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 Permanent Link:
 http://ufdc.ufl.edu/AA00029846/00001
Material Information
 Title:
 On Waring's problem with cubic functions
 Creator:
 Yates, Richard Lee, 1931
 Publication Date:
 1957
 Language:
 English
 Physical Description:
 iii, 49 leaves. : ; 28 cm.
Subjects
 Subjects / Keywords:
 Arithmetic progressions ( jstor )
Cubic functions ( jstor ) Cubic polynomials ( jstor ) Integers ( jstor ) Mathematical congruence ( jstor ) Mathematical theorems ( jstor ) Mathematics ( jstor ) Necessary conditions ( jstor ) Numbers ( jstor ) Prime numbers ( jstor ) Dissertations, Academic  Mathematics  UF Functions ( lcsh ) Mathematics thesis Ph. D Partitions (Mathematics) ( lcsh )
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 bibliography ( marcgt )
nonfiction ( marcgt )
Notes
 Thesis:
 Thesis  University of Florida, 1957.
 Bibliography:
 Bibliography: leaf 47.
 General Note:
 Manuscript copy.
 General Note:
 Vita.
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 University of Florida
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 Copyright [name of dissertation author]. Permission granted to the University of Florida to digitize, archive and distribute this item for nonprofit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.
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Full Text 
ON WARING'S PROBLEM WITH
CUBIC FUNCTIONS
By
RICHARD LEE YATES
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
June, 1957
ACKNOWLEDGMENTS
The author wishes to acknowledge his indebtedness to the members of his supervisory committee and in particular to Dr. E. H. Hadlock for suggesting the topic of this dissertation and for his continued interest and assistance in its preparation. The contributions of others, direct or indirect, were appreciated.
ii
TABLE OF CONTENTS
Page
ACKNOWLEDGMENTS. . . .. . . 11
PART
I, INTRODUCTION . . . . . . 1
II. SOME NECESSARY CONDITIONS FOR THE
REPRESENTATION OF ALL INTEGERS . . 3
III. SPECIFIC RESULTS WITH E = 1. . . . 7
IV, A GENERALIZED REPRESENTATION THEOREM 1 V. SOME SPECIFIC RESULTS. . . . . 28 VI. SOME FORMS NOT USABLE IN THEOREM 1 . 43 BIBLIOGRAPHY . . . . . . . . 47
BIOGRAPHICAL SKFeCH, ... . . . . 48
CERTIFICATE OF APPROVAL. . . . 49
il.
PART I
INTRODUCTION
In 1770 E. Waring made the conjecture that every
positive integer is a sum of nine cubes, is a sum of nineteen fourth powers, and is a sum of a limited number of nth powers. The study of the various portions of this conjecture comprises part of what came to be known as Waring's Problem. However Waring's Problem has been generalized to include the representation of all, or almost all, or certain classes of positive integers as sums of a finite number of values of a given function. This paper will be concerned with the representation of positive integers as the sum of a finite number of cubic functions. In particular formulae will be derived which give all such possible functions.
For this purpose we shall need in Part III a lemma1 and a theorem2 of L. E. Dickson.
Lemma: Let t be an integer and consider any set of
integers > 0
(A) f(t), f(t+l), f(t+2), * *.
Make the hypothesis that every integer i for which
h < i < g is a sum of kl or fewer numbers (A),
allowing repetitions. Let f(j+l)f(3) < gh
(j = t, e ", Ml1), where the integer M exceeds t.
Leonard Eugene Dickson, Modern Elementary Theory of Number,(Chioago,) 1939, p. 140..
L_., P. 131.
2
Then every integer which exceeds h+f(t) and is < g+f(M)
is a sum of k or fewer numbers (A).
Theorem; Let F(x) > 0 for all integers x t. Then 24
every integer > 35"2 is a sum of nine values of F(x) for integers x > t provided that D, C, and t
satisfy the eight mild inequalities:
D < 927+36t if 8 < t < o
D 2 908+36t if 16< t < 9
DE 881+36t if 24 < t < 17
3D < 27 if 1 < t < 4
2(6D1) < 216 if D > 0 and t < 0
Dt < 1(224+29) if D < 0 and t < 0
3
C < 61*2 if 1 < t < 4
7D < 5*211 if 1 < t < 4,
The symbol (a,b) will be used to denote the greatest common divisor of a and b, while alb denotes a divides b. The symbol (nip) is that of Jacobi in which p must be prime to 2n in order for the symbol to have meaning. We shall use max(al, a2,' ", an) to signify the maximum of the n elements al, a2,' an, and similarly min(al, a2,' an) to signify the minimum of the n elements al, a2,' an' The function c(m) as used in Part IV is Euler's Sfunction which denotes the number of positive integers not exceeding m which are relatively prime to m. Finally the statement "a is an Si" will mean that the integer a can be represented as a sum of i values of the particular function under discussion.
PART II
SOME NECESSARY CONDITIONS FOR THE REPRESENTATION OF ALL INTEGERS
In order that all positive integers may be represented as a sum of k values of the function
(1) F(x) = E(x3x)+Dx+C the function should have nonnegative integral values for all integers x > t and represent 0 and 1 for two such integers. We wish to investigate the necessary conditions that must be placed on the coefficients E, D, and C for these requirements to be fulfilled.
Assume that E is positive and odd. Let
(2) F(x1) = 0 and F(x2) 1i. Then E(x3x1)+6Dx+6C = 0 and E(x2 3x2)+6Dx2+6C m 6 whence E(x23x13x2+x1)+6D(x2x1) = 6 and
(3) (x2x1) [E(x 22+X 2x+X21)+6D] = 6.
Let x2x1 = d. Then d16 and hence d = +1, +2, +3, or +6. Substituting
(4) x2 = x1+d in (3) we have dE(3x12+3xld+d21)+6dD = 6 whence 3dEx 2+3d2Exl+d3EdE+6dD6 = O. Solving this quadratic in lwe have 3d2E+72dE72d2DE+12d2E23dE2
1 6dE
3
4
whence
(5) 3d2E+Y where
(6) y2 = 72dE72d2DE+12d2E2 3d 4E2
We then have two cases,
a) Assume d odd; then y must be odd and a multiple of 3. Let
(7) Y = 3(2n+1). Then by (6)
(8) D = 72d+1 a2 E2d34E2 2 72d2E
whence substituting (7) we have
(9) D 24.E+4d2E2_.4E2_=.2. 2n24d2E
By (1), (2), and (5) F(x E27d6E327d4E2y9d2Ey2j
6dE 6dE + C =0 whence
(10) C = 64d'4E2D2~16d2DEy+27 d6E' d4E2 +gd2Ey2 3. 08d4E3 +36d2E2y
1296d3E2
By (8) and (10)
(11) 648d3E27216dEyTd4E2 y+2y3
C = 2963 1296d E2
5
By (5) and (11)
(12) 12 2E22d4E2nrd4E2724dEn?1 2dE+8n3+,12n2+6n+l
c =
24d3E2
We then have four subcases.
I. Let d = 1. Then from (9) and (12)
D =8E+E2n24n and 9E
12E 2E2n E2724Enrl2E+gn3+12n2+6n+1.
C =.
24E2
II. Let d = 1. Then
D E2gEnn2an and
  gg 12E2+E2+ 2E2nv24En12ET8n3T712n276nF1
C _21E24E2
III. Let d = 3. Then
D g4E15E n 1 and 72E
324E27162E2nTg 1E2T 2EnT3 6E+n3+12n2+ 6n+1+
648E2
IV. Let d = 3. Then
D = 24E+15E2 +4n+4n+1 and 72E
324E2+162E2n+81E 2 72En? 6EMnL 2n 276nl
648E
b) Assume d even; then by (5) it follows that y must be a multiple of 6. Let
6
(13) y = 6n. Then D = 72dE+ 12E2 4E2 2 2 4S+4d.2E2d 2 2n2
72d2E 24d2E
From (S), (10), and (13) 6d3E2 12dEnd4E2n+4n3 12d3E2
We then have four additional suboases.
V. Let d = 2. Then
D = n2 and
TE
12E2r6En?4E2n+n3
VI. Let d = 2. Then E+n2 and
12E2r6En+4E2nTn3
24E2
VII. Let d = 6. Then
D = 12E96E2n2 and 72E
324E29En?3 24E2n+n3
648E
VIII. Let d = 6. Then
D = l2E+96E2 +n and 72E
324E2 1SEn+324E2nTn3
S=648E
PART III
SPECIFIC RESULTS WITH E = 1
We shall now investigate particularly the cases
of the preceding part with E = 1 using the same numbering for the eight cases.
I. D 2
2
C = 2n3+n2_n C =1C
u 6 u
where Cu and C1 are the values of C using upper and lower signs respectively.
We see that D(n) = D(n1), Cu(n) = Cl(n1), and
Cl(n) = Cu(n1). But if n < 0, then n1 O; hence we do not need to use any n < 0. We see also that D and C will be integers for any integral value of n. Hence by assigning nonnegative integral values to n, we get specific forms which may satisfy our original conditions.
D = 2
2
C  7n 2+2 C = 1C
u 6 1 u
We see that D(n) = D(nl), Cu(n) = Cl(nl), and Cl1(n) = Cu(n1) so that again we do not need to use any n < 0. Also D and C will be integers for any integral value of n. Again we get specific forms by assigning nonnegative
7
integral values to n.
