 6,^59 = 1. Example 4 . [case (4) of Table 10] Pi = 3, P2 = 5, P3 = 11, (P3IP1) = (11I3) =+1. (P3IP2) = (11I5) =+1, (pi!p2) = (3 1 5) = 1. D = 3511 = 165, tn = 2X, + 1 = 1,079 and u, = ^ Ibld . ^Cayley, op. cit .. XIII, p. ^^30. Â•^Legendre , op. cit . , Table X,
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7^ whence Xj_ = 11 Â•(7)" ^ a square. By (105), 7^. ^ * ^1 6 is a Bolution of 35x2 . ^^^2 = Checking: 15(6)^ 11(7)^ = 5^ 539 = 1. Example 5 . [case (5) of Table 10] ?! = 5, P2 = 11. P3 " 131, (P3IPL) = (I31I3) = +1, (P3IP2) = (I31I1I) = +1. (P1IP2) = (3ll) = +1. D = 311131 = ^,323, t;j_ Â» 2X^ + 1 = 263 and u^^ " k, whence X]_ = 131'(l) ^ a square. By (I05), y^ = 1, 2 2 x^ = 2 is a solution of 3*llx 131y = 1. Checking: 33(2)2 _ 131(1)2 = 1, Example 6 . [case (6) of Table 10] Pi Â» 3, P2 11. P3 = 17, (P3IP1) = (17I3) =+1. (P3IP2) (17I11) =+1, (PiiP2) = (3ll) = +1. D = 31117 = 561, t^^ = 2X3^ + 1 = 522,7^5 and U;j_ = 22,072,^ whence Xi = 17* (12^)^ ^ a square. By (I05), y^ = 12^, x^ = g9 is a solution of 3*11x2 17y2 = 1. Checking: 33(^9)^ 17(12^+)^ = 261,393 261,392 = 1. Example 7 . [case (7) of Table 10] Pi = 7, P2 = 11. P3 = 13, (P3iPl) = (13I7) = +1, (P3IP2) " (13I1I) = +1, (P1IP2) = (7li) = 1. D = 711*13 = 1,001, t;j_ = 2X1 + 1 = 1,060,905 and u, Â» 33,532,2 whence X, = 13(202)2 ^ ^ square. By (IO5), Legendre, pp. oit ., Table X, 2cayley, op. cit .. XIII, p. ^+20.
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75 = 202, = 53 is a solution of 7*llx^ 13y^ = 1. Checking: 17{^>3)^ 13(202)^ = 530,^53 530,^452 = 1. An illustration of Corollary ?! for equation (72), 2 2 P1P2P3X y =1, Is the follov;ing Example . Let pj^ = 5, pg = 13 , Pj = 17, all odd primes of the form Then D = 5*13 '17 = 1,105 in (1), t^ Du^ = 1, tL = 2X3_ + 1 = 1,623,132,289 and u^^ = ,^0^ ,k^2,^ Hence = (2gif,g?52)2. By Corollary g and relations (g)^ and (g)j^, yj^ = 2gi+,g22, X;j_ = 857 is a solution of (117) 1105x2 y2 = 1. Checking: 1105(857)^ (2Sh?^S2)^ Â«= 811,566,1^^5 811, 566, 1^^ = 1. We note that In Table 3 the excludents (73);j^, (7^)^, {7<^)^ all hold In this example, thus by Lemma K equations (73) (73) have no solutions. Since In Theorem 1 one of (72) (78) has a solution, equation (72) has a solution, which in this case is (II7) .
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CHAPTER IV CONDITIONS FOR ONE OF 6^x^ d^y^ = 1 TO HAVE A SOLUTION WHERE ^^.'^l ^' ^ PRODUCT OP POUR DISTINC^r ODD PRIMES Lemma 7 . Let D = p^p^P^Pj^, in equation (1), p P t Du = 1, where p^^ < P2 < Pj < Pl^. a^^Â® distinct odd primes. Let (1) have the fmidamental solution tj^ + u^j^V^ where tj_ = 2X"j_ +1. If one of the excludents listed in Table 11 to the 2 2 right of the equations (119) (13^) of (6), 5j_x d^y = 1, holds, then that equation has no solution in integers x, y. Proof . Since D is a product of s = ^ distinct odd a U /primes, p^, p^, P^ , Pi^, there are 2 =2 =16 positive divisors of D. These are dig) 1, Pi' Pg* P1P2. P1P3. p^pi^. P2P3. PgPi^' P3Plj.. P1P2P3. V2^. P1P3PV ^2^ 34, ^2^3^^For t^ = 2X^+ 1 and uj^, the fundamental solution of (1), the equation (6) of Theorem 1, vrhere 6i*d_ = D = P]_P2P3Pl4.Â» sixteen possible forms, namely equations (119) (13^). By Theorem la equation (II9) and one of (120) (13^) always 76
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77 m 0) H O W ra g Â•H P m H^ CM^ r^Â— p< ft ft ft ^ JdÂ— ft^ft^ft^ft t^ft K\ft OJft CVIft ft r<>ft f<>ft CVJft fvj Tuft Hft Hft Hft ft CVIft Hft Hft H I ft I ft I ft Â• ft O H OJ H ft ft Pi I jdft ^ ft ft ft ft ft Â— Â— OJ Â— r<> Â— ^ CVJ ft ft ^ ft ft I ft I ft I t<\ ^ CVJ >. OJ ft OJ ft CVJ ft Â— ft Â— ft CVJ t<> st OJ ft ft ^ ft ft I ft I ft I ft ^*^ Â— "h Â— H ft ft I OJ ^ H CVJ ft Hft ft ft H CVJ Hft CVJ ft ft I ft I ^ H ^ H Hft Hft ft Pi ^ Â— Â— ^ ft jijft ft I ft I n 11 II It It II OJ OJ ft I %^ ft CVJ CVJ OJ CVI ^ H ft ft CVJ CVJ CVJ ft OJ ft CVJ CVI ft Â« ft OJ ft OJ ft ft H ft CVJ CVJ ft I Â« ft ft ft ft Â« CVJ CVJ ft ft ft OJ ft H ft CVJ. ft ft i ft KNOJ ft Â« H ft ft I X s& ft CVJ OJ ft X H j=h ft ft H Â• H 1 II II OJ >Â» to 4i ft fl 0) 1 :s CVJ H X % (U ft H H H ft Â«Â«; CJN O H H CVJ CVJ H H H CVJ CVI OJ H OJ Lc^ M3 OJ OJ CVI H H H CVJ V Â•p H s Â« Â•p g I o a o o H H
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7Â« CQ CM ft ft ft ft ft ft ft rift % ft CM ft ft ft ^<^ ft ft OJ 1 ft 1 ft 1 ft ft H ft ft P. ft 1^ 1 ft H ft H ft ft H Â— ft ^ Â— ft Jft (Vi p, oj ft ft T\j p, oj I ft I ft ft Â— _ cvi OJ ft ft H ft ft ^ ft t"^ ft ft Â— ft Â— rv^ ft ^ ft . . ft H ft H ft H ft H ft I ft I ft Â» ^r^ ft OJ ft OJ ft ft Â— ft Â— ft Â— Â— OJ Â— Â— sir ru ft K\ ft rh ft ft H ft H ft H H ft H ft H ft ft I ft I ft I OS O H OJ r<^ ^ OJ K\ 1^ r\ iA t\
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79 arise from (6) and have Integral solutions x, y. We note the following about (120) (13^). Let y^. Then equation (120) oannot arise. 2 For if it did then by Corollary 7 and (30)^^, = y^, a contradiction to the assumption that ^ y^. Let (121) arise from (6) with 5]_= p]_ and = P2P3PI4.. By Theorem 1 or Corollary 9 there exist integers x, y satisfying (121). It follows from (121) that 2 2 Pq^x s 1 (mod pg), P]_x 5 1 (mod p^), (135) 2 2 P3_x 5 1 (mod pij.) , and PgPjPi^y s 1 (mod p^) . (4 Define the integers p', p" , p"' , p^ by p'Pj^ = 1 (mod pg), p^p^ = 1 (mod p^), (136) p"i3L s 1 (mod Pi^), and p^^^ p^p^p^^^ = 1 (mod p^). Then (135) may be written as p'Pj^x^ = p' (mod pg), P^Pj^x^ s p" (mod p^), J^P^x^ = p"' (mod pj^) , and p^^^ P2p5Pi.y^ = P^***^ (mo^l p^), which in view of (I36) reduce to 2 2 X 5s p' (mod pg), X = p" (mod p^) , (137) x^ s P*" (mod p), and y^ s P^^^ (mod p^^) .
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BO By (136)^ vre may write (p'IPg) ~ (P ~ (Pxlp^)* Hence if (p^lpg) = (p'IPg) ~ ~1Â» then (135)^ Is not solvable and, therefore, neither is (121). Similarly (121) is not solvable if either (p^^lp^) = 1> or (pj_pi.) = 1, or (Â•P2P3PI4.IP1) ~ Thus the four excludents listed opposite (121) in Table 11 are established. By like reasoning the four excludents listed opposite equations (122) (13^) in Table 11 are derived. We will make use of the following lemmata. Lemma 6 . The conditions where p , p , p, , p. are distinct odd primes and 1, J, k, X J xC I I =1, 2, 3, k, follow If and only if either (139) ^PilPj) = ^Pi'Pk^ ^ (PilPj) = +1 and p^ = 1 (mod k) , or (lUo) (pjp.) = (PJPJ = (PJP,) = +1 and i J i k i I Pj s P^ s Pj s 3 (mod k) , or (1^1) (Pj_Pj) = ^PilPij^ =(P^Pj) = +1 eJid exactly two of Pj , P^, P s 1 (mod k) .
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&1 Proof , If (1385) holds, consider (PjP^PjIp^) = +1, which may be written as (Pj I p^^ ) (p^l Pj^ ) ( P j 1 P^ ) = +1, or ilk2) i(pj^l)(pjl)+i(Pil)(Pkl)+i(Pll)(Pl) By (I3g) and (1^+2), i(p,l)(p.+P^,P53) (+1)(+1)(+1)(1) J K V = +1, or i(p,l)(p +P. P.3) (1) ^ ^ ^ Â« =+1, which holds if rnd only if either p^^ = 1 (mod or (pj+Pi^Pj3)/2 = 0 (mod 2). Thus Pj+Pj^Pj = 3 (mod 4), which holds If and only If either p^ s P^^ = Pj s 3 (mod k) or exactly two of p^ , p^^, Pj = 1 (mod ^) . Hence (139) (1^1) are established. Conversely, if either (139) or (1^10) or (1^+1) holds we may vrrite Pl^r P.1^PkP}"3 . = (+l)(+l)(+l)(l)"^ if J = Hence (I3S), This completes the proof of the lemma. Lemma 9 . The conditions
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S2 (1^3) (PilPj) = (PiiPk^ = (PilPj) = ^PjPkPj'Pi^ " where p^^, Py p^, and pj are distinct odd primes, and i, J, k, 5 =1, 2, 3, k, follovj if and only if either ilkk) (p^Ipj) = (Pilp,^) = (Pilpj) = +1 and s 3 (mod or (1^5) (PilPj) = (Pilpk^ " (P^iPj) = +1 and Pj a Pjj s Pj =1 (mod 4) , or (PilPj) = (PiIpij) = (PliPj) = +1 a^id exactly two of Pj, p^^, Pj s 3 i^od ^) , Proof . If (1^3) holds, then the relation (PjPkPjlPi^ ~ "'1 "^^y ^Â® v^ritten as (1^7) (PjPi)(PklPi),(PjlPi) = +1Also, since (p^lpj) = +1, it miay be written as (pj^+l)/2(p.l)/2 (PjIPi)(l) = +1; or upon multiplying each side by (pjip^^), (Pi+l)/2(p.l)/2 (l4g) (1) ^ ^ = (pjIpi).