III* D 2 2na2
C 2n3+3n257n+52 C 1
u 162 1 u
Setting 2nn2= O(mod 18) gives n l(mod 3). Hence let n = 3k+l. Then
2
C = 2 k+k2k C = 1C
u 6 C1 lCu
Then D and C will .be integers for any integral value of k. Also we see that D(k) = D(kl), Cu(k) = C1(k1), and Cl(k) = Cu(k1) so that we do not need to use any k < 0. Thus we again get specific forms by assigning nonnegative integral values to k.
IV. D 10++2
Then we must have n2+n+l O(mod 9) whence we have (2n+1)2 =26(mod 9) which is an impossibility. Hence we get no results in this case.
V. D 4112
Cu = ~ 2 1 u
Setting 4n2 O(mod 8), we get nE2(mod 4). Hence let n = 4k+2, Then
9
D = 2k22k
C 8k3+12k2+k C = 1C
u 3 C1 u lC
Then D and C will be integers for any integral value of k. Also we see that D(k) = D(kl), Cu(k) = Cl(k1), and Cl(k) = Cu(k1) so that again we do not need to use any k < 0. Thus we again get specific forms by assigning nonnegative integral values to k.
VI* D =
C = 122 C1 = 1C
Setting 4+n2= O(mod S) gives n =2(mod 4). Hence let n = 4k+2. Then
D = 12k2k2
C = Sk3+12k2+7k
u 3 C1 = 10u
Then D and C will be integers for any integral
value of k. We see that D(k) = D(kl), Cu(k) = C1(kl), and Cl(k) = Cu(k1) so that again we do not need to use any k < 0. Thus we again get specific forms by assigning nonnegative integral values to k.
VII. D = 4+2 72
Then we must have 84+n20 whence n2 6(mod 9) which is clearly impossible. Hence we get no results in this case.
10
VIII. D, 1082 72
Cu = J1 C1 ICu
Setting 10+n2EO(mod 72) gives n6(mod 12). Hence let n = 12k+6. Then
D = 22k2k2
C = 9+11k12k2k3 =
Then D and C will be integers for any integral value of k. Also we see that D(k) = D(kl), Cu(k) = Cl(k1), and Cl(k) = Cu(k1) so that again we do not need to use any k < 0. Thus we again get specific forms by assigning nonnegative integral values to k.
We shall now show that in every case we have D < 1 and that D = 1 only in case I with n = 0. But by (1) the two forms that this gives are I(x3x)+x and \x3x)+x+l
6 6
both of which have been covered by Dickson.1 Thus we shall assume D < O in seeking further forms not covered by Dickson.
2 2
In case I we have D = 1 21 But + > 0. Thus
2 2
D < 1 and D = 1 when n = 0.
In case II D = 1 22A. But 231 > 0. Thus D < 1.
2 2
In case III D = k2+k But > 0 so that D < 0.
2 2
lIbid. p. 141.
In case V we have D = (2k+2k2) and since 2k+2k2 > 0 we see that D < 0.
In case VI D = 1(2k+2k2) and since 2k+2k2 > O, we have D < 1.
Finally in case VIII we have D = 2(2k+2k2). Then since 2k+2k2 > 0 we have D < 2.
Note: Since D < 1, the first five of the eight mild inequalities of the theorem of Dickson quoted in Part I are always satisfied.
Since F'(x) = 2mx2 I +D and F"(x) = x, we see that
2 6
if D # 1 so that 1 > 6D the function F(x) has a maximum at x, / and a minimum at D Also F(x) Xo 3 3 3 is an increasing function for x < x0 and x > x3 and a decreasing function for x0'< x < x3.
If d # 1 and x2 > x1 > x3, then from (4) we have
d > 1. Hence there exists an integer xl < x4 < x2 for which
0 = F(x1) < F(x4) < F(x2) = 1, contrary to F(x) being an integer for all integral values of x. If d < 0 and xi > x2 >x3, then 0 = F(xl) < F(x2) = 1 contrary to F(x) being an increasing function for all x > x3.
Hence since we must have F(x) >_ 0 for every x > min(xl, x2), we must have F(x ) > 0. This gives
S+ 0 whence C and with
(i +C > 0 whence C> and with
27  27
12
D # 1 we have
(14) c2 (16D)3 243
Using the values of D and Cu from III in ( 14) we have 729k4+1134k3531k2+36k+4 < 0 which is impossible for integral k > 0. Using the values of D and C1 from III in
(14) we have 729k4+1728k3+441k21584k968 < 0. But since k I 0 this is possible only for k = 0 which gives the form 1(x3x)+l. But since this form arises in case I with n = 1,
6
case III gives no new usable forms.
Using the values of D and Cu from V in (14) we have 1296k4+1944k3+441k2 +36k+l & 0 which is clearly impossible for nonnegative k. Using the values of D and Cl from V in (14) we have 1296k4+3240k3+2385k2+198k242 < 0 which is satisfied only by k = 0, But this gives the form (x3x)+l which arises in case I. Hence case V gives no new usable forms.
Using the values of D and Cu from VI in (14) we have 1296k4+3240k3+3465k2+1764k+343 < 0 which is clearly impossible. Using the values of D and C1 from VI in (14) we have 1296k4+1944k3+1521k2+630k+100 < 0 which is clearly impossible also. Hence case VI gives no usable forms.
Using the values of D and Cu from VIII in (14)
we have 11664k4+23976k3+14265k2+738k+10 < 0 which is also
13
impossible. Using the values of D and C1 from VIII in
(14) we have 11664k4+22680k3+12321k2+2520k+1225 S 0 which is obviously impossible also. Hence case VIII also gives no usable forms.
Using the values of D and Cu from II in (14) we have 8ln4+4g6n3+1197n2+1764n+1372 < 0 which is clearly impossible. Thus in case II we can get usable results only for D and C1.
Consider now case I where d = 1. Here we have
x3 2~T 6D) and xI = 3+V/172D If C < 0, then since
3 6 6 "
D < 0 we have F(x ) = (6D1)/(l6D +C < 0. Hence we
3 27
must have xl > x This gives 3j+/V172D > 2/3T1 D)T which is obviously impossible for the lower sign. We thus have (using the upper sign) 3+/8l72D > 2/3(16D) which gives D > 11/9, i. e. D = 0 or D = 1. But D = 1 does not arise while D = 0 for n = 1 from which we have the two forms (x3x) and (x3x)+1 both of which have been covered by Dickson.1 But C1 < 0 for all values of n except n = 0 and n = 1 both of which we have already considered. Hence with these exceptions we get no usable forms for C1 in case I.
Thus except for the two forms l(x3x)+l and
6
14
'(x3x)+x+l which already have been considered, we get distinct usable forms only from D and Cu in case I and from D and C1 in case II. For 0 < n < 7 in case I and
0 < n < 6 in case II we duplicate the results of Dickson.1 However for larger values of n we get new usable forms.
For instance, letting n = 8 in case I, we get the form l(x3x)35x+196 for which we shall show that all integers are the sum of nine values.
With x > 2 we have F(x) = 0,1,7,11,21,31,41,62,66, 95,105,127,161,231, etc. Then 04, 716, 1826, 2836, 3846, 4850, 5256, 5960, 6291, and 93121 are S4. By adding 1, we get 5,17,27,37,47,51,57,61, and 92 as S5. Hence 05, 757, and 59121 are S5. By adding 41, 103132 and 134162 are S5. By adding 62, 124153 and 155183 are S5. By adding 95, we get 154, 157186, and 188216 as S5. By adding 105, 167196 and 198226 are S5. By adding 127, 18921g and 220248 are S5. By adding 161, 223252 and 254282 are S5. Also 253 = 231+21+1 is an S Hence with the exception of 6 and 58, 0282 are S5.
Apply the lemma of Dickson quoted in the introductory part with h = 59, g = 282, and k = 6. The conditions hold if
Ibid.
15
(15) (2M1)2 < 1+8(ghD). The value of the right side is 2065 whence (15) holds if M = 23. Hence 59 through 292+F(23) are S6. But 6 and 58 are Sg also. Thus six summands suffice from 0 through gl = 282+F(23) = 1697.
Apply the lemma again with h = 0, g = gl = 1697, and k = 7. The value of the right side of (15) is 13,857 whence (15) holds if M = 59. Hence seven summands suffice to g2 = 1697+F(59) = 34,148.
Apply the lemma again with h = 0, g = g2 = 34,148, and k = 8. The value of the right side of (15) is 272,665 whence (15) holds if M = 261. Hence eight summands suffice to g3 = 34,14g+F(261) = 2,988,329.
Apply the lemma again with h = 0, g = g3 = 2,988,329, and k = 9. The value of the right side of (15) is 23,906,913 whence (15) holds if M = 2400. Hence nine summands suffice to g4 = 2,988,329+F(2400) = 2,306,904,125.
The remaining three mild inequalities stated in the theorem of Dickson quoted in the introduction are satisfied trivially for this form. Hence by the theorem every integer : 35"224 = 587,202,560 is the sum of nine values of F(x). Hence every integer > 0 is the sum of nine values of F(x) = 1(x3x)35x+196 for integers x > t where 16 < t < 2.