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Â«3 Similarly since (PilPi^^ (PilPj) = +1. (Pi+l)/2.{pi,l)/2 (1) ^ = (Pk'Pi^ (li^9) (p,+l)/2(Pjl)/2 (1) ^ ' = (PjlPl)Substituting (li+g) and (l^+S) into (l'+7) and combining, we get (Pi+l)/2(p.+Pj,+Pp3)/2 = (1) = +1. which holds if and only if either p^ = 3 (mod k) or (Pj+Pk+P03)/2 = 0 (Â°Â»Â°^ 2). Thus Pj+Pk+P} = 3 (mod 4), which holds if GJid only if either = = pj ss 1 (mod k) or exactly two of p^ , Pj^, Pj s 3 (mod k) . Hence (iW (li^6) are established. Conversely, if either (1^^^^) or (1^+5) or (1^+6) holds, we may write (Pj^ I Pj) ( P^ I %X P^ Ipj ) = +1, or (p.+l)/2.(p +p +P,3)/2 (PjlPi)(Pl,Pl)(PjlPl)(l) ^ ^ * ^^^^ so that (PjPi,P!Pi)(+l) = +1. Hence (lil3) . This completes the proof of the lemma. Lemma 10 . Let p^ , Pj , p^^, Pj be four distinct odd primes where i , J , k, 0 = 1 , 2, 3 , ^. The conditions (150) (PiPjlPk) = (piPjl^l^ = (PkPjIPi) = (PkPjlPj^ =
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84 hold if and only if (151) ^PiPjlPk^ " (PlPjIPj^ ' (PkP}IPi) " +1 and one of il^2) (163) hold (mod k) , where (152) Pi= Pj= Pk^ Pj= 1. Pi= Pj=3Pk^ P= 1, (153) 3Pi=3Pj=3Pi^Pj= 1. (159) 3Pi=3Pj= Pi^3pj= i. (15^) Pi= Pj= Pk=3Pj= 1, (160) Pi=3Pj=3Pi,= Pj= 1, (155) 3Pi=3Pj=3Pij= Pj= 1, d^l) 3Pi= Pj= Pi^Pj= 1. (156) Pi= Pj=3Pk=3P{= 1. (162) Pi=3Pj= PkH3P{= 1, (157) 3Pi=3Pj= Pk= Pj= 1. ^163) 3Pi= Pj=3Pi^ pj= l But the conditions (164) (p^PjIpi,) = (piPjIpo) = ^PkPÂ«'Pi^ (Pj^pjIpj) = ^1. hold If and only if (I51) and one of (165) (l6g) hold (mod 4) , where 1(165) Pi=3p^ Pi^ PF ^' ^^^'^^ Pis3P^3Pi^3PjS 1, (166) 3PiS Pjs3Pi^3p,= L (i6j?) 3Pi5 P^ P^^ Pp Proof . Since p^ , p^ , p^, Pj are four distinct odd primes where 1, k, i=l, 2, 3, 4, there are 2^ = I6 possible arrangements of the least positive residues (mod 4) of these four primes, namely, (152) (163) and (I65) (l6g) above. Let (151) hold and consider (152) (159). In each p^ = Pj (mod 4), hence
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(l69) PiPj = 1 (Â°Â»o^ ^) (llPl) = (iIpj). Thus the first two members of (151), or (P^PjiP^^ ^ ^^l^j'^l^ 1, may be written as and (Pjlp^Pj) (pjlPi)(PjlPj) = +1. respectively, so that (170) (Pk'Pi^ = ^Pk'Pj^ ^PjlPi^ " (PjlPj). Prom (151)3 , 1= (PkPilPi) = (lPi)(PklPi)^PjiPl). which by (169)2 and (I70) may be written as 1 = (lpj)(Pi,IPj)(PilPj) = ^PkPj'Pj^Hence when (15I) and one of (15?) (159) hold, (P^PjlPj) = Again let (I5I) hold end consider the remaining arrangements (I60) (163) and (I65) (l6g). In each = 3n (mod ; hence (171)
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36 Now multiplying (151>i and (l^Dg ^"^IPk^ (lP) respectively, we obtain (Pj^PjlPi,) = (iIpj,) (PiPjIpj) = (ipÂ§)These, by (17l)i, may be written as (172) (PiclPiPj^ = (llPlc^ and (pjlPiP^) = (lP{). Multiplying {172)1 and (172)2 by (p^^lPj) and (PjIPj) Â• respectively, V7e get (173) (PklPi)=(PklPj^^lPk^ (P5lPl)=(P0iPj)(llP5)Now multiplying corresponding terms of (173}i and (173)2 together we obtain (17^) {PkPj'Pi^ " (PkPlPj)^llPkP^. Hence multiplying corresponding terms of (171)3 and (17^) produces (1751 (PkPIPi) = (PkP0lPj^^"^PkP^Since by (151)3, (PkP!Pi) = ^1' (^75) reduces to (176) 1 = (Pl,PjlPJ)(liPkP{)Â• Consequently in (176), (PkPjlPj^ = ^"^^^ Pi^^l S 3 (mod k); and (PkPjIPj) = "1. only if Pj^pj 5 1 (mod U) . Hence, remembering that p^ = 3Pj (mod k) , = +1 when
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(151) holds simultaneously with one of (I60) (I63); and, (P^PjlPj) = 1 ^^^^ il5'^) holds simultaneously with one of (165) (162). Conversely, if (150) holds, consider (I50)i^, or (P^PjlPi^ ~ ''^> ^Â® written as { Por (pjl)/2(pi^+Pj)/2 ^ (177) (PjlPkP5)(i) Now from (150)i and (150)2, (173) tPilPk^ " ^Pj'Pk^ (PlIPj) = (PjlPj). Hence (177) niay be written as (p.l)/2(pi.+pj)/2 (179) (PilPkPi)(l) * =+1. If we let Pjj 3 Pj (mod k) in (179). It may be written as (p,l)/2.(2p^)/2 (Pl,PjIPi)(l) ^ ^ =+1, or upon multiplying both sides by (lp^), (Pll)/2 (PkPj iPi)<l) " (lPl) 1
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This last reduces to (pil)/2 (IgO) (1) = (ilPi), since by (150)3. (p^PjlPi) = +1note that (ISO) holds if and only if = Pj (mod , Hence when (150) holds, = Pj and = Pj (mod k) , one of (152), (153), (156), and (157) holds. If we let p^ = 3pj (mod k) , then by (150)3 and (179) (Igl) 1 = (P^PjlPi) = (PilPkPj)Hence p^, Pj = 1 or 3 (mod 14). Thus when (150) holds and Pk = 3Pj (mod K), one of (15^), (155), (15^). (159), (160) (163) holds. Now If ilSh) holds, consider or (P^PjlPj^ ' 1, which may be written as (p.l)/2.(Pj,+P})/2 (lg2) (PjlPi,P,)(l) ^ ^ Â« = .1. By {16k) ^ and UGk) ^, relations (17^) hold, and hence (182) may be written as (p.l)/2(p,+Pj)/2 (1^3) (PilPkPj)(i) ^ 1^ Â« = 1. Now let p^ = Pj (mod in (1^3), then it may be written as (Pi,PjiPi)(l) J = 1, or after multiplying both sides by (ijp^^), we get
PAGE 95
g9 (p.l)/2 Hence, since by {l6k)^, or (P^PjIPi^ ~ (1) =(llpi). This last holds if and only if p^^ = 3Pj (mo
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90 or {186) pj = = P = 3 (mod k) and (pjIPk) = (PjlPj) = (PklP) = +1. or (1J57) Pj = p^ s Pj = 3 (mod k) and (PjIp^) = (pj1p{) = (PklPj) = +1. then either equation (121), (123), (125), or (127) has a solution in Integers x, y according as i = 1, 2, 3, or with J, k, 5 = 1, 2, 3, ^ where J < k < J. Solutions are given by (IgS) yi = (Xl,uA) and 2x^y^ = u^^. Proof . By hypothesis and Theorem la one of (120) (13'+) of (6) has a solution in integers x, y. By Lemma 7, since ^ equation (120) has no solution. Let i = 1, J = 2, k = 3, I = ^ in conditions (139) of Lemma S, then (1^9) (PilPg) = (P1IP3) = (P1IP4) = +1 and P3_ = 1 (mod k). Hence by Lemma S, ( PgP^Pl^. I Pi ) = +1; and thus by Lemma 7 the necessary conditions for (121) to have a solution are satis2 fied. Now all excludents of Table 11, except X^ y^, csji be expressed in terms of one or more of the six Legendre symbols,
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91 (PiIps). (P1IP3). (PilPlt). ^P2iP3^' (p^iPi^), and (p^lpi^). Hence when conditions (189) holi, the values of (pglPj), (PjIPlj.)' (P3lPl)Â» least positive residues modulo k of pg, p^, and PI,, can be arranged in 2^ = 64 different arrangements or cases as shown in the six columns headed "Conditions" in Table 12. From taking each of the sixtyfour oases in turn we determine those excludents of Table 11 which are consistent with conditions (1^9). The procedure is the same as that used in developing Table 4. The results are recorded in the last column of Table 12, omitting the column which designates the particular excludent which holds, as was done in Tables 4, 5, and 6. We see from Table 12 that equations (122) (13^) have no solutions under the conditions (189) and those of any one of the fourteen cases: 2: c, e, g, h, 4: c, e, g, h, 6: c, e, g, h, g: d, e, which are recorded in Table 17 as cases (1) (l4). By Theorem la, since y^, one of the equations (121) (134) has a solution in Integers x, y. Hence when conditions (IS9) and those of any one of the cases (l) (l4) in Table 17 hold, and ^ y^, equation (121) of (6) has a solution. Conditions (I89) and those of the first tvrelve cases of
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92 TABLE 12 CONDITIONS TO EXCLUDE EQUATIONS {122) W^k) WHEN y^, (P3^1p2) (P1IP3) = (P1IP4) = +1Â» Pi S 1 (mod k) Case Conditions I ^3* (mod Pl+i 0 (P2lP3)= (p2lPi) = (P3lpi4.) = Equations excluded^ l(a Kb l(c l(d Ke l(f Kg l(h 2{a 2(b 2(c 2(d 2(e 2(f 2(g 2(h (133). 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "A B C D and E F 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 denotes denotes denotes denotes (13^). denotes denotes +1 +1 +1 +1 1 1 1 1 +1 +1 +1 +1 1 1 1 1 equations equations equations equations +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 +1 1 1 +1 1 +1 +1 1 +1 1 1 +1 1 +1 +1 1 none B C D C B C C P Q A H A I A A equations equations (130), (132), and (133). Q denotes equations H denotes equations (13^). I denotes equations (122) (124). (125)(12g), (131)(13^). (123) (13^). , , , , (123), (124), (127)(130), (123)(126}, (129)(132). (122), (124), (126), (127), (122), (12i^)(12g), (130)(13^) (122)(12J), (126)(130), (132) (122)(127), (129)(133). (Table 12 is continued on the next page.)
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93 TABLE 12 Continued Conditions 2= 1 P3= I (mod k) (P2lP3) = Tp^IPliT TpTIpi^ Equations excluded 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 +1 Â•41 +1 +1 1 1 1 1 +1 +1 +1 +1 1 1 1 1 +1 +1 +1 +1 1 1 1 1 +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 +1 1 1 +1 1 +1 +1 1 +1 1 1 +1 1 +1 +1 1 +1 1 1 +1 1 +1 +1 1 p J K H L I M M J Q A K A L A A F 0 M N K I L K J denotes equations (122), (12k), (125), (12^), ^Tlen^es^eiuiiions (122)(125). (127)(131). (133). ^""l^llll^^ equ8.tions (122)(126), (12g).(132), and K denotes equations (122), (123), (125)(129), (131)(13jj).^^^^^^ equations (122), (123). (126), (123), (129), (132), and (13^). (Table 12 is concluded on the next page.)
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1 9^ TABLE 12 Concludeg Case 3 Conditions (mod k) (p^Pl^) = Equations excluded 6(a) 6(b) 6(0) 6(d) 6(ej 6(f) 6(g) 6(h) 7(a) 7(b) 7(c) 7(d) 7(e) 7(f) 7(g) 7(h) g(a) g(b) g(c) g(d) g(e) g(f) 2(g) g(h) 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 +1 +1 +1 +1 1 1 1 1 +1 +1 +1 +1 1 1 1 1 +1 +1 +1 +1 1 1 1 1 +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 +1 +1 1 1 n +1 1 1 +1 1 1 +1 1 +1 +1 1 +1 1 1 +1 1 +1 +1 1 +1 1 1 +1 1 +1 +1 1 N M A H A L A A J Q I K M N H I I K K A A I M M Table 17 establish (IgU) and (1Â£55), while (1^9) on^ the oondltlons of the thirteenth and fourteenth cases in Table 17 establish il&k), (1^6), and (127) for i = 1, J = 2, k = 3, \ = By symmetry, Interchrmging the subscripts 1 and 2 in the cases (l) ilk) of Table 17, we get the necessary and sufficient conditions for equation (123) to have a solution. Also by symmetry, interchanging the subscripts 2 and 3 in the conditions by which (123) has a solution, we obtain those by v/hlch (125) has a solution. Again by symmetry,
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95 Interchanging the subscripts 3 and k in the conditions gotten for (125) to arise, we obtain those by which (127) has a solution. Thus (ig^) (187) are established whereby equation (121), (123), (125) or (127) has a solution according as i = 1, 2, 3, or if, j, k, J = 1, 2, 3, ^ where J < k < I. By Theorem 1 solutions are obtainable from and (8)2, n = 1 = G, thus establishing (lJ?g) and completing the proof of the theorem. Theorem g . Let D = I>2y>2'^j'^l\^^^^^ ^1 ^2 ^ ^3 distinct odd primes. Let tj^ + "iV^ ^Â® fundamental solution of (1), t^ Du^ Â«= 1, where t^ = 2X^ + 1. If \^ y^, Pj = Pjj = Pj =3 (mod , and either (191) (PjiPk) = (PjIPj) = ^Pk'Pp "^^ Pi s 3 (mod k) , or (192) (PjIp^) = (PjlP{) = (PiclPj) = Pi S 3 (mod if), then either equation (121), (123), (l^^), or (I27) has a solution in integers x, y according as i = 1, 2, 3, or if where J, k, J =1, 2, 3, if and J < k < i . Solutions are given by
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1 96 (193) yi = (^i.u^A) and 2x^y^= u^. Proof . By hypothesis and Theorem la, one of (120)(13^) of (6) has a solution In Integers x, y. By Lemma 7, slnoe ^ y^, equation (120) of (6) has no solution. Let 1 = 1, J = 2, k = 3, i = ^ in conditions ilkO) of Lemma 3, then (19^) (p^lPg) = (PiiP3) = (PilPi^) = +1 and P2 = Pj s Pii s 3 (mod k) . Hence by Lemma ^5, (PjP^Pi^.! P]_) = +1; and thus by Lemma 7 the conditions necessary for (121) to have a solution are satiafled. Now, all excludents of Table 11, except f y^, can by expressed in terms of one or more of the six Legendre symbols, (p^IPp), (p^^lp^), (P3_lPi^.), (P2IP3). (p2lPi.^Â» (p^iPlj.)Kence when conditions (19^) hold, the values of (PglP^)* (PglPi^.^ (p Ipj^), and the least positive residues modulo 4 of pj_ can be arranged in 2 = lb different arrangements or cases as shown in the four columns headed " Condi t lone" in Table I3. From taking each of the sixteen different cases in turn we determine those excludents of Table 11 which are consistent with conditions (19^). The procedure is the same as that used in developing Table 12, but only the equations of Table 11 excluded are recorded in the last column of m
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97 o IÂ— Â• w III Q. III 1 1 III OJ rvi n 1 + cn II O M ft 1 J? ft W II o ft ~H o ft EH cn II o M ft M i ft o o CO U o Â•P CQ O a o o ?! H O X o ft ft _ft ai ft ft ai ft ft o B CO cd o CM CVJ CVl H H H CVJ fVJ CVI CVJ CM CVJ H HH H H fH CM CM CM (l)a} CQ 03 CO Ca Â— CC OQ CQ csJa' ccaJcvjCMcSoj cs OOOOCMCMOOO iH H tn Â«M OQ %k 05 CM CVI O CVJ CM H H Vl V( *M ^
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1 9g Table I3. We see from the table that equations (122) (13^) have no solutions by the conditions (19^) and those of one of the four cases 1(d), 1(e), 2(d), and 2(e). We observe that oases 1(d) and 1(e) have already been derived in Theorem 7 as cases g(d) and S(e) of Table 12. By Theorem la, since ^1 ^ ^1' (121) (13^) has a solution. Thus when the conditions (19^) and those of either case 2(d) or case 2(e) hold, equation (121) of (6) has a solution. These two additional sets of conditions are recorded in Table 17 as cases (15) and (16), which establish (190) (192) for i = 1, j=2^k = 3^j=I., By the same argument as in Theorem 7 using the property of symmetry with regard to the subscripts, (190) (192) are established completely. By Theorem 1 solutions of (121), (123), (125), or (127) are obtainable from (g)nL and (2)2, n = 1 = 0, establishing (193) and completing the proof of the theorem. Theorem 9 . Let D = PjP2'P7^V>i^. where p^ < P2 < p^ < Pj^ are distinct odd primes. Let t^ + u^ VD be the fundamental solution of (1), t2 Du2 = 1, where t]_ = + 1. If X^ 7^ j^, (p^IPj) = ^PilPk^ ^ (Pj^Ipj) +1, at least two of (195) (PjIpjj), (PjlPj), (p^lP}^ ' 1 (196) Pj^ s 3 (mod H) , and exactly two of Pj , p^^, Pj s 1 (mod k) , then either equation (121), (123), (12S), or (127) has a
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99 solution In integers x, y aocording as i Â»= 1, 2, 3Â» or ^, J, k, H = 1, 2, 3, ^ where J < k < I. Solutions are given by Proof . By hypothesis and Theorem la one of (120) (13^) of (6) has a solution in integers x, y. 3y Lemma 7Â» since Xj^ ^ yj^, equation (120) of (6) cannot arise. Let i = 1, J = 2, k = 3, i Â» if in conditions (l4l) of Lemma 2, then Hence by Lemma S, ~ +lj and thus by Lemma 7 the necessary conditions for (121) to have a solution are satisfied. Conditions (19^) present three different sets of assumptions, i. e., (197) 2 2 ^1 ~ ^^lÂ»^l/^^ ^^1^~ ^1* (19^5) (P3^IP2) = (p^Ip^) = (PilPij.)= +1 and exactly two of p^, p.,, p^^ s 1 (mod \) Â» (199) (p^^iPg) = (p^Ip^) = (P]_IPi^) = +1 Pg = P3 3 3Pj^ s 1 (mod \) , (P1IP2) = (P1IP3) = (PilP!) +1 and and (200) P2 = 3P5 = Pi). = 1 (mod ^) , and
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Â• 100 (PliP2) = ^Pi'P3^ (PliP]^.) = +1 and (201) s P^ s Pi^ = 1 (mod k) . 2 Nox7 all excludents of Table 11, except jj^, can be expressed in terms of one or more of the six Legendre symbols (Pj^lPg). (P1IP3). (PiiPl^.), (P2lP3^ (P2iPl.^ ^PjlV]^) ' Hence when (199), (200) and (201) hold, the values of (p^lpj), (p^jpj^), (p jp^), and the least positive residues modulo k of p^ can be arranged in 2 = 16 different arrangements or cases as shown in the four columns headed "Conditions" in Tables 1^4, 15, and 16 which follov/. From taking each of the sixteen oases in turn we determine those excludents of Table 11 which are consistent with conditions (199), (200), and (201). The procedure is the same as that used in developing Table 12, The results are recorded in the last column of Tables 1^, 15, and 16. We see that the cases 1(a) (h) of each of the three Tables 1^, 15, and 16 are the same as the cases 2(a)(h), ^(a)(h), and 6(a)(h) respectively of Table 12, which were derived in Theorem J. Omitting these, vie note there are twelve additional sets of conditions, four in each of the Tables ik , I5, and 16 by which equations (122) (13^) have no solutions. These conditions are: (199) and those of ajiy one of the cases 2: c,e,g,h of Table 14, (200) and those of any one of the oases 2: c,e,g,h of Table 15, (201) and those of any one of the cases 2: c,e,g,h of Table I6.