6
Similarly with n = 7 in case II we get the form 1(x3x)29x+148. With x > 4 we have F(x) = 0,1,7,9,23,
6
42,49,86,135, etc. Then 04, 712, 1428, 3034, 3670, 7276, and 7886 are S4. By adding 1, we see that 5,13, 29,35, and 71 are S5. Adding 23, 5993 are S5. Adding 49, we have 85119 and 121125 as S5. Adding 86, we have 116120, 122156, and 164172 as S5. Adding 135, 149163 and 171205 are S5.' Hence with the exception of 6, 0205 are S5.
Apply the lemma of Dickson quoted in the introductory part with h = 7, g = 205, and k = 6. The conditions of the lemma hold if (15) holds. The value of the right side of (15) is 1817 whence (15) holds if M = 21. Hence 7 through 205+F(21) are S6. But 6 is also an S6 so that six summands suffice to gl = 205+F(21) = 1284.
Apply the lemma again with h = O, g = gl = 1284, and k = 7. The value of the right side of (15) is 10,505 whence (15) holds if M = 51. Hence seven summands suffice to g2 = 1284+F(51) = 22,053.
Apply the lemma again with h = O, g = g2 = 22,053, and k = 8. The value of the right side of (15) is 176,657 whence (15) holds if M = 210. Hence eight summands suffice to g3 = 22,053+F(210) = 1,559,576.
17
Apply the lemma again with h = 0, g = g3 = 1,559,576, and k = 9. The value of the right side of (15) is 12,476,841 whence (15) holds if M = 1766. Hence nine summands suffice to g4 = 1,559,576+F(1766) = 919,462,065.
Again we have the mild inequalities trivially satisfied so that by the theorem of Dickson stated in the introduction every integer > 35*224 = 587,202,560 is the sum of nine values of F(x). Hence every integer > 0 is the sum of nine values of F(x) = 1(x3x)29x+14 for integers x > t
6
where 15 < t < 4.
Similarly with larger values of n we may get additional forms for which it can be shown that all integers are a sum of nine values of any one form.
PART IV
A GENERALIZED REPRESENTATION THEOREM
We shall now consider the representation of certain sufficiently large integers as sums of eight values of the function f(x) = Dx+ 4x3x) with x > 0 and with E > 0 and
6
D > 0. Let
(16) G = A+B+C and H = A3+B3+C3 with A, B, and C arbitrary positive integers and let s be any positive integer.
With every ternary quadratic form there is associated a set of progressionsI given by 2k(gm+a), piki(pim+bi) where the pi (i = 1,2,' ,n) are odd prime divisors of the determinant and (pi,bi) = 1. Then it is possible to find a least
n
k < nilPi such that the set of progressions km+r with
i=l 3
(j = 1,2,* *,t) contains the progressions associated with the form.
We first define a condition relating to a regular
ternary form Ax2+By2+Cw2 and the function f(,x) = Dx+E(x3x),
6
Condition V: Given the regular ternary form
Ax2+By2+Cw2 whose excluded progressions are contained in the progressions km+rj for j = 1,2,* *,t and (k,3) = 1
1B
Burton W. Jones "A New Definition of Genus forTernary Quadratic Forms, Amergoan Mathematical Society Transactions, XXXIII (1931), p. 99.
IS:
19
and not contained in any set of progressions with a smaller value of k. This is equivalent to saying that the excluded progressions are contained in the number classes which are = rl,r2,* *,rt(mod k) with (k,3) = 1 where k is the smallest possible value for the modulus. Then it is possible to prove that, for any z and the proper choices of d, r, and a prime p, that M2$ rl,r2,' ,rt(mod k) where
(17) p3EM = s2Dz E(z3+3t2zz)2DGp3 ~p9+p3.
By examination of the 102 regular ternary forms,
we see that excluding those for which k would not be prime to 3 gives a k dividing 1600 in every case. Hence the distinct prime factors of k can be only 2 and 5.
Lemma 1: With k as defined in Condition V and (E,3k) = 1, there exist primes p with the properties:
(18) (p,3kE) = 1
(19) (p,E6D) = 1
(20) (31p) = 1
(21) (3E(E6D)Ip) = 1
(22) pad(mod k) with d=2 or 3(mod 5) if EED(mod 5)
and 51k, otherwise d arbitrary.
Proof.Property (20) will be satisfied if we take p 2(mod 3). Putting this with (22) we have by the Chinese Remainder Theorem
20
(23) ph(mod 3k) where h2(mod 3) and h Ed(mod k).
The product of (20) and (21) gives (E(6DE)Ip) = 1. Let N = E(6DE). Then we wish to show that we can always choose a prime ph(mod 3k) such that (Nip) = 1. N cannot be a square since N = K2 with K an integer gives 6ED = K2+E2 whence E2+K20O(mod 3) and E2=K2(mod 3). But this is possible only if EKEO(mod 3) contrary to (E,3) = 1. Hence if N = SN' where S is the largest square contained in N, we see that N' # 1 while (Njp) = (N'jp).
Let N' = plP2.Pn where the pi (i = 1,2,'* ,n)
are distinct primes. Then since (E,3) = 1, we have (N,3) = 1 and (N',3) = 1. Hence Pi # 3 for i = 1,2,* *,n. Since (E,k) = 1 we have similarly that pi # 2 if k is even. Hence (N',3k) = 1 or 5. We thus have two cases to consider,
a) If pi 3 5 (1 = 1,* *,n), we write (N' p) = (plN')(1) 2 2 = (1) 2 2 (pp l)(p p2)..(pjpn)" Thus we can choose a p which is a quadratic nonresidue of Pi and a quadratic residue of pi for I = 2,3,*" ,n if either pSl(mod 4) or N'=_ l(mod 4); while if p N'_ 3(mod 4) we can choose a p which is a quadratic residue of all the Pie This gives us p~j(mod N') and putting this with (23) we have by the Chinese Remainder Theorem p :q(mod 3kN')
21
where qJ(mod N') and q=h(mod 3k). But this congruence is satisfied by an infinite number of primes,
b) Next suppose pl = 5. When N'0(mod 5) then
N = E(6DE)O 0(mod 5) and either E.O(mod 5) or E'D(mod 5). First if N' = pl = 5, then (N'lp) = (51P) = (p15). If 51k, then EfO(mod 5) since (E,k) = 1. Thus if 51k, then by
(22) we have p= 2 or 3(mod 5) and (N' p) = (p15) = 1 with pSh(mod 3k). If 5Xk, then (N',k) = 1 so that if we choose pE2 or 3(mod 5) we can put this with (23) and have by the Chinese Remainder Theorem pq(mod 15k) where qh(mod 3k) and q=2 or 3(mod 5).
If N' = 5p2P3'Pn', then (N' p) = (1) 2 2 (pIN') = (1) 2 2 (p 5)(P1p2).**(Ppn). Then if either pl1 (mod 4) or N'E_ 1(mod 4) we choose p a quadratic residue of p for i = 2,3,* ,n; while if p=N'=3(mod 4) we choose p a quadratic nonresidue of p2 and a quadratic residue of pi for i 3,4,. *,n. This gives pi=g (mod N'/5). Thus if 51k, then by (22) we have p=2 or 3 (mod 5) and (p15) = 1 so that we have (N'jp) = 1 for p=q(mod 3N'k/5) where qh(mod 3k) and q g(mod N'/5). Similarly if 51k, then we choose p=2 or 3(mod 5) and have pq(mod 3N'k) where q Sh(mod 3k), q=2 or 3(mod 5), and q g(mod N'/5). But this congruence is satisfied by an
22
infinite number of primes.
Thus (20), (21), and (22) are satisfied simultaneously by an infinite number of primes chosen as previously described. But since 3kE and E6D are constants, any prime p > max(3kE, E6D) must be prime to both. Thus all five conditions are satisfied simultaneously by an infinite number of primes.
Lemm 2: Let p be a prime having the properties of Lemma 1, s a given integer prime to p, d as in Lemma 1, and r an arbitrary integer. Then there exist positive integers t and z such that
(24) 3Et2 E6D(mod p)
(25) t= r(mod k)
(26) s 2Dz+_(z3+3t2zz)(mod p3)
3
(27) 0 < t < kp; kp < z < p3+kp.
Proof.Since (p,3E) = 1, there exists an integer K such that 3EK l(mod p). Hence (3EKlp) = 1. Then from
(21) we have 1 = (3EK p)(3E(E6D)jp) = (K(E6D) lp). Hence there exists an integer tI such that t l2 K(E6D)(mod p) whence 3Etl2 3EK(E6D)EE6D(mod p) and (24) is satisfied by t1. Then by the Chinese Remainder Theorem (since (k,p) = 1) there exists an integer t2 with 0 < t2 < kp such that tl t2(mod p) and t2Hr(mod k). Hence congruences (24),
(25), and (26)1 are satisfied simultaneously by t = t2.
23
Next if (z,p) = 1, then 2DZ+=(z3+3t2zz) is also prime to p since (24) holds. Suppose that also (u,p) = 1 and 2Dz+~( z3+t2zz) 2Du+~(u3+3t2uu)(mod p3) with
3 3
(28) kp S z < p3+kp; kp < u < p3+kp. Then (z3u3)E+(zu) (3Et2+6DE)_ O(mod p3) whence (zu)[E(z2+zu+u2)+3Et2(E6D)] EO(mod p3). If the second factor is a multiple of p, then by (18) and (24) we have
z2+zu+u20(mod p) whence (2z+u)2= 3u2(mod p). But this is impossible since (u,p) = 1 and (31P) = 1. Therefore zEu(mod p3) whence zu = Mp3 and (28)1 plus the negative of (28)2 gives kp(p3+kp) < zu < p3+kpkp or p3 < zu < p3. But this says that M = 0 and hence that z = u. We have thus shown that the f(p3) integers
(29) 2Dz+E(z3+3t2zz) with kp < z < p3+kp are prime to p and further that no two of them are congruent to each other modulo p3. Hence every integer s which is prime to p is congruent modulo p3 to one of the integers
(29). Therefore congruence (26) has a solution which satisfies (27)2.
Lemma 3: When s is sufficiently large there exists at least one prime p having the properties (18)(22) of Lemma 1 which does not divide s and which satisfies:
(30) Pp9 s < ~i
(31) +5Ep9.2DGp EHp9+EGp3 < Ep9
(32) 9 3Ep9 E(p3+kp)3(p3+kp)(2D+Ek2p2)_2DGp3+kp Hp9
+Ep3 > 0
3
where 0 and H are given by (16).
Proo.Assume that there exists a number class of primes p:d(mod k) with (d,k) = 1 which satisfy conditions
(18)(22) of Lemma 1. It is known that the number of primes ps d(mod k) with (d,k) = 1 which satisfy the inequalities x < p < x+ax increases indefinitely with x.1 Let
x = [ 5]9 and a = r2H 9 1. Then for s sufficiently large there are at least ten primes satisfying the conditions (18)(22) of Lemma 1 which satisfy
(33) E6s 9 [ 6 9
E(2H+5) < p E (2H+3
Their product is greater than E6=S 9 6s 1 > a E(2H+5) E2H+ 95
for s sufficiently large. Hence for s sufficiently large atleast one of the ten primes does not divide s for if all ten distinct primes divide s, then their product must divide
1Edmund Landau, "Uber eine Anwendung der Primzahltheorie auf das Waringsche Problem in der elementaren Zahlentheorie," Mathematisohe Annalen, LXVI (1909), pp. 103104.
25
s contrary to their product being greater than a. Hence if a > C1 for some integer C1, there exists at least one prime which does not divide s and which satisfies (33) and the conditions (1S)(22) of Lemma 1. But (33) is equivalent to (30).
Again (31) is equivalent to Ep9 > 2p3G(E6D) which is satisfied by any p sufficiently large. But we can increase p by increasing s. Hence if a > C2 for some integer C2, then all primes satisfy (31).
Similarly (32) is equivalent to
Ep9 > 6Ekp7+12Ek2p5+2p3E(4k3G)+6D(G+1) +2kp(6DE) which is satisfied by any p sufficiently large. But since we can increase p by increasing s, if s > C3 for some Integer C then all primes satisfy (32).
Hence for s > max(C1, C2, C3) there is at least one prime which satisfies the conditions of Lemma 3.
Theorem 1: Let s be a given integer and Ax2+By2+Cw2 a regular ternary form satisfying Condition V. If one of the sets of values of z and p determined in Lemmas 1, 2, and 3 satisfies the congruence
(34) s 2Dz+2DGp3(mod E), then s can be represented as the sum of eight values of f(x) = Dx+t(x3x) with x > 0.
Proft.We first determine p by means of Lemmas 1
26
and 3. By Lemma 2 there exist integers t and z which satisfy congruences (25) and (26). Thus we have s 2Dz+!(z3+3 t2zz)(mod p3) whence
(35) a = 2Dz+(z3+3t2zz)+p3M. Let M1 be determined by
(36) M = 2DG+ p +M' 3 3
Then we have
(7) a =2Dz+(z+3t2zz)+2DGp+Rp G3 +p3m1
Since (E,p) = 1 and by (34) and (37) we see that MI50 (mod E). Let
(3j) MI = EM2*
From inequalities (27) and E > 0 we obtain E > L cp,
3 3
2Dz > 2D(p3+kp), E3 > gp3+kp)3, and finally
Et2z > E(kp)2(p3+kp) whence
(39) 2Dz (z+3t2zz) > (p3+kp)3(p3+kp)(2D+Ek2p2)+kp But also since E > 0, D > 0, z > 0, and z3z > O, we have
(40) 2Dz E(z3+3t2zz) < 0.
3
Hence from (30), (39), and (40) we obtain
(41) Z3Ep9 Pp (p3+kp)3_(p3+kp)(2D+Ek2p2)+,kp < s2Dz
9(z3+3t2zz) < tiEp9 whence by (37) and (41) we have
(42) ? Ep9 (p3+kp)3(p +kp)( 2D+Ek2p2)+kp2DGp3 Hp9
27
+Op3 < p3M1 2' 5gp92DGp3 Hp9+Op3.
Then by inequalities (31), (32), and (42), when p is sufficiently large, we have 0 < p3M1 < Ep9 whence
(43) 0 < M2 P6
Since Condition V holds, there exist integers X, Y, and Z such that
(44) N2 = AX2+BY2+CZ2.
We thus obtain from (35), (36), (38), and (44) s = 2Dz+ (z3+3t2zz)+p3M
= f(z+t)+f(zt)+p3(2DG+Hp6 )+p3M1
3 3
= f(z+t)+f(zt)+p3(2DG+E p6_ G)+p3E(AX2+BY2+CZ2)
= f(z+t)+f(zt)+f(Ap3+X)+f(Ap3X)+f(Bp3+Y)+f(Bp3Y)
+f(Cp3+Z)+f(Cp3Z).
From (43) and (44) it follows that 0 < AX2+BY2+CZ2 < p and hence that IXI, IYI, and IZI are each less than p3. From (27) it follows that 0 < t < z. Hence z+t, zt, Ap3+X, Ap3X, Bp3+Y, Bp3Y, Cp3+Z, and Cp3Z are all positive and thus that s is the sum of eight values of the cubic function f(x) = DxZ(x3x) with x > 0.
PART V
SOME SPECIFIC RESULTS
Theorem 2: Let s be a sufficiently large given
integer. If one of the sets of values of z and p determined in Lemmas 1, 2, and 3 satisfies the congruence s =2Dz+26Dp3(mod E), then s can be represented as the sum of eight values of f(x) = Dx+,(x3x) with (E,30) = 1 and x > 0.
We shall show that by using the regular ternary form x2+2y2+10w2 Condition V is satisfied. The excluded progressions are all of the form 25k(25n+5) and 8n+7 and are thus (with k = 40) included in the progressions 40n, 40n+5, 40n+7, 40n+lo, 40n+15, 40n+20, 40rr23, k4n+25, 40n+30, 40n+31, 40n+35, and 40n+39. Hence we need to show r21,5,7,10,15,203,3,25,,31,35,39(mod 40). This we do by showing M2t 7(mod 8) and by (38) that MI = EM2 O(mod 5) and by using the fact that E is prime to 5. Table 2 shows Mlt O(mod 5) while Table 3 shows that M2t 7(mod 8). The choices of d and r are given in Table 1. (To save space Table 1 has been written in two separate parts module 5 and 8. For example rl16(mod 40) is given as rl(mod 5) in Table la and r_0(mod 8) in Table lb.)
As an example of the use of Table la, consider the case D_=El and s O(mod 5). Table la with (22) and (25)
28
29
then tells us to take pd=2(mod 5) and t=r=l(mod 5). Also with the form x2+2y2+10w2 we have G = 13 and H = 1009. By (17) of Condition V we have p3M1 = pEM2. s2Dz E(z3+3t2zz)2DGp3 p9+9Gp3(mod 5).
1 2 3 3 3 Multiplying through by 6 we get p3M1E s+3Dz+3E(z3+3t2zz)+3DGp3+3EHp9+2EGp3(mod 5). Replacing D, E, s, p, t, G, and H by their above values, we have 3M1 3z3+4z+4(mod 5) whence MI1 z3+3z+3(mod 5). Then taking zEO,1,2,3,4(mod 5) we have Ml=2,3,4(mod 5) so that Ml O(mod 5). Similarly Table lb is used to show that we have M2t 7(mod 8).
Table la: (Note that d 2 or 3(mod 5) whenever we have ED(mod 5).) The values of E, s, r, and d are given modulo 5.