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101 H pa CVI I d o g H III III H P) 1 III OJ At rv * VI 1 Â—J 11 + CO s; II o H EH ft H ft II 3 a ft ~H o ft fH CO II O M Eh ft H H g ft O o CQ o m O d O o H O 0) W CVI H H "vd"kd^ roi^ r<v^ r<>,f<>^^ r<>^ r'N'^ OJ CVI OJ CVI I I II Q) ca o CVJ OJ cu OJ 1 1 1 1 Q) 0) (1) CQ ^ m c5 OS CVI ce CVI ca CVJ OJ O O CVJ O OJ O CVI OJ H H H H o o o o o
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102 H W Et CVJH Ik H << H OJ CM CO O H i o o O H Ei H O O 'Ci o s H III P* III ft III OJ ft + ft ft ft ft OJ ft ft (0 o Â•H OS W H O O H Â•P Â•H o o rift ft ft OJ ft ft ai ft ni ft o s 03 Ci o CV) C\J H H OJ H (D H E^ OJ H (D H cG Eh OJ OJ H H H H CVJ H H OJ H (]} H 43 CO Ei ^ r>r\ ro,Â»^, Â— r'vÂ— > r<^ Cd^H'dHÂ«MHH Cd^HdHiMrHH I 0) CO cc CM O CVJ H I CO 'v 03 CM O CM I I CO CM CM O CM CM Cm Â— ^ o o O O
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103 VX5 H Ik 'O o H g /< H 1 Â— P* (II _l 11 CL IN 1 CVJ P4 (\l CVJ H + 05 It O H ft 1 H W ft U II O X ft pa H o ft II o M C\J &^ ft 14 Â»^ ft o o CO a o H ?! w OQ o H n o o H O 0) ft ft ft ft OJ OJ C\J CVI H H H H
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10'+ These tvielve sets of conditions are recorded In Table 17 as cases (17) (25). By Theorem la, since ?^ y^, one of (121) (13'+) has a solution. Hence when the conditions of any one of the cases (17) (2^) in Table 17 hold, equation (121) has a solution, thus establishing (195) and (196) for 1 = 1, J = 2, k = 3, J^'i. 3y the same argument as above in Theorem 7 using the property of s^'^rametry with regard to the subscripts, (195) and (196) are established completely. By Theorem 1 solutions of (121), (123), (125), or (127) are obtainable from (g)]_ and (5)2, n = 1 = G, establishing (197) and completing the proof of the theorem. Since Table I7 is the result of the assuniptlons and proofs of Theorems 7 9 , we have proved the following summarizing Theorem 10 . Let D = pj^PjP^Pii where pj_ < P2 < < Pi}, are distinct odd primes. Let t^ + u^\/D be the fundamental solution of (1), t^ Du^ = 1, where t^^ = 2X^ + 1. If X;j_ f y^, (202) (p^lPj) = (PilPk^ = (PilPj) = +1. and either (203) exactly two of p^, p^, Pj s 1 (mod 4) and at leest two of ((p p ), (p,p,), (p, !p,) = 1, or (20i) Pj = p^ = Pj 5 3 (mod 4) and (PjIPij) = (PjIpj) = ^Pk'Pj^ '
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105 TABLE 17 CONDITIONS BY WHICH E(4UATI0N (121) HAS A SOLUTION WHEN Conditions Table Case ^1' 1 (P2lP3)~ (Pglpl^)^P^IPl^)where found (mod 4) 1 1 1 1 "I ~X **x 12 2 1 1 1 7 ^ "Â•X ~x 12 3 1 1 1 3 ~x ""X +1 12 4 1 1 1 3 Â—x ~x Â• X 12 5 1 1 3 X "X Â• X 12 g 1 1 3 X Â— X 4.1 'X Â•X 12 7 1 1 3 1 Â—X "Â•X ~x 12 1 1 X Â— X Â— X "* X 12 9 1 3 1 1 +x Â—X "Â•X 12 10 1 3 1 X X tX *x 12 11 1 3 1 1 1 1 +1 12 12 1 3 1 1 1 1 1 12 1 3 3 3 +1 1 +1 12 1 3 3 3 1 +1 1 12 3 3 3 3 +1 1 +1 13 16 3 3 3 1 +1 1 17 3 1 i 3 +1 1 1 \^ IS 3 1 1 3 1 +1 1 19 3 1 1 3 1 1 +1 V\ 20 3 1 1 3 1 1 1 \^ 21 3 1 3 1 +1 1 1 15 22 3 1 3 1 1 +1 1 15 3 1 3 1 1 1 +1 15 1^ 3 1 1 1 1 1 15 3 3 1 1 +1 1 1 16 1^ 3 3 1 1 1 +1 1 16 27 3 3 1 1 1 1 +1 16 2g 3 3 1 1 1 1 1 16 then either equation (121), (I23), (12^), or (1^7) has a solution in Integers x, y according as i = 1, 2, 3, or 4, J, k, I = 1, 2, 3, ^ xvhere J < k < {. Solutions are given by (205) y^ = and 2x^yj_ = u^.
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lo6 By the argi^ment of symmetr.y with regard to the subscripts, oases (1) (12) and (17) (2S) in Table 17 establish (202) and (203), while oases (I3) (lb) produce (202) and (20^1). We next derive the necessary and sufficient conditions for eqEuations (122), (12^), (126), and (123) to have solutions. We will make ur,e of Lemma 9. Theorem 11 . Let D = ^^2,^2^'}^^ where p^ < p^ < pj < Pj are distinct odd primes. Let tj^ + u^VD be the fundamentrl solution of (1), t^ Du^ Â« 1, where = 2X^ + 1. If y^, (206) ^PilPa^ = ^'Pi1p3^ Â° (p^Ipi^) = +1, and the conditions of any one of the cases (1) in Table 23 hold, ecuaticn (122) of (6) has a solution in integers x, y. Solutions are obtainable by relation (205). Proof . By hypothesis and Theorem la, one of (120) (13^1) of (6) has a solution In integers x, y. .By Lemma 7, since ?^ y^, equation (120) has no solution. Let i = 1, j = 2, k = 3, } = ^ in conditions ilkk^) of Lemma 9, then (P^Ip^) = (p^lp^) = (P^Ipi^) = +1 and (207) P^ = 3 (mod k) . Hence by Lemma 9, (PgPjPi^lp^^) ~ "^^i sind thus by Lemma 7 the necessary conditions for equation (122) of (6) to have a solution are satisfied. By (207)^^, (207) can be vrritten as
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107 (PglP^) = ^P3iPi) =
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log TABLE 1& CONDITIONS TO EXCLUDE EQUATIONS (121) AND (123) (I3J+) WHEN A 1 d v2 r y^f n ^ =: P^) (P3lp^) = (PliPl) ' 41 rt = ^x, P^ = Conditions Equations Case 2 ( 3 mod 4 4 ) (r, In ^ = *^2' ^3 V Pjl PI).' ( P, 1 P), ) = excluded 1(a) 1 1 1 +1 +1 +1 B Kb) 1 1 1 +1 +1 1 C 1(c) 1 1 1 +1 1 1 A 1(d) 1 1 1 +1 1 +1 D 1(e) 1 1 1 1 +1 1 A 1(f) 1 1 1 1 +1 +1 E Kg) 1 1 1 1 1 +1 A Kh) 1 1 1 1 1 1 A 2(a) 1 1 3 +1 +1 +1 B 2(b) 1 1 3 +1 +1 1 C 2(c) 1 1 3 +1 1 1 A 2(d) 1 1 3 +1 1 +1 D 2(e) 1 1 3 1 +1 1 A 2(f) 1 1 3 1 +1 +1 E 2 g) 1 1 3 1 1 +1 A 2(h) 1 1 3 1 1 1 A 3(a) 1 3 3 +1 +1 +1 C 3(b) 1 3 3 +1 +1 1 C 3(c) 1 3 3 +1 1 1 A 3(d) 1 3 3 +1 1 +1 A 3(e) 1 3 3 1 +1 1 A 3(f) 1 3 3 1 +1 +1 A 3(g) 1 3 3 1 1 +1 A 3(h) 1 3 3 1 1 1 A 1 3 1 +1 +1 +1 B 4(b) 1 3 1 +1 +1 1 C 4(c) 1 3 1 +1 1 1 A 4(d) 1 3 1 +1 1 +1 D ^A denotes equations (121), (123)(134). B denotes equations (121), (124), (126), (12??), (129), (131), and (133). C denotes equations (121), (124)(129), (I3I)(13^). D denotes equations (121), (1^3), (124), (126)(131), (133), and (154). E denotes equations (121), (123)(126), (12S)(133).