D l(mod 5)
1 1,2 2,3 1,2 2,2 2,2 2 0,2 2,2 2,1 2,1 2,3 3 i,i 1,3 0,3 0,1 1, 4 0,1 II ii1 1,2 1,4
30
D2(mod 5)
rd
E 0 1 2 3 1 4 1 1,1 0,1 1,3 0,3 0,3 2 1,2 1,3 2,3 2,2 1,2 3 0,1 1,2 1,1 1,4 1,1 4 0,2 2,1 2,2 2,3 2,2 DE 3(mod 5)
rd
ti 0 1 2 3a1 4 1 0,2 2,1 2,3 2,2 2,1 2 0,1 1,1 1,4 1,1 1,2 3 1,2 1,2 2,2 2,3 1,3
4 1,1 O,3 1,1 1,3 0,1 D 4(mod 5)
r,d
1 0,1 1,4 1,2 1,2 1, 2 1,1 1, 0,1 0,4 1,3 3 0,2 2,3 2,1 2,1 2,2 4 1,2 2,2 1,3 1,2 2,3
31
D 0(mod 5)
1 1,1 1,2 0,2 0,1 1,1 2 1,1 0,1 1,2 1,1 0,2 3 1,1 0,2 i,I 1,2 0,1 4 i,i 1i,1 0,1 0,2 1,2
Table 1b: Suppose s is even, that is s 2n. Then
(17) becomes after multiplying by 9
(45) pEM2= aZDz3E(z3+3t2 )2Dp4Ep(mod 8). Take p1 and tEO(mod 8). Then (45) becomes M2= 2n2Dz3Ez3+3Ez2D4E = 2(nDzD2E)3E( z3z)(mod 8) and since z3z is even we have M2$ 7(mod 9). The case with a odd is covered by the following table and Table 3. The values of E, s, r, and d are given modulo 8. DE1 or 5(mod 8)
1 1,1 1,1 1,3 1,1 3 1,1 1,1i 1,3 1,1 5 1,1 1,1 1,3 1,1 7 1,i 1,1 1,3 1,1
32
D=O, 2, 4, or 6(mod ~)
r, 1
E 3 L 5 .r) 7
5 1,1 1,3 1,1 1,3 7 1,1 1,3 1,1 1,3 D 3 or 7(mod 8)
E 5 7
3 1,3 i,i i,i i,i 3 1,3 1,1 1,1 1,1 5 1,3 1,1 1,1 1,1 7 1,3 1,1 1,1 1,1
Table Q: The values of s, E, and M1 are given modulo 5.
DE 1(mod 5)
0 1 2 4
1 2,3,4 1,5,4 1,2,3 1,2,4 1,3,4 2 1,2,3 1,2,4 1,2,3 2,3,4 1,2,4 3 2,3,4 1,2,4 1,2,3 1,3,4 1,2,3 4 1,3,4 1,2,3 2,3,4 1,3,4 1,2,3
33
DE 2(mod 5)
0 2 3
1 1,3,4 1,2,3 2,3,4 1,3,4 1,2,4 2 1,3,4 1,2,4 1,2,3 1,2,3 1,2,4 3 1,2,3 1,2,3 1,2,4 1,2,4 1,3,4 4 1,2,4 1,3,4 2,3,4 2,3,4 1,2,3 DE 3(mod 5)
0o . 2 3 4
1 1,3,4 1,3,4 1,2,3 1,2,3 1,2,4 2 2,3,4 1,2,4 1,3,4 1,3,4 2,3,4 3 1,2,4 1,3,4 2,3,4 2,3,4 1,3,4 4 1,2,4 1,3,4 1,3,4 1,2,3 2,3,4 D 4(mod 5)
0 1 2 4
1 1,2,4 2,3,4 1,204 1,3,4 2,3,4 2 1,2,3 2,3,4 1,24 12,4 1,3,4 3 2,3,4 1,3,4 1,2,3 2,3,4 1,3,4 4 1,2,3 1 4 2,3,4 2,3,4 1,2,4
34
DE (mod 5)
0 o 3 1 4
1 2,3,4 1,3,4 2,3,4 1,3,4 1,2,3 2 1,3,4 1,2,3 1,2,3 1,2,4 1,3,4 3 1,2,4 12,4 1,3,4 2,,4 2,3,4 4 1,2,3 2,3,44 1,2,3 1,2,4
Table 3: The values of E, M2, and s are given modulo 8.
DE1 or 5(mod 8)
11  7 .
1 0,2,3,4,6 0,2,4,5,6 0,1,2,4,6 0,1,2,4,6 3 0,2,3,4,6 0,2,4,5,6 0,1,2,4,6 0,1,2,4,6 5 0,2,3,4,6 0,2,4,5,6 0,1,2,4,6 0,1,2,4,6 7 0,2,3,4,6 0,2,4,5,6 0,1,2,4,6 0,1,2,4,6 D=o 2, 4, or 6(mod 8)
1 0,1,2,4,5,6 0,1,,4,5,6 0,1,2,4,5,6 0,1,2,4,5,6
7 0,1,2,4,5,6 0,1,2,4,5,6 0,1,2,4,5,6 0,1,2,4,5,6
35
D =3 or 7(mod 8)
1 0,1,2,4,6 0,1,2,4,6 0,2,3,4,6 0,2,4,5,6 7 0,1,2,4,6 0,1,2,4,6 0,2,3,4,6 0,2,4,5,6 5 ,,,6 ,1,,,6 ,2,4,6 0,256 7 0,1,2,4,o,4 ,6 0,2,3,4,6 0,2,4,5,6 Theorem 2 then follows by an application of Theorem 1.
Theorem *: Let s be a sufficiently large given integer satisfying one of the conditions:
i) s is odd
ii) D is odd and s4 or 12(mod 16)
iii) D is odd, s0(mod 16), and DEE+2(mod 4). If one of the sets of values of z and p determined in Lemmas 1, 2, and 3 satisfies the congruence sB2Dz+8Dp3 (mod E), then s can be represented as the sum of eight values of the function f(x) = Dr+(x3x) with (E,6) = 1 and x > 0.
We shall show that by using the regular ternary form x2+y2+2w2 Condition V is satisfied. The excluded progressions are all of the form 4k(16n+14). Hence we need to show M2 0,4,8,12,14(mod 16). Since p must be large, we can assume p # 2. Then we have p3EM2 = a2Dz E(z3+3t2zz)gDp3 Ep9+4Ep3 whence
2 3 3 3
36
EM2 = ps2Dpz+5pE(z3+3t2zz)+gD+2Ep24E(mod 16) and since E is odd we obtain
M2= E3ps2DE3pz+5p(z3+3t2zz)+8DE3+2p24(mod 16).
Take D= l(mod 16). Then we have
M2 = E3ps2E3pz+5p(z3+3t2zz)+8E3+2p24(mod 16). Then take
El, sl, tO, and pal(mod 16). Then we have M2. 5z3+9z+7 l1,3,5,7,9,11,13(mod 16).
Increasing a by an even integer 2k will add 2pk to
M2 which will still give a set of seven odd values and hence will not contain any of the excluded values. Thus any odd s is satisfactory for EDEl(mod 16).
Since E is odd, any increase must be a multiple of 2, say 2n. But this adds 6E2psn+12En2ps+8n3ps12E2npz
24En2pz16n3pz+48E2n+96En2+64n3 = 2pn( 3E2s+6Ens+4n2s6E2z12Enz8n2z )+16n(3E2+6En+4n2) to the value of M2. Since this value is even, it will again give a set of seven odd values which are thus distinct from the excluded values. Thus with DEl(mod 16) for any odd E and odd s we choose tO and pEl(mod 16) and have M2 0,4,8,12,14(mod 16).
But if we increase D by one we add 8E32E3pz
= 2E3(4pz) to M2. But this is even and hence will again give a set of seven odd values. Thus for any D, any odd E, and any odd s we choose tSO and p=1(mod 16) and have
37
M20 0,4,8,12,14(mod 16).
Before considering even s = 2n, we shall make three pertinent remarks.
1) For any D, E, and s = 2n we have with t o(mod 16) M2 2K+5pz3+llzp(mod 16) and with t2(mod 16) M2= 2K+5pz3+7pz(mod 16) where K = E3pnE3Dpz+4E3D+p24. In either case for any odd p we have M2O= 0,2,4,6,,10,12,14 (mod 16). Thus we can never show M2fO0,4,S,12,14(mod 16) with tO or 2(mod 16) and s even.
2) With any D if we increase E by 2k (which we must do since E is always odd), we have added J = 2pk(3E2s+6Eks+4k2s6E2Dz12EkDzgk2Dz)+16kD(3E2+6Ek+4k2) to the previous value of M2. Since s = 2n, we have j = 4pk(3E2n+6Ekn+4k2n3E2Dz6EkDz4k2Dz)+16kD(3E2+6Ek+4k2). If k = 4, we have JE O(mod 16). Hence increasing E by 8 gives duplicate results.
3) Increasing D by an even integer 2h adds
16E3h4E3pzh to the previous M2. But this addition is
 4E3pzh(mod 16).
Take D2 and EEl(mod 16) and s = 2n. Let
a = 2p2+2pn+12. Then with tEl(mod 16) we have M2= 5pz3+6pz +a(mod 16) and with tE3(mod 16) we have M2 5pz3+14pz+a (mod 16). And in either case we have M2E aa+p,a+3p,a+4p, a+5p,a+7p,a+8p,a+9p,a+llp,a+12p,a+13p,a+15p(mod 16) and
since a is even we always get either M2 0,1,3,4,5,7,8,9, 11,12,13,15(mod 16) or M2= 1,2,3,5,6,7,9,10,11,13,14,15 (mod 16). Hence at least one of the excluded values always arises. Thus in this case there are no values of p and t such that we obtain M20,4,8,12,14(mod 16). With DE2 and E3, 5, or 7(mod 16) we duplicate these results By remark 2) we see that the results will also be duplicated for E9, 11, 13, or 15(mod 16).