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1 109 TABLE IS Continued Conditions Eauations excluded ^ o A (r ^3' nod l^) (p lp,)= ( P^l Pi. ) = ( p., 1 Pi. ) " "rye; 1 J 1 1 + 1 1 A X X J 1 1 +1 +1 E X X 1 1 1 fl A Mv n; X X J X 1 1 1 A Â•z J 1 X +1 +1 +1 D T X 7 J +1 +1 1 A J T X 7 J +1 1 1 A z J T X X +1 1 +1 D pie; 1 J X X J 1 +1 1 A ; X J 1 X X J 1 +1 +1 A y V g; z J X X 1 1 +1 A I J 1 X X 1 1 1 A DV a; I J X 1 X +1 + 1 +1 B o\ O f I J T X X +1 + 1 1 C O V C y J T X 1 X +1 1 1 A X J T X X +1 1 +1 D X J X 1 X 1 +1 1 A o\i 1 X J T X X 1 +1 +1 E (^( eT\ 0\g) X J T X 1 X 1 1 +1 A ov n / X J X X 1 1 1 A / ks.) X J :t J 1 X +1 +1 +1 E X J X 1 +1 1 A 7(c) 3 3 1 +1 1 1 A 7(d) 3 3 T X +1 1 XT ~x A 7(e) 3 3 1 1 +1 1 A 7(f) 3 3 1 1 +1 +1 E 7(g) 3 3 1 1 1 +1 A 7(h) 3 3 1 1 1 1 A g(a) 3 3 3 +1 + 1 +1 C g(b) 3 3 3 +1 + 1 1 C g(c) 3 3 3 +1 1 1 E 2(d) 3 3 3 +1 1 +1 A 2(e) 3 3 3 1 + 1 1 A 2(f) 3 3 3 1 +1 +1 D 2(g) 3 3 3 1 1 +1 D 2(h) 3 3 3 1 1 1 E
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110 equivalent to (20S), and the oonditions of any one of the thirtysix cases (page I07) hold. Conditions (207) and these cases are summarized in Table 23 as cases (l) (36). By Theorem 1 solutions of (122) are obtainable from (6)j_ and i^) 2 n Â» 1 = Q, which produces (205). Next let i = l, j = 2, k = 3, \ a^lin conditions (1^5) of Lemma 9, then (P1IP2) " (P1IP3) = (P^lPi^) = +1 and (209) P2 = P^ = P4 = 1 (mod i) . Hence by Lemma 9, (P2PjPl.! P]_) ~ "^^J thus by Lemma 7 the necessary conditions for (122) of (6) to have a solution are satisfied. By (209)j, we may vjrite (209) as (P1IP2) = (PiiP3) = (P]_iPi) = +1 and (210) P2 s Pj = 1 (mod k) , Nov; all excludents of Table 11 can be expressed in terms of one or more of the six Legendre symbols (pj^lpg)! (p]_lpj), (pj_iPl^), (P2IP3), (PglPii.). and (p^lp^). Hence when (210) holds, the values of (p^ip^), (PglPi^), (p^lPi^), and the least positive residues modulo h of can be arranged in 2^ = 16 different arrangements or cases as shovm in the first four columns of Table 19. From taking each of the sixteen cases
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Ill TABLE 19 CONDITIONS TO EXCLUDE EQUATIONS (121) AND (1?3) (l^k) WHEN = (p^lp^) = (p^Ipi^) Â« +1. P, s P3 ^ s Pj^ 2 1 (mod k) III. II ...1 w Conditions p s (P2lPi) = ^1 (mod excludedÂ® 1 1(a) 1 +1 +1 +1 none Kb) 1 +1 +1 1 B 1(0) 1 +1 1 1 C 1(d) 1 +1 1 +1 n 1(e) 1 1 +1 1 C 1(f) 1 1 +1 +1 E Kg) 1 1 1 +1 C Kh) 1 1 1 1 C 2(a) 3 +1 +1 +1 P 2(b) 3 +1 +1 1 G 2(c) 3 +1 1 1 A 2(d) 3 +1 1 +1 H 2(e) 3 1 +1 1 A 2(f) 3 1 +1 +1 I 2(g) 3 1 1 +1 A 2(h) 3 1 1 1 A (133), (129), (13^). (131). (133). A denotes B denotes C denotes D denotes and (13^). E denotes F denotes (131), and G denotes equations equations equations equations equations equations (133). equations H denotes equations (133), and (13^). I denotes equations (121), (123)(15^). . (125)(12g), (131)(13^0. (123)(13'+). (123), (12i^), (127)(130), (1?3)(126), (1P9)(132). (121), (12^), (126), (12g), (121), (12if)(129). (131)(121), (123), ilPk), (126)(121), (123)(126), (12g)
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112 in turn we determine those excludents of Table 11 which are consistent with conditions (210). The procedure Is the s?me as that used in developing Table 12. The results are recorded in the last coluinn of Table 19, We see from the table that equations (121) and (123) (13^) have no solutions when (210) and the conditions of any one of the cases 2: c, e, g, and h hold. But we observe that these are the same conditions derived and recorded in Table lf5 as cases 1: c, e, g, and h respectively. Thus no new sets of conditions are found using (210) whereby (121) and (123) (I3U) have no solutions. We next let 1 = 1, j = 2, k = 3, \ = i+ in the conditions (1^6) of Lemma 9, then Hence by Lemma 9, (PpP^Piilp^) = "Â•"IJ thus by Lemma 7 the necessary conditions for equation (122) of (6) to have a solution are satisfied. Conditions (211) imply three different sets of conditions, 1. e., (211) (P3_1p2) = (P1IP3) = (PilPi^.) = +1 and exactly two of ip^* P] = 3 (mod k) , (212) (213) (P^IPg) = (P1IP3) = (PilPi^.) = +1 3P2 = 3P3 s Pi^ 3 1 (mod k) , (P^Ip^) = (P^Ip^) = (P3^lPi,.)= +1 3P2 = P3 3P2^ = 1 (mod ^) , and and
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113 and (P^Ip^) = (P]^!?^) = (P]_IP^) = +1 and (214) Pg 3 3p^ 35 3P4. s 1 (niod if) . By (212)i, (213)1^ and {2lk)^, conditions (212) {2lH) may be written as (215) (216) and (217) (p^lpg) = (PiiP^) = +(PxlPl) = "^1 3P2 s 3P^ = P1. a 1 (mod ^0 , (p^lp^) = *(P;j_P3) = (PilP2) = +1 and 3P2 S P3 s 3Pij. 2a 1 (mod , +(P2,1p2^ = (P1IP3) = (PilPl;) = +1 and Pg Â» 3P3 a 3P4 s 1 (niod respectively. Now all excludents of Table 11 can be expressed in terms of one or more of the Legendre symbols (p^lPj), (P;j_p^), (P]_IPi^), (P2iP3)Â» (P2lPi<.^ and (p^lpi^^). Hence when (215), (2l6), ejid (217) hold, the values of (P2ip3), (P2lpi.). (pjlPi^)* and the least positive residues modulo k of p^ can be arranged in 2^ = I6 different arrangements or cases as shown in the four columns headed
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"Conditions'* in Tables 20, 21, and 22 which follow. Prom taking each of the sixteen cases in turn determine those excludents of Table 11 vjhich are consistent \vlth conditions (215), (2l6), and (217). The procedure is the same as that used in developing Table 12. The results are recorded in the last column of Tables 20, 21, ajid 22. We observe at this point that since conditions (215), (2l6), and (217) are derived from conditions (212), (213), and (2l4) respectively, and also that conditions (203) are derived from (207), the results of cases 2 (a) (h) of each of the Tables 20, 21, and 22 are exactly the same as the results in Table 13 of cases 7 (a) (h), 5 (a) (h), and 3 (a) (h) respectively. Hence there are twelve new sets of conditions, four in each of the Tables 20, 21, and 22, by which equations (121) and (123) (13^) have no solutions. These are conditions; (212) and those of any one of the cases 1: b,d,e,g of Table 20, (213) and those of any one of the cases 1: b,c,f,g of Table 21, (21^) and those of any one of the cases 1; c f of Table 22. These tv/elve sets of conditions are recorded in Table 23 as cases (37) (^S). By Theorem la, since y^, equation (122) has a solution when conditions (206) and those of any one of the cases (37) iW in Table 23 hold. By Theorem 1 solutions of (122) are obtainable from {S)^ and (3)2, n = 1 = G, which are relations (205), This completes the proof of
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115 'd o Â£3 M H H >^ III I 1^ ft CM III H w CM ft P 1^ H + ft ft II ft ft II CM ft H ft Equations excluded^ u ft ft + llfl++l+ll+l++l II ft CM ft ;ions + +II++II++II + +II Condi t II ft ~~CM ft Hrir{r{t^Hr^HHHr\fitir^ttri + + + +IIII++ + +IIII 5" III H Ki ft o ri ii H H ii H H H f<>KNK\KNKNÂ»^r
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116 O B H III + Â• Â— 1 H M t III 1 CVl II B III H ft Q 1^ <; H H + H ai II H EH CO ft /Â—I o H ft EH < 1 II ft ~H ft w o II EH CO OJ Â»Â— * ft O H H r( ft M 1 O O Gu Â•d (D Â•d H o Â« o OS o Â•H pa 11 y Â— ^ ft pH ric~{Hr\r^riHHHriHr^r^r^riri + li+l++l+ll+l++Â» it tions ft ft + +ll++ll++fl + +lÂ» Condi n ft OJ ft + + + +IIII+ + + +IIII Â—J HI H 'd ft o s ii ri H t\ r{ 1^ r<>r^i^r (0 Cd G5 ?S
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117 OJ EH CM H Ik ^ /< o Q H III 1 C\I III OJ Q ft <Â«J Â•ft H + H CvJ 11 H Â—1CO ft" S5 O H ft Ch ^5 1 II w ft u P, M O ti PH CO nj P< O M H P. hi T O o 13 H O X <^r
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o Â© Ih ^<^ ^> ^^ ^^ ^^ ^^ (0 n iH CU r'V:*iTvUJ r^^o c7\o H CM r
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119 +> O CM W CQ Â•H P H Â§ o o i m o S H 8 Pa ""cvj C\J VOMDVÂ£)VsO hr^hl Â— r^r^tO X<.Ht\riHriiirir{T\HriH P^HHHHHHHrHHHHCVJCVlOJCMCMCVlCVJCMCVlOJCMCU I l+t I l+l+l+l l+l+l Â•++Â» + 1 + HHHHr{rir^rir\Hr\Hrir
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120 the theorem. The necessary aiad sufficient conditions obtained in Theorem 11 ajnd listed in Table 23 by vrhich (122) has a solution can be used to obtain the necessary and sufficient conditions by which equations (12^), (126), and (123) have solutions. By the property of symmetry vrlth regard to the subscripts, these conditions are gotten by interchanging the subscripts : 1 and 2 in the results for (122) to obtain those for (12^), 2 and 3 in the results for (12^1) to obtain those for (126), 3 and k in the results for (126) to obtain those for (12??). The above discussion aiid Theorem 11 prove the following Theorems 12 and I3. Theorem 12 . Let D = Pj^pgP^Pli whe.^e pj_ < P2 < P^ < P2. ^^^ distinct odd primes. Let tj_ + uj^V^ tie the fundamental solu tion of (1), t^ Du^ = 1, Inhere = 2\j^ + 1. If ^ y^, (P^iPj) = ^PilPk^ ^ (P^iPj) = +1, (21g) Pi s 3 (mod ^) , and either at least t\io of (p^pj^), (p^ipj), ^Pj^lPj) = 1, (219) and at least one of Pj , r>^, pj = 1 (mod H) ,