But by remark 3) we see that increasing D by 2h
adds an integer = 4E3pzh(mod 16) to the previous M2. This does not affect the even values we are getting. Consequently if D and s are both even, there are no values of p and t such that M2;O,4,,12,14(mod 16).
a) Consider DEEl(mod 16) and s = 2n. Let
b = 2p2+2pn+4. Then we have for te l(mod 16) M2 E5pz3+8pz+b (mod 16) and for t=3(mod 16) M2E5pz3+b(mod 16) which gives in either case M2=b,b+p,b+3p,b+5p,b+7p,b+8p,b+9p, b+llp,b+13p,b+15p(mod 16). And with nE2 or 6 and pEl (mod 16) we have M25 1,2,3,5,7,9,10,11,13,15(mod 16). Thus with s4 or 12(mod 16) we take tEpl(mod 16).
b) Consider DEl and E)=3(mod 16) and s = 2n. Let
c = 2p2+6pn+4. Then for tal(mod 16) we have M2= 5pz3+4pz+c (mod 16) and for tE3(nod 16) we have M2= 5pz3+12pz+c (mod 16). In either case we obtain M2 c,c+p,c+3p,c+5p,
39
c+7p,o+gp,c+lp,c+ll 13p,+15P(mod 16). Then with n5O(mod 16) we get M2S=1,3,5,6,7,9,11,13,15(mod 16) with p=l(mod 16). With n=2 and p1=l(mod 16) we get M2E1,2,3,5,7,9,11,13,15 (mod 16). With n=6 and pil(mod 16) we get M 2=1,3,5,7,9,10,11,13,15(mod 16). Hence with s~0, 4, or 12 (mod 16) we take tp1l(mod 16).
c) Consider DE1 and EE5(mod 16) and s = 2n. Let d = 2p2+10pn+4. Then for t 1(mod 16) we have M2= 5pz3+d (mod 16) and for t3(mod 16) we have M2 5pz3+8pz+d(mod 16). But since d = b+8pn, this duplicates the results of case a). Hence with s=4 or 12(mod 16) we take t=p=l(mod 16).
d) Consider DS1 and E=7(mod 16) and s = 2n. Let e = 2p2+14pn+4. Then tE1(mod 16) gives M2=5pz3+12pz+e (mod 16) while tE3(mod 16) gives MN2 5pz3+4pz+e(mod 16). In either case we obtain M2= ee+p,e+3p,e+5p,e+7p,e+9p, e+llp,e+13p,e+15p(mod 16). Then with nEO and pEl(mod 16) we have M2 31,3,5,6,7,9,11,13,15(mod 16). With n=2 and p~l(mod 16) we have M2E1,2,3,5,7,9,11,13,15(mod 16). With n=6 and pEl(mod 16) we have M2=1,3,5,7,9,10,11,13,15 (mod 16). Hence with sO, 4, or 12(mod 16) we take tEp=l(mod 16). Then by remark 2) we have corresponding results for DE1 and EE9, 11, 13, or 15(mod 16).
e) Consider D3 and E.l(mod 16) and s = 2n. Let f = 2p2+2pn+4. Then tl(mod 16) gives M2E5pz3+4pz+f
(mod 16) while t3(mod 16) gives M2= 5pz3+12pz+f(mod 16). In either case we obtain M2= f,f+p,f+3p,f+5p,f+7p,f+9p, f+llp,f+13p,f+15p(mod 16). With nEO and pEl(mod 16) M2E 1,3,5,6,7,9,11,13,15(mod 16). With n=2 and pil (mod 16) M2= 1,3,5,7,9,10,11,13,15(mod 16). With nE6 and pBl(mod 16) we have M2 l1,2,3,5,7,9,11,13,15(mod 16). Thus with s=O, 4, or 12(mod 16) we choose tpE1l(mod 16).
f) Consider DEE 3(mod 16) and s = 2n. Let
g = 2p2+6pn+4. With t~l(mod 16) we have M2E5pz3+gpz+g (mod 16) and with tE3(mod 16) we have M2. 5pz3+g(mcd 16). In either case we have M2Eg, g+p,g+3p,g+5p,g+7p,g+p,g+9p, g+llp,g+13p,g+15p(mod 16). With n 2 or 6 and p= 1(mod 16) we obtain M22 1,2,3,5,7,9,1C,11,13,15(mod 16). Hence with
s 4 or 12(mod 16) we choose t=p=l1(mod 16).
g) Consider D= 3 and EE 5(mod 16) and s = 2n. Let h = 2p2+lOpn+4. With tEl(mod 16) we have M2= 5pz3+12pz+h (mod 16) and with t =3(mod 16) M2 5pz3+4pz+h(mod 16). In either case we have M2= h,h+p,h+3p,h+5p,h+7p,h+9p,h+llp, h+13p,h+15p(mod 16). With nBO and pl(mod 16) we have M2 1,3,5,6,7,9,11,13,15(mod 16). With nE2 and p 1 (mod 16) we have M2=1l,3,5,7,9,10,11,13,15(mod 16). With nE6 and p_1l(mod 16) we have M2=1 ,2,3,5,7,9,11,13,15 (mod 16). Thus with s=O, 4, or 12(mod 16) we choose
tSpl(mod 16).
41
h) Consider D=3 and E=7(mod 16) and s = 2n. Let I = 2p2+14pn+4. Then t=l(mod 16) gives M2= 5pz3+i(mod 16) while tS3(mod 16) gives M2= 5pz3+gpz+i(mod 16). But since I = g+8pn, this duplicates case f).
By remark 2) we have corresponding results for EG9, 11, 13, or 15(mod 16) and DE3(mod 16).
i) Since in each of the preceding cases we have
M2 5pz3+2(Vpz+W)(mod 16), excluded even values can arise only for even z. Setting z = 2z' in remark 3), we see that an integral amount =8E3pz'h(mod 16) is added to M2 when D is increased by an integer 2h. With h = 2 this added integral amount is O(mod 16). Hence increasing D by 4 gives duplicate results.
Thus for even a we have M20O,4,,12,14(mod 16) for D odd and either:
a) s 4 or 12(mod 16) or
b) si0(mod 16) and DEE+2(mod 4).
In either case we choose tpEl(mod 16). Theorem 3 then follows by means of Theorem 1.
Similarly, using other regular ternary forms, we may obtain other representation theorems by making use of Theorem 1.
By a similar proof based on the regular ternary
form x2+2y2+5w2 all of whose excluded progressions are of
the form 5m, but without the general Theorem 1, L. K. Hua1 proved a representation theorem stated as follows:
Theorem: Let s be a given integer. If one of the sets of values of z and p determined in Lemmas 1,
2, and 3 satisfies the congruence s2Dz+16Dp3
(mod E), then s can be represented as the sum of
eight values of f(x) = Dx+~x3x) where E is prime
to 15.
However we can show that this and Theorem 2, although they may overlap for some sets of s, D, and E, do not duplicate each other in every case. For example, Hua's theorem will not hold if s2Dz16Dp34 O(mod E) for all pairs of p and z determined in Lemmas 1, 2, and 3. But the two conditions s2Dz16Dp3 O(mod E) and s2Dz26Dp3 0 (mod E) can hold simultaneously if 10ODp3A O(mod E) which is always the case if E and D are relatively prime since (E,30) = 1 and (E,p) = 1. Hence there can exist sets of s, D, and E for which Hua's theorem will not hold while Theorem 2 does hold.
By similar arguments we can show that each of the Theorems 1 and 2 and Hua's theorem covers cases excluded by the other two.
1Lookeng Hua, "On Waring Theorems with Cubic Polynomial Summands," Mathematische Annalen, CXI (1935), p. 624.
PART VI
SOME FORMS NOT USABLE IN THEOREM 1
In contrast to the theorems of the preceding two
parts it is not possible to prove a representation theorem from Theorem 1 using certain of the 102 regular ternary forms. Specifically we have the following three theorems.
Theorem 4: No regular ternary form which has the progressions 4k(9n+7) in its set of excluded progressions can be used to prove a representation theorem by means of Theorem 1.
First we show that a theorem cannot be proved using the regular ternary form x2+y2+w2. For to do so we would have to show M2i 0,4,7(mod 8). With x2+y2+w2 we have p3EM2 = s2Dz z3+3t2zz)6Dp3Ep9+Ep3 whence assuming p # 2 (which we can do since p must be a large prime) and thus that p is odd, we have EM2 ps+6pDz+5Ep(z3+3t2zz)+2D (mod 8) from which by assuming E odd (which it must be since (E,k) = 1 and k = 8), we have
(46) M2 Eps+6EpDz+5p(z3+3t2zz)+2ED(mod 8).
Taking D=EEl(mod 8) and sSO(mod 8) we have
M2 5pz3+pz+7pt2z+2(mod 8). For distinct results we must have t=O, 1, or 2(mod 8). With t=O(mod 8) we have M2~ 5pz3+pz+2E= 2,2p+2,4p+2,6p+2(mod 8) and for any choice
43
of p we have M2 0,2,4,6(mod 9). With tl1(mod 9)Ewe have M2 5pz3+2 2,p+2,3p+2,5p+2,7p+2(mod 8) and for any choice of p we get M2S1,2,3,5,7(mod 8). With t2(mod 8) we have M2 5pz3+5pz+2 2,2p2,4p+2,p+2,6p+(mod ) and for any choice of p we have M2n 0,2,4,6(mod 8). Hence this choice of D, E, and s fails.
But from (46) increasing s by an integer n gives a new M = M2+Enp and we see that for any value of t the addition of a multiple of p still gives one of the sets of values containing excluded number classes. Hence we are unsuccessful for EDEl(mod 8) and any value of s.
But increasing E by an even integer 2n (since E
must remain odd) gives a new M2 w M2+2nps+12npz+4n = M2+2m where m = nps+6npz+2n. But adding an even integer to M2 leaves the sets of four values with at least one excluded value unchanged. Hence we are unsuccessful for DSl(mod 9) and any s and odd E.
But by (46) increasing D by an integer n gives a new M' = M2+6Enpz+2En = M2+2m where m = 3Enpz+En. Thus again we get the same sets of four values including at least one excluded value. Hence we fail for any combination of D, s, and odd E and cannot use the form x2+y2+w2 to prove a representation theorem by means of Theorem 1.
Next let Ax2+By2+Cw2 be any regular ternary form
45
which does not represent 4k(Sn+7). In order to prove a representation theorem using this form we must be able to prove M2 0,4,7(mod 8) in addition to any other values arising from an excluded progression.
We have (again letting G = A+B+C and H = A3+B3+C3) p3EM2 = az)z (z3+32zz)2DGp3 Hp94Gp3. Then by assuming p # 2 and E odd, we have
(47) M2 = Eps+6DEpz+5p(z3 +3t2zz)+6DEG+5H+3G(mod 8).
But we showed in connection with the form x2+y2+2 that (with M" signifying the M2 of the form x2+y2+w2) M"= Eps+6DEpz+5p(z3+3t2zz)+2ED(mod 8) gives in all cases 2
either M2=0,2, 4,6 or MELa,l,3,5,7(mod 8) where a is even, But from (47) we have (with g = 5H+3G) M2N M2ED(3G1)+g (mod 8) whence M2E; (0,2,4,6)+g or M2= (8,1,3,5,7)+g(mod 8) where 8 = a+2ED(3G1) is even. Hence whether g is even or odd we get a set of values including at least one excluded value and hence we cannot prove a theorem using this form. This theorem excludes 20 of the 102 regular ternary forms.
Theorem 5, No regular ternary form which has the progressions 4k(Sn+3) in its set of excluded progressions can be used to prove a representation theorem by means of Theorem 1.
In order to prove a theorem using such a form, we
would have to be able to show M29 0,3,4(mod 8). The remain
der of the proof is identical with that above since we get the same sets of four values including at least one of the excluded values in every case. This theorem excludes six additional forms.
Similarly we can prove:
heoglre : No regular ternary form which has either 4k(8n+5) or 4k(gn+l) in its set of excluded progressions can be used to prove a representation theorem by means of Theorem 1.
These exclude ten and three additional regular ternary forms respectively.
These three theorems together with the exclusion of those regular ternary forms for which k would not be prime to 3 leave exactly fifteen regular ternary forms for possible use in proving specific representation theorems from the general Theorem 1.
BIBLIOGRAPHY
Dickson, Leonard Eugene. Modern Elementar Theory of Numbers.
Chicago: The University of Chicago Press, 1939.
"A New Method for Waring Theorems with Polynomial
Summands". AmeriaA Mathematical SocietY Transactions,
xxxvI (1934). pp. 731748.
"Waring's Problem for Cubic Functions". Aerui an
Mathematical Soolety Transactionsa, XXXVI (1934. pp.
112.
Hua, Lookeng. "On Waring Theorems with Cubic Polynomial
Summands". Mathematische Annale, CXI (1935), pp. 622628.
James, R. D. "The Representation of Integers as Sums of
Values of Cubic Polynomials". Ameroan Journal of
Math tlos, LVI (1934). PP. 303315.
"The Representation of Integers as Sums of Values
of Cubic Polynomials. II". American Journal of Mateatica, LIX (1937). PP. 393398.
Jones, Burton W. "A New Definition of Genus for Ternary
(Quadratio Forms". Amoeioan Mathematical Society Transactins, XXXIII (1931). pp. 92110.
Landau, Edmund. "Uber eine Anwendung der Primzahltheorie
auf das Waringsohe Problem in der elementaren Zahlentheorie". Mathematisohe Aipalen, LXVI (1909). pp, 102105.
Webber, Cuthbert G. "Waring's Problem for Cubic Functions".
Amergan Mathematical Soiety Transactions, XXXVI
(1934). pp. 493510.
47
BIOGRAPHICAL SKETCH
The author was born in Red Oak, Iowa, on June 15, 1931. He moved to Florida in 1944 and graduated from Palm Beach High School in 1949. In June, 1952, he received the Bachelor of Science in Education, g lR Q jAe, from Florida Southern College where he was valedictorian. In June, 1954, he received the Master of Arts in Mathematics from the University of Florida where he held a teaching assistantship for two years. He taught one year at Belle Glade High School before returning to the University of Florida in June, 1955. He again held a teaching assistantship for one year and has been the holder of a fellowship from The Southern Fellowships Fund during the past year. He is a member of Kappa Delta Pi and the Mathematical Association of America.
48
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.
June, 1957
Dean, College of Arts and Sciences
Dean, Graduate School
SUPERVISORY COMMITTEE:
Chairman
49

Full Text 
29
then tells us to take p E d = 2(mod 5) and t = r= 1(mod 5)
Also with the form x2+2y2+10w2 we have G = 13 and H = 1009.
By (17) of Condition V we have
= P3EM2= s2Dz y(z3+3t2zz)2DGp3 tp9+Gp3(mod 5)*
Multiplying through by 6 we get
p3tf = s+3Dz+3E(z3+3t2zz)+3DGp3+3EHp9+2EGp3(mod 5). Re
placing D, E, s, p, t, G, and H by their above values, we
have 3M1= 3z3+4z+4(mod 5) whence Mj_= z3+3z+3(mod 5) Then
taking z = 0,l,2,3,4(mod 5) we have 2,3,4(mod 5) so that
M^j O(mod 5) Similarly Table lb is used to show that we
have S),
Table la: (Note that d= 2 or 3(mo
have EED(mod 5).) The values of E, s, r, and d are given
modulo 5
D = 1(mod 5)
0
1
2
3
4
1
1,2
2,3
1,2
2,2
2,2
2
0,2
2,2
2,1
2.1
2,3
3
1,1
1,3
0,3
0,1
1,1
4
0,1
1,1
1,1
1,2
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INGEST IEID EGXYEP59O_PUKJOO INGEST_TIME 20150325T20:19:06Z PACKAGE AA00029846_00001
AGREEMENT_INFO ACCOUNT UF PROJECT UFDC
FILES
31
Table lb: Suppose s Is even, that is s = 2n. Then
(17) becomes after multiplying by 9
(45) pEKpE s2Dz3E(z5+3t2z2)2Dp4Ep(mod g),
Take pEl and t = 0(mod g), Then (45) becomes
M2= 2n2Dz3Ez^+3Ez2D4E = 2(nDzD2E)3E(z^z)(mod g)
and since z^z is even we have g). The case with
s odd is covered by the following table and Table 3 The
values of E, s, r, and d are given modulo g,
DE 1 or 5(mod g)
E^O
1
1
. 7
1
1,1
1,1
1,3
1,1
3
1,1
1,1
1,3
1,1
5
1,1
1.1
1,3
1,1
7
1,1
1,1
1,3
1,1
46
der of the proof Is Identical with that above since we get
the same sets of four values including at least one of the
excluded values in every case. This theorem excludes six
additional forms.
Similarly we can prove;
Theorem 6: No regular ternary form which has either
4k(Sn+5) or 4k(gn+l) in its set of excluded progressions
can be used to prove a representation theorem by means of
Theorem 1.
These exclude ten and three additional regular ter
nary forms respectively.
These three theorems together with the exclusion of
those regular ternary forms for which k would not be prime
to 3 leave exactly fifteen regular ternary forms for possible
use in proving specific representation theorems from the
general Theorem 1,
13
impossible. Using the values of D and from VIII in
(14) we have Il664k^+226g0k3+12321k2+2520k+1225 < 0 which
is obviously impossible also. Hence case VIII also gives
no usable forms.
Using the values of D and from II in (l4) we
have gln^+4g6n^+1197n2+1764n+1372 < 0 which is clearly
Impossible. Thus in case II we can get usable results
only for D and C^.
Consider now case I where d = 1. Here we have
X, = V.3.l.Ol. and X, = 3V^172D. if c < 0, then since
3 3 6
D < 0 we have F(x^) = ( +c < 0. Hence we
must have x^ > x^. This gives 3+V8172D > 2V316D)
which is obviously impossible for the lower sign. We thus
have (using the upper sign) 3+ V372D > which
gives D > 11/9, i. e. D = 0 or D = 1, But D = 1 does
not arise while D = 0 for n = 1 from which we have the two
forms g(x^x) and ^Kx^x)+l both of which have been covered
by Dickson.1 But C1 < 0 for all values of n except n = 0
and n = 1 both of which we have already considered. Hence
with these exceptions we get no usable forms for in
Thus except for the two forms 4(x^x)+l and
6
case I
36
EM^E ps2Dpz+5pE(z^+3t2zz)+8EH2Ep24E(mod l6) and since
E is odd we obtain
M^= E^ps2DE^pz+5p(z^+3t2zz)+5DE^+2p24(mod l6).