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121 or (220) or (221) or (222) or (223) = s 3Pj = 1 (mod H) , (p^lPj) =+1. and (p^ll^) = (PjlPj) 3Pj S Pi^ S 3Pj =1 (mod ^) , (PjIPj) = +1, and (PjIPj^) = (p^Ipj) 3Pj = 3Pk 5 Pj =1 (mod , (Pjlpi^) = +1, end (PjIpj) = (p^iPj). P, = P, = P. 3 3 (mod k) and (Pjlp^) = (PjlPj) = ^Pk'Pj^' then either equation (122), {12k), (126), or (12S) has a solution in Integers x, y according as i =1, 2, 3, or ^ vjhere j, k, t = 1, 2, 3, 4 and J < k < J. Solutions are obtainable by (22^) = (X]_,u^A) and 2xj_y^ = u^. Using the argument of symmetry with regard to the
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122 subscripts In Table 23: the twentyeight cases (1) (9), (11), (13) (1^), (20), (21), (23) (2Â«), (30), and (32) {3k) establish (21g) and (219); the cpses (10) and (12) establish (21g) and (220); the oases (19) and (22) establish (21^?) and (221); the cases (29) and (31) establish (21g) and (222); the cases (35) and (36) establish (218) and (223). Theorem 13 . Let D = p^p^p^p^^ where p^ < p^ < P^ < Pi^ are distinct odd primes. Let tj^ + u^V^ t>e the fundamental solution of (1), t^ Du2 = 1, where t^ = 2Xj_ + 1. If y^, (225) (PiIpj) = (PilPk) = (PilP{) = +1. Pi S 1 (mod 4), and either (226) 3Pj a 3Pij = Pj si (rnod k) and (PjIPj^) = t(Pjpj) Â» i:(pj^ipj ). or (227) 3Pj = Pic 2 3P} si (mod 4) and Â±(PjiPu) = (PjIPjj) =t(PicIP(). or (22g) Pj = 3Pic = 3Pjj si (mod k) and Â±(PjPlc) = +(PjiPjj) = (PijIPj), then either equation (122), (12^+), (126), or (12g) has a solution in Integers x, y according as 1 = 1, 2, 3, or 4where J, k, { =1, 2, 3, 4 and J < k < . Solutions are obtained by (22^1). Using the argument of symmetry with regard to the
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123 subscripts in Table 23: the cases (37) (^O) establish (225) and (226), the cases i^l) (^^) establish (225) and (227), the cases (^5) iW establish (225) and (223). There remains to derive the necessary and sufficient conditions by which each of the equations (129) (13^) has a solution. We will make use of Lemma 10 already established. Theorem 1^ . Let D = p^p^p^Pi^ where p^ < p^ < P^ < P] are distinct odd primes. Let tj_ + u]_VD be the fundamental solution of (1), t^ Du^ = 1, where t^ = 2X^ + 1. If 7^ y^, and in Table 24 the conditions of sjiy one of the cases (3) (g) and (11) (2^^) hold, equation (129) of (6) has a solution in integers x, y which Is given by (229) yx = (X^.u^A) and 2x^y^ = u^. Proof . By hypothesis and Theorem la, one of (120) (134) of (6) has a solution in Integers x, y. By Lemma J, p since yL, equation (120) has a solution. Let 1=1, j=2, k=3, l=4in Lemma 10. Then let (151) and each one of (I52) (I63) hold in turn, so that by Lemma 10, (p^PjilPj) = +1. Then by Lemma 10 and Lemma 7 there are tv;elve sets of necessary conditions by which (129) of (6) has a solution. Consider one set of these conditions, najnely (151) and (153), which are, respectively,
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12k(M H H iH 1 ' ^ 1 11 O rt ft 1 tl (M CM H Â— ' ft 1 CVJ '^ ft H H ai ft H II CO o H ft rt < CVJ ft B H w ft o X W o Eh CO o H Eh M Q O o F c Â• It ft 1Â— 1 >l iH rH H H Â«Â— 1 1 + + 1 1 1 H O Â« tl "^CVJ ft H ft HHHH H HH 1 +H H 1 HH II ft OJ PÂ« HHHHHHHH H + 1 1 + + 1 + 1 + Conditions tl ft H ft _iÂ— 1Â— l_lrIHrIH H + 1 1 + + 1 + 1 + II II ft CM ft HHHHHrIHH H + 1 + 1 + 1 + 1 + II ft ft ^^^^r^rir\r\ H + I + 1 + Â• + Â• III ft H H r^f^K\f^H H 1^ III ft ^ til o CVJ ft Â— ' III H ft H H KnKnH H r<^K\H H CVJ K\=}LTMJ r^Â«o cr\ H H I I o H H xc to CVJ ru I I H H CVJ (\J H H Â»Â— Â» 160 CV) '' Â— ^ H VÂ£) CVJ H ^Â— ' , ^ in OJ O H CQ O "d to Â•p H P! O Â®
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125 0) o u I w CP <: EH (0 o o o tl ft ft ft ft ft ~~CVJ ft _ft H ft _ft OJ ft ft ft til ft III ^ ft xS III o c\J S ft III P. 9 CO 5 HHH HHH H H H III I +1 I I t I o Â— o Â— o Â— oHHH HHH H H H +IH I +!+Â• Â» Â• Â• Â• ll++l+l+l+l+'+' I I+ + 1+1* + 1+ 1+ 1+ I I+I+I+I+I+I+I+' l+l+l+l+l+l+l+l r^H HHH KNt^H H r^r^r^r^H H KnH H ksknH H km^H HHH r^r^ H r^KNH H r^KM^r^H H r^i^H H H r^hf>H H r^KsH H K>r<\H H r< 11 H +1 1 11 JdII ft OJ KN Jdft ft ft ft II ft Cm OJ Â•H ft II P ~~r c. ft OJ Si w ft 4J H Kl ^ ft 0) o Â•
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126 since i 1, J = 2. k = 3, 5 = ^, (230) (pLP2iP3) = (plPjIpi^.) = (P3PliPi) = +1 and (231) P^ s Pj as 3 Pi = 3 ^'^od k) . Relation (230), and (230) imply that (232) (p^lp ) = (pglp,) and (pj_ipi^) = (Pglp^) respectively; while (230)^, which by (231), may be written (.llp^)(p,p^)(Pi^_lp^) = (l)(pLlp3)(pLlPij.) = +1, implies that Hence from (232) ajid (233) it follows that (23'+) iv^lp^) = (P2IP3) = (PilPl) = (P2lPi^.)Â• Thus from (230) (23^), and therefore from (I5I) and (153), there follow two sets of necessary conditions by which (129) has a solution. These are (233) (p^Ip^) = (plIp!). (235) PjL s Pg = P3 = Pi,. 5= 3 (mod k) , (P1IP3) = (P2IP3) = (PilPlj.) = (PglPi}.) = +1. and
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127 (236) (p^lp^) = (P2IP3) = (P^iPl^.) = ^V^\V]^) = +1. Similarly we obtain from (230), (232) and each of (152), and (15^) (163), two sets of necessary conditions by which (129) of (6) has a solution. These in addition to (235) and' (236) produce twentyfour oases wiilch are recorded in the first ten columns of Table 2M. Now, all excludents of Table 11, except \ ^ can be expressed in terms of one or more of the six Legendre symbols (pj^iPg). (p^^lp^)* (p^lPii). (p^lp^), (PglPl)* (p^Ipl}.)* Hence when the conditions of each of the twentyfour cases hold in turn vxe have only the values of (p^iPg) (P^lPi^.) luaas signed. Prom taking each of the tv:entyfour cases in turn we determine those excludents of Table 11 which are consistent with each case and also what value (p^iPj) (p^IpLi.) ^^^t have so that equations (121) (122) and (I30) (13^) iiave no solutions. The values obtained for (p^lp^) ond (p^lpj^) are recorded in Table 2^. For example, by the conditions of (235) v/hich are found in case (3) in Table 2^, the excludents [121) j^, (122)5, (123)2. (12^)2. (125)1^2' ^^2;?)L^2' ^^5Â°^1,2,3,1 that is = 1. Hence by (235) and Lemma 7, equations (121) (125), (123), and (13O) have no solutions. Further, excludents (131)2, (133)2. (13^)l. if (p^lPi^) = +1; while excludents (131)2+. (133)i+, and (Ij^Og ^old if (p^lpi;) = 1.
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12g Thus, since (p^lPi.) must = +1 or 1, equations (I3I), (133) Â» end (13^^) are also excluded by (235) and Lemma 7. Similarly, since excludent (132)2_ holds if (p^lPg) = 1Â»* excludent (132)2 holds if (p3_iP2) = +1; (PilPg) ^^^^ = "^^ equation (132) is also excluded by (235) and Lemma 7. Finally, since excludents (126)^ ^ and (127), p cannot hold, but excludents (126)^ ^^ and (127)^ i hold if (p^lp^^) = +1, we see that equations (121) (12g) and (I30) (13^) are excluded by (235) and the additional assumption that (p^jpi^) = +1. We add this condition to those of (235) and record it in the proper column in Table 2^ for case (3), The fact that (p^lp^) may = +1 or 1 is also recorded. Kence the conditions of case (3) state that equations (121) (12g) and (130) (13^) have no solutions if and only If p^ = P2 = = Pis 3 (mod k) and (PllP3)=(P2lP3)=^PllPif^=^P2i'''i^^"*^Pl'P2^"^P3'%^ " This fact is recorded in Table 2k in the column headed "Equations excluded" by the use of the capital letter "A." In like manner it caji be shovm that when conditions (236) hold and (p^lpi,.) = 1, equations (121) (12S) and (130) (13^) have no solutions. This fact is recorded in Table 2^1 under case {k) , noting that (p, Ipp) may = +1 or 1,
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129 and by placing the letter "A" in the last column. Similarly, when conditions (I5I) and one of (152), (15^) (I63) hold, vilth i = 1, j = 2, k = 3, \ =4, the values of the characters (PllP2^ (PjlPij.) fixed, if necessary, so as to exclude equations (121) (12g) and {130) (13^). These additional conditions are recorded in Table 2^ for the appropriate case. The last column may not always contain the letter "A," since by the conditions in the first ten columns of Table 2^1for a particular case, none of the corresponding excludents of a particular equation in Table 11 may hold. Thus from Table 2kwe see that equations (1?1) (12g) and (130) (13^) have no solutions by the conditions of cases (3) (g) and (11) (2^). Kence by Theorem la, since ^1 ^ ^1' (129) of (6) has a solution. By Theorem 1 a solution is obtainable from i^)^ and {S)^, n = 1 = G, thus establishing (229). This completes the proof. We observe that by the property of symmetry with regard to the subscripts in Table 24, the necessary and sufficient conditions by which each of the equations (I30) (134) has a solution caja be obtained. These conditions are gotten by interchanging the subscripts: 2 and 3 in the results for (129) to obtain those for (I3I), 3 and k in the results for (I3I) to obtain those for (133) Â» 1 and 3 in the results for (I33) to obtain those for (I30), 2 and 3 in the results for (I30) to obtain those for (132),
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130 3 and k In the results for (132) to obtain those for il^h) , The above discussion and Theorem l^lprove the follov:lng summarizing theorem vrherein the letter "S" denotes the set of Legendre symbols, (p^Ipjj), (PjlPic^ ^p^lPj). ip^lv^) , where i, J, k, I = 1, 2, 3, ^, and p^, , p,^, pj are distinct odd primes. For S = +1, we shall mean that each symbol in S has the value +1; similarly for S = 1. Theorem 1^ . Let D = p^p^p^Pj^^ where p^ < p^ < P^ < p^j. are distinct odd primes. Let t^ + U]_V'D be the fundamental solution of (1), t^ Du^ = 1, where t^ = 2X^ .+ L If \ ^ vf* and either s Pj = \ H Pj s 3 (mod k) and (237) (p^IPl^) = (PjlP^^ = (PilPj^ = '^Pj'^}^ " (P^IPj^ or Pj^ = Pj s 1 (njod k) , P^ a 3Pj (ffio^ ^) either (23s) S = +1 and (p, ip.) =
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131 = Pj = 3 i^od k) , p,^ s 3Pj (mod k) and either (239) S = +1 and (Pj^lPj) = 1, or S = 1, or 3Pi J= 3Pj 3 Pj^ 3 Pj s 1 (mod k) , (2il0) (p^IPj^) = ^PjIPk^ = (PllPj^ = (p^lPj) and (Pj^lPj) = or = 3p^ (mod , Pj^ = 3Pj (mod k) , and either (21^1) S = +1 and (p^lPj) = ^Pk'Pj^ " Â°^ S = 1 and one of (p^p^), ^P^JPj) = "^l* then either equation (129), (130), (131), (132), (133), (13^1) has a solution in integers x, y according as 1 = 1, J = 2, k = 3, 5 = ^. or 1 = 3, j = k = 2, 5 = 1, or i = 1, J = 3, k = 2, J = ^, or i = 2, J = ^, k = 3, Â« = 1, or i = 1, J = ^. k = 2, \ = 3, or i = 2, J = 3. k = ^, { = 1. Solutions are given by (229), ^1 ~ ^\>^i^^^ ^Vi ^1*
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132 Using the argument of symmetrj'with regard to the subscripts in Table 2^: cases (3) and {H) establish (237), cases (5), (6), (13) and {ik) establish (23^), cases (7), (g), (15) and (lb) establish (239), cases (11) and (12) establish (2^0), and cases (17) (2^1) establish (2^11). By interchanging the subscripts in each of the Tables 17, 23, and 2kaccording to the results in the theorems associated with each, we obtain the follovring useful Table 25. Each row of the three parts of the table can be substituted for the conditions at the head of the Tables 17, 23, and 2^4respectively to produce the necessary and sufficient conditions by which the equation indicated at the end of each row has a solution. 2 2 ExamDle of equation (I30), p^pj^x P^Pg^ ~ ^Â» where conditions of (23^)2 of Theorem 15 hold. Prom (232)2, i = 3, J = ^, k = 2, I = 1, p^ s p^^ = 1 (mod k) , P2 5 3Pi (mod k) , (p^p2) = (pi^.p2) = (P3IPL) = (piPx) = 1. and (p^lpi,.) = +1. Choosing p^ = 5 sind p^ = 11 to satisfy (238), it follows that (p^lll) = (pill) = (P3I5) = (Pl;l5) = 1. Hence
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133 CO o ^ ] (vf^ LP\ ^ Co '&0 CM CVI CVJ CVJ CM CM ci H H H H H H Â» w w ^ ^ ^ ^ Jd^ JÂ± Pi Pi Pi p< p< p< r'^ c\j CVI r<> CVI CM p4 P, Pi Pif P< p4 CVJ Q ^ <^ P< Pi PÂ« Pi Pi Pi H H H H H H Pi Pi p4 Pi Pi Pi C\J ^ ^ H (0 1^ ai CM ^Â«^ CM ai Pi Pi Pi Pi Pi Pi CO O u ^ H rt H CM H H H El P. Pi Pi Pi Pi Pi >> (D ' ' ' ^ ' ^ 0) W w CO H H Eh ,c . CC !^ EH ^ ^ Ft ^ ^ ^ M =h ^ ^ JÂ± c. rv Q* Ml >H M Qi d. CL Ml >^ Ml o o fa fa CVJ tiM Oh Pi Pi Pi Pi 1 1 1 Eh H rr\ CVJ CM r<> CVJ oj CO M< >H VH o. o. >M VM M* m f\I 4CO Pi Pi Pi A Pi Pi III o CO Ml Hi rt. o. Ml Ml Ml CVJ KN ^ H Pi Pi Pi Pi Pi Pi Q 1 1 1 HEA sf j:Â± Cii p. Pi Pi Pi Pi CQ CU OJ 0) Pi Pi Pi Pi Pi Pi B *H H H rÂ» H H H Pu. Pi Pi Pi Pi Pi Pi CVJ N> ^ Pi Pi Pi Pi Pi Pi CVJ 0) H CS EH t. O fa H o CM H H H H H X Â— Â«. , . , , , , H H H Pi Pi Pi Pi Pi CU CM Pi Pi Pi Pi Pi 1 1 Â— J Â—4 I > Pi Pi Pi Pi Pi H H CU CVI Pi Pi Pi Pi Pi ' Â— ^ Pi Pi Pi Pi Pi jdPi Pi P. Pi Pi ^ ^ ^ ^ H H H Pi Pi Pi Pi Pi H H OJ CV) Pi Pi Pi Pi Pi CU '"cvj Pi Pi Pi Pi Pi Pi Pi P. Pi Pi Pi Pi Pi P. P. H H CM CU P. Pi Pi Pi Pi H H H Pi Pi Pi Pi Ci, OJ CVJ CVI Pi Pi Pi Pi Pi jÂ± Pi Pi Pi Pi Pi H H OJ OJ Pi Pi Pi Pi Pi
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13^ = 1 (mod , Pl = 1 (inocL k) = +2 (mod 5) , and p^^ = +2 (mod 5) , p^ = N (mod 11) Pl = N (mod 11) , where N Is a quadratic nonresidue of 11. Therefore p^ Â» 13 and pj^ = 17 satisfy the above congruences. Nov;, equation (1), t^ 12,155u^ = 1, has the fundamental solution = tj_ = 2X]_ + 1, U]^ = vrhlch was calculated by the writer. Hence = 5 11 '2^ ^ a square. By (229) , = 1, y^ = 2 is a solution of Checking; 221(1) ^ 55(2)^ = 221 220 = 1. The following examples Illustrate the theorems of this chapter. Since the four smallest odd primes are 3, 5, 7, and 11, 3'57ll = 1155. Also, since the illustrations would be similar to those used with regard to Theorem 5, only pertinent date are given. Except where indicated all solutions were calculated by the writer. Example 1 . [Equation (121), p^x pgP^Pj^y = 1] Pi = 3, P2 = 11. P3 = 13, Pl = 37, so that (P1IP2) = (311) = +1, (P1IP3) = (3I13) = +1, (PilPij.) = (3l37) Â« +1, (P2IP3) = (III13) = 1, (P2lPl.) Â» (III37) Â» +1, and (p^lPi^) = (13i37) = 1; D = 3*1113'37 = 15,273, t3_ 2X^+1 = 10,533 and Uq_ = Sk, whence Xj. * H *13 *37 '1 ?^ a square. Hence conditions (202) and (203) of Theorem 10 are satisfied
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135 for 1=1, J=P, k=3, Those are the conditions of case (26) of Table 17. Thus equation (121), _ ll1317y2 = 1^ has a solution. By (205), = ^2, = 1 Is a solution. Checking: 3(^2)^ 5,291(1)^ = 5,292 5,291 = 1. Exam\ile 2 . [Equation (126), p^Pz^k^ " "^y ~ ^^ Pi = 3, P2 = 5, = 11, Pi = ^7, so that (P3IP1) = (I1I3) = +1, (P3IP2) = (1115) = +1. (P3IP4) = (lllii7) = +1, (P1IP2) = Ol5) = 1, (PiiPii) = (31^7) = +1, and (pglpi^) = (51^7) = 1; D = 3511U7 = 7,755, t^ = 2Xj_ + 1 == 1,409 and uj_ = I6, whence = 11 g^ ^ a square. Hence conditions (21?5) and (219)_ of Theorem 12 are satisfied for 1=3, J=l, k=2, j=4. These are the conditions of case (21) of Table 23. Thus equation (126), 3 5 '^7x2 lly^ = 1, has a solution. By (22^1), x^^ = 1, yj_ = S is a solution. Checking: 705(1)^ 11(g)= 705 70^ = 1. Example ^ . [Equation (127), p^^x^ p^p^p^y^ = l] Px = 5, P2 = 29, P3 = ^3, Pi. = 1,559, so that (Pl^IPl) = (1559I5) = +1, (P4IP2) = (1559I29) = +1, (Pi^ip3) = (15591^+3) = +1, (P1IP2) = (5i29) = +1, (P1IP3) = (51^3) = 1, and (pgjp^) = (291^^3) = 1; D = 529^3*l559 Â« 9,720,365, t;L = 2X,_ + 1 = 12,^^71 and uj_ = whence \^ = 5'29J43l2 ^ a
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136 square. Hence conditions (202) and (203) of Theorem 10 are satisfied fori=^, k=2, 5=3. These are the conditions of case (1?) of Table 17. Thus equation (127), 1559x2 5 29 ^37^ =1, has a solution. By (205), = 2, y_ = 1 is a solution. Checking: 1559(2)^ 6235(1)^ = 155,525,2^1 155 ,525 ,Â«'^0 = 1. 2 2 ExaniT)le k . [Equation (13g), pj^ppp^x piy = l] Pi = 3, P2 = 19, P3 = 109, Pi = 15^3, so that (Pi^lPl) = (1553I3) = +1. (Pij.lP2) = (1553I19) = +1, (PiIP3) = (I553I109) = +1, (PiiPg) = (3I19) = 1, (PiiP3) = (3I109) = +1, and (pglp^) = (I9I109) = l; D = 3'191091553 = 9,64c^,739, tL = 2X_ + 1 = 12, '+25 and u^ = k, whence = 1553*22 7^ a square. Hence conditions (225) and (226) of Theorem 13 are satisfied for i= J = 1, k = 2 , } = 3 . These are the conditions of case (39) of Table 23. Thus equation {12f^) , 3* 19* 109x2 1553y^ = 1, has a solution. 3y (22'+), = 1, y,_ = 2 is a solution. Checking; 6213(1) ^ 15^^3(2)2 = 6213 6212 = 1. Example 5 . [Equation {128), p^P2P^x2 p^y2 = l] p^ = 3, P2 = 5, P3 = 11, P4 = ^1, so that (Pl^IPl) = (^ll3) = +1. (Pi^lPj) = ('+li5) = +1, (PjiP3) = (i+lill) = +1, (p^ipg) = (3l5) = 1, (P1IP3) =(3lll) = +1.
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137 and (P2IP3) = (5lll) = +1; D = 3v5lli+l = 6765, = + 1 = 329 and uj^ = ^, whence = kl,2 f square. Henoe conditions (225) and (227)2 Â°^ Theorem 13 are satisfled for 1=4, J=l, k=2, i=3. These are the conditions of case (43) of Table 23. Thus equation (12g), has a solution. By (225), x^^ = 1, = 2 Is a solution. Checking: 165(1)^ 41(2)^ = 165 l64 = 1. Example 6 . [Equation (129), V^^ P^Pi^Y^ = 1] Pi = ^Â» P2 P3 % so that (P1IP3) = (5I17) = 1, (P2IP3) = (I1I17) = 1, (p^iPi^) (5I23) = 1, (P2iP]^) = (I1I23) = 1. (PilPg) = (5lll) = +1, and (p^lpi^) = (17I23) = 1; D = 5111723 = 21,505, = Zh^ + 1 = 7039 and u^ Â» whence X^ = 17*23 '3^ 7^ a square. Hence conditions (2^1)2 of Theorem I5 are satisfied for 1 = 1, j=2, k=3> i=4. These are the conditions of cese (22) of Table 2^. Thus equation (129), 5*llx^ 17*23y^ = 1, has a solution. By (229), xj_*=g, y^ = 3isa solution. Checking: 55(^)^ 391(3)^ = 3520 3519 = 1. Exanple 7 . [Equation (132), PgPi^.^ V^p^Y ~ 1] ?! = 3, P2 = 29, p^ = 61, Pi^ = 101, so that (P2IP3) = (296l) = 1, (pi^lp^) = (101I61) = 1, (pjiPi) =
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I3g (2913) = 1, (Pi^lPi) = (loi3) = 1, (PglPi^) = (29iloi) = 1, and (p^1p3_) = (6lj3) = +1; D = 32961101 = 53d,007, tj^ = 2 2X^ + 1 = 5257 8jnd = 2, whence = 3*6l*45^ a square. Hence conditions (233)2 of Theorem I5 for 1=2, J=4, k=3, j = 1 are satisfied. These are the conditions of case (6) of Table 2^. Thus equation (132), 29101x2 3 6172 = 1, has a solution. By (229), X]_ = 1, y^, = ^ is a solution. Checking; 2929(1)^ 1^3 (^)^ = 2929 2923 = 1. Example g . [Equation (133), VjP^x p^P^y " 1] Pi = 3, Pg = 67, = 6$, pi^ = 7951, so that (P1IP2) =(3l67) = 1, (Pl^lpg) = (7951I67) = 1. (P1IP3) = (339) = 1, (Pi^lP3) = (7951I39) =Â» 1, (p^lPi^) =(3.17951) = 1, and (P2IP3) = (67I39) = 1; D = 367397951 = 1^^2,235,^39, ^1 + 1 = ^7,705 and U3_ = 4, vihence X^ = 67*39 2^ ^ a square. Hence conditions {239)2 Theorem I5 are satisfied for i = 1,J = ^, k = 2, J = 3. These are the conditions of case (3) of Table 2h, Thus equation (133), 37951x^ 6739y2 ^ 1, has a solution. By (229), = 1, yj_ = 2 is a solution. Checking: 23,353(1)^ 5963(2)^ = 23,353 23,352 = 1. Example 9 . [Equation (133), P]_Pij.x P2P3y l3 p^ = 19, Pg = 61, p, = 39, pi^ = 127, so that
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139 (p^lPg) = (19i6l) = +1, (pi^lpg) = (I27!6l) = +1, (P3_p^) = (191^9) = 1, (Pi^lP3) = (I27ig9) = 1, (p^lPi^) = (19I127) = 1, and (pgip^) = (611^9) = 1; D = 19*6lg9l27 = 13,100,177, = 2XL + 1 = i^3,^33 and = 12, whence = 6l'g9'2^ 7^ a square. Hence conditions (2^0) of Theorem I5 are satisfied for 1=1, j=4, k = 2, j=3. These are the conditions of case (11) of Table 2^. Thus equation (133), 19*127x^ 6lg9y^ = 1, has a solution. By (229), 3Â» Y^. ~ 2 is a solution. Checking; 2^+13(3)^ 5^29(2)^ = 21,717 21,7l6 = 1. 2 2 Example 10 . [Equation (133), P]_PlX PgP^y ' l3 P^L = 3, Pg = 59, P^ = 73, Pij. = 57^3, so that (p^iPg) = (3l59) = +1. (Pi^lPj) = (57^3159) = +1, (P1IP3) = (3i73) = +1, (PiIP3) = (57^3173) = +1, (PilPi^) =(3l57^3) = l, and (P2IP3) = (59l73) = 1,* D = 3597357^3 = 7^,205,303, t^ = 2X^ + 1 = 3'+, ^57 and u^ = vrhence = ^9 '73 '2^ ^ a square. Hence conditions (239)^, of Theorem I5 are Sci tisfied for i = l, j = i, k = 2, {=3. These are the conditions of case (15) of Table 2^1. Thus equation (133), 357^3x^ 59 737^ = l, has a solution. By (229), = 1, y^^ = 2 is a solution. Checking; 17,229(1)^ ^^307(2)^ = 17,229 17,225 = 1.
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i4o Example 11 . [Equation (133), Pj_Pi,x^ V2'^3^^ ^ ^1 ^' 5, P^ = 7, Pij. = 13, so that (PllPg) = (3l5) = 1, (P1IP2) = (13l5) = 1. (P1IP3) ^ (3i7) = 1, (pi^lp^) = (I3I7) = 1, (p^lpi^) = (3!l3) = +1, and (P2I.P3) = (5l7) = 1; D = 35713 = 1365, t^ = 2X^ + 1 = 25,271 and = SSk,'^ vihence = 5*7'19^ ?^ a square. Hence conditions (241)2 of Theorem 15 are satisfied for i = 1, j = ^, k = 2, { =3. These are the conditions of case (20) of Table 2k. Thus equation (133), 313x2 5*7y2 = 1, has a solution. Bj^ (229), = 1&, = 19 is a solution. Checking: 39(13)^ 35(19)^ = 12. 636 12,635 = 1. Example 15 . [Equs.tion (13^), PgPjX^ VjV^y^ = 1] Pi = 3, Pg = 7, P3 = 223, = 2051, so that (PglPif) = (720gl) = +1, (p^lpj^) = (223120851) = +1, (P2IP1) = (7l3) = +1, (P3IP1) = (223i3) = +1, (P2IP3) = (7I223) = +1, and (pup3_) = (20Slj3) = 1; D = 3 Â•7Â«223 20.^1 = 9,7^5,323, t^ = 2X^ + 1 = 12,^^27 and u^ = k, whence = 3 2031 1^ ^ a square. Hence conditions (239)j_ of Theorem 15 are satisfied for i=2, j=3, k=^, \ =1. These are the conditions of case (15) of Table 24. Thus equation (13^), "Â•Cayley, op. oit . . XIII, p. k20 .
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1^1 7*223x^ 3*20gly= 1, has a solution. By (229), = 2, = 1 is a solution. Checking: 156l{2)^ 62^13(1)^ = 62^^ 62^13 = 1. Exan^le of equation (I32) , p^p^^x' PiP^y ~ ^> v/here conditions (237) of Theorem I5 hold. Prom Theorem I5, 1 = 2, j=^, k = 3, ojid } = 1. Then (237) reduces to 3 Pi^ = Pj = = 3 (mod k) , and (P2IP3) = (Pi^lP3) = (PglPi) = (Pi^.lPi) = (p^Ip^). Choose IPj^ 3 " 7 satisfy (237). Hence (7IP3) = (Pi^lpj) = 1 = (Pil3) = (P3I3). Therefore, P^ s 3 (mod , P!^ s 3 (mod K) , P3 = 1Â» 2, ^ (mod 7), and Pi s 1 (Â™o^ 3), p^ 2 (mod 3) , Pl4.sK (mod pj ) , where N is a quadratic nonresidue of p^ . Kence p^ = 23 and Pl^ = 43. The primes p^ = 3, P2 = 7, P3 23, and pj^ = satisfy the conditions (237) so that the equation 301x^ 69y^ =1, has a solution, if the value tj^ of the equation (l), 20,7^9". =1, is odd. We check the preceding statement
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Ik2 by the following congruences which show by Theorem la that if none of the remaining equations (120) (I3I), (133), and (13^) has a solution, then it follovrs that (I32) has a solution; in particular, equation (120) has no solution in view of Corollary 2. The vjork to check the preceding statement is displayed in TABLE 26 CONGRUENCES SHOWING INSOLV ABILITY, EQUATIONS (121) (131), (133), AND (13i+) Equation Impossible congruences (121) 3x2 Â„ 7.p3.43y2 =z 1 x2 2 (mod 7) (122) 7 23 ^3x2 3y2 1 y2 2 (mod 23) (123) 7x2 _ 3.p3.l^.3y2 1 x2 10 (mod 23) (12i) 3Â•23Â•i^3x2 7y2 1 y2 1 (mod 3) (125) 23x2 3.7.U3y2 1 x2 2 (mod 3) (126) 3 7 43x2 23y2 s: 1 y2 ? (mod 7) (127) i^3x2 37'23y2 rx 1 x2 3 (mod 23) (125) 3 Â•723x2 ii3y2 cs 1 y2 1 ( mod 3) (129) 37x2 . 23l3y2 1 x2 2 ( mod 23) (130) 23*^3x2 37y2 1 y2 2 (mod ^3) (131) 323x2 7 ^372 1 x2 1 ( mod 7) (133) 3^3x2 7 '2372 1 x2 3 (mod 7) il3^) 723x2 3'l^3y2 1 x2 2 ( mod 3)
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BIBLIOGRAPHY Arndt, P. "Untersuchungen, uber einlge unbestimmte Qlelchirngen zweiten Grades und n^er die Verwandlimg der Quadra twurzel aus elnem Bruche in elnem Kettenbruch, " ArohlY der Mathematlk und Physlk . XIII ( Grief swald, IW), pp. 211 249. Cay ley, Arthur, Collected Mathematical Papers . Vol. XIII, Cambridge: The University Press, 1^97, PP. ^+30 ^b?. Dickson, Leonard Eugene. History of the Theory of Numbers . 3 vols. New York: G. E. Stechert and Co., 193^. . Modern Elementary Theory of Numbers . Chicago: The University of Chicago Press, 1939. , Studies in the Theory of Numbers . Chicago: The University of Chicago Press, 1930. Legendre, AdrienMarie , Theorie des Nombres . e
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BIOGRAPHICAL SKETCH John William Oreiner was bora In Philadelphia, Pennsylvania, on February 26, 19l'+. He attended schools there until 1929, when he enrolled in Mount Herraon Preparatory School for BoySj.Hount Herraon, Massachusetts, froTi which he was graduated. Cum Laude, in June, 1933 Â• After one year of study at Colgate University, Hamilton, I'Jew York, he worked at odd Jobs in New York City for two years and then became associated v^ith the National City Bank of New York imtil May, 19*+2. Prom then imtil September, 19^3 Â» writer was a production man with the Phelps Dodge Copper Refinery, Inc., Queens, L. I., New York. From September, 19^3, until December, 19^5 > he served in the United States Navy as a Physical Training Specialist. In February, 19^6, he returned to Colgate University from which he received tvjo degrees in the Physical Sciences, the A. B. (with honors), January, , and the M. A. (vrith distinction), January, 1950. During his graduate year at Colgate University he was an instructor in the Physical Sciences and in the College of Education. From September, 19^9, to August, 1950, the writer taught Science at the Hudson Falls, N. Y. Public Schools.
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lk3 At present he Is an Assistant Professor in Physical Sciences at the University of Florida, where he has taught since September, 1950. During the academic year I95556, viille on leave of absence, he held a Graduate Fellowship In Mathematics at the University of Florida. He Is a member of the Mathematical Association of America, Alpha Chi Sigma, and the Cum Laude Society, q^f JX
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This dissertation was prepared under the direction of the chairman of the candidate's supervisory committee and has been approved by all members of that committee. It was submitted to the Dean of the College of Arts and Sciences and to the Graduate Council, and was approved as partial fulfillment of the requirements for the degree of Doctor of Philosophy. February 1, 195^ Dean, College of Arts and Sciences SUPERVISORY COMTilTTEE: Chairman
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This dissertation was prepared under the direction of the chairman of the candidate's supervisory committee and has been approved by all members of that committee. It was submitted to the Dean of the College of Arts and Sciences and to the Graduate Council, and was approved as partial fulfillment of the requirements for the degree of Doctor of Philosophy. February 1, 195^ Dean, College of Arts and Sciences SUPERVISORY COMTilTTEE: Chairman