Take DH l(raod l6) Then we have
Mg= E^ps2E^pz+5p( z^+3t2zz)+SE^+2p24(rood l6). Then take
ESI, sil, ti 0, and p = 1( mod l6). Then we have
m2= 5z3+9z+7= l,3,5,7,9,ll,l3(mod 16).
Increasing s by an even Integer 2k will add 2pk to
which will still give a set of seven odd values and hence
will not contain any of the excluded values. Thus any odd
s is satisfactory for E=DEl(mod l6).
Since E is odd, any Increase must be a multiple of
2, say 2n. But this adds 6E2psn+12En^ps+2n^ps12E^npz
24En"pzl6n^pz+4<3E2n+96En2+64n^
= 2pn(3E2s+6Ens+4n2s6E2z12EnzSn2z)+l6n(3E2+6En+4n2) to
the value of Mg. Since this value is even, it will again
give a set of seven odd values which are thus distinct from
the excluded values. Thus with DIE 1( mod l6) for any odd E
and odd s we choose t=0 and pil(mod 16) and have
M2/ 0,4,3,12,l4(mod l6) .
But if we Increase D by one we add BE^2E^pz
= 2E^(4pz) to Mg. But this is even and hence will again
give a set of seven odd values. Thus for any D, any odd
E, and any odd s we choose t = 0 and pEl(mod l6) and have
30
D = 2(mod 5)
0
1
2
3
4
1
1.1
0,1
1.3
0,3
o,3
2
1.2
1.3
2,3
2,2
1.2
3
0,1
1.2
1.1
1.*
1.1
4
0,2
2,1
2,2
2,3
2,2
D 3(mod 5)
\s
0
. 1
2
3
4
1
0,2
2,1
2,3
2,2
2,1
2
0,1
1.1
M
1.1
1,2
3
1.2
1.2
2,2
2,3
1,3
4
1,1
o,3
1,1
1.3
0,1
D E 4(mod 5)
0
1
2
3
4
1
0,1
iA
1,2
1,2
1,1
2
1,1
1,1
0,1
0,4
1,3
3
0,2
2,3
2,1
2,1
2,2
4
1,2
2,2
1,3
1,2
2,3
BIBLIOGRAPHY
Dickson, Leonard Eugene. Modern Elementar:/ Theory of Numbers.
Chicago: The University of Chicago Press, 1939
. "A New Method for Waring Theorems with Polynomial
Summands ". American Mathematical Society Transactions,
xxxvi (193V). pp. 731^s.
. "Waring1s Problem for Cubic Functions". American
Mathematical. Society Transactions. XXXVI (193^). PP.
112.
Hua, Lookeng, "On Waring Theorems with Cubic Polynomial
Summands". Hathematlsche Annalen. CXI (1935). PP# 622
62g.
James, R. D. "The Representation of Integers as Sums of
Values of Cubic Polynomials". American Journal of
Mathematics. LVI (193*0. PP. 303315.
. "The Representation of Integers as Sums of Values
of Cubic Polynomials. II". American Journal of Kathe
g&SJLfig, UX (1937). PP. 3933937
Jones, Burton W. A New Definition of Genus for Ternary
Quadratio Forms". American Mathematical Society Trans
actions XXXIII (1931). PP. 92110.
Landau, Edmund. "Uber eine Anwendung der Primzahltheorie
auf das Waringsche Problem In der elementaren Zahlen
theorie". Mathematlsche Armale^. LXVI (1909). pp. 102
105.
Webber, Cuthbert G. "Waring's Problem for Cubic Functions".
Mericafl MatftamtjcaA s
(193*0. pp. 793510.
**7
4
whence
(O 3d2Ey where
X1 Sd
(6) y2 = 72dE72d2DE+12d2E23d4E2.
We then have two cases.
a) Assume d odd; then y must be odd and a multiple
of 3 Let
(7) y = 3(2n+D.
Then by (6)
(g) D = 72dE+l?d2E2^5d4E2y2
72d2E
whence substituting (7) we have
(9) D 24dE+4d2E2d4E212n2l?n3
24d2E
By (1), (2), and (5) w, )
1 bl *16* V
* $&) c o
(10) c = 64gd4E2DT2l6d2DEv+27d6E^T27d4E2y+qd2Ey2:ry3.logd4E3
136d2E2y^
1296d3E2
By (g) and (10)
(11)
64gd^E2:T2l6dEyT'd4E2y+i2y^
1296d^E2
C =
PART I
INTRODUCTION
In 1770 2. Waring made the conjecture that every
positive integer is a sum of nine cubes, is a sum of nine
teen fourth powers, and is a sum of a limited number of n
powers. The study of the various portions of this conjecture
comprises part of what came to be known as Waring's Problem,
However Waring*s Problem has been generalized to Include the
representation of all, or almost all, or certain classes of
positive integers as sums of a finite number of values of a
given function. This paper will be concerned with the repre
sentation of positive Integers as the sum of a finite number
of cubic functions. In particular formulae will be derived
which give all such possible functions.
For this purpose we shall need in Part III a lemma"*"
p
and a theorem of L. E. Dickson.
Lemma: Let t be an integer and consider any set of
Integers > 0
(A) f(t), f(t+l), f(t+2), .
Make the hypothesis that every Integer 1 for which
h < 1 < g is a sura of k1 or fewer numbers (A),
allowing repetitions. Let f(j+l)f(j) < gh
(J = t, *, Ml), where the integer M exceeds t.
"^Leonard Eugene Dickson, Modern Elementary Theory
of Numbers.(Chicago.) 1939, p. l4o.
5lbX4., p. 131.
1
19
and not contained In any set of progressions with a smaller
value of k. This is equivalent to saying that the excluded
progressions are contained in the number classes which are
= r^Tg,* *,rt(mod k) with (kf3) = 1 where k is the small
est possible value for the modulus. Then it is possible to
prove that, fpr any z and the proper choices of d, r, and a
prime p, that M2^rl,r2'* *,rt(mod k) where
By examination of the 102 regular ternary forms,
we see that excluding those for which k would not be prime
to 3 gives a k dividing l600 in every case. Hence the dis
tinct prime factors of k can be only 2 and 5
Lemma 1: With k as defined in Condition V and
(E,3k) = 1, there exist primes p with the properties:
(18)
(19)
(20)
(21)
(p,3kE) = 1
(p,EoD) = 1
(3 Ip) i
(3E(e6d)p) = i
(22) p = d(mod k) with d = 2 or 3(mod 5) if E=D(mod 5)
and 5k, otherwise d arbitrary.
Proof.Property (20) will be satisfied if we take
p = 2(mod 3). Putting this with (22) we have by the Chinese
Remainder Theorem
PART III
SPECIFIC RESULTS WITH E = 1
We shall now Investigate particularly the cases
of the preceding part with E = 1 using the same numbering
for the eight cases.
I. D = 2nn2
2
Cu cj. iou
where C^ and C^ are the values of C using upper and lower
signs respectively.
We see that D(n) = D(nl), cu(n) = C^(nl), and
C^n) = Cu(nl). But If n < 0, then n1 > 0; hence we do
not need to use any n < 0. We see also that D and C will be
Integers for any integral value of n. Hence by assigning
nonnegative Integral values to n, we get specific forms
which may satisfy our original conditions.
11 d = Statn2.
2
cu = aaffi*3 cx = icu
We see that D(n) = D(nl), Cu(n) = C.^n1), and
C^(n) = Cu(nl) so that again we do not need to use any
n < 0. Also D and C will be integers for any integral value
of n. Again we get specific forms by assigning nonnegative
7
5
By (5) and (11)
(12) 12d^E2=F2dlE2n:P'dl4'E2TP4dEnTl2dE+gn^+12n2+6n+l
C = *
24d^E2
We then have four subcases.
I. Let d = 1. Then from (9) and (12)
D = gE+E24n24nl and
gE
12E2=T2E2n:PE2:F24En:ri2E+gn5+12n2+ 6n+l.
c = __^= = =
24e
II. Let d = 1. Then
D = E2gE4n24nl and
SE
12E2+E2+ 2E2nT24En;fl2ETgn^T12n2:r6nTl #
III.Let d = 3. Then
n 24E15E24n24nl and
v 72E
C =
3 24E2Tl62E2nT,glE2T7 PEn^ 6E+gn^jl 2n2+6n+l .
64gE2
IV. Let d = 3 Then
>2, ,.2
tv 24E+15E+4n+4n+l and
U ~ 72E
C
3 24E2+l62E2n+glE2;T72En:r3 6ETgn3:TL Pn2^^!
64gEc
b) Assume d even* then by (5) It follows that y
must be a multiple of 6. Let
14
^(x^x)+x+l which already have been considered, we get
6
distinct usable forms only from D and In case I and
from D and In case II. For 0 < n < 7 I*1 case I and
0 < n < 6 In case II we duplicate the results of Dickson,^
However for larger values of n we get new usable forms.
For instance, letting n = 8 in case I, we get the
form x^x)35x+196 for which we shall show that all ln
6
tegers are the sum of nine values.
With x > 2 we have F(x) = 0,1,7,11,21,31,41,62,66,
95,105,127,161,231, etc. Then 04, 716, 1826, 2836,
3846, 4850, 5256, 5960, 6291, and 93121 are S^. By
adding 1, we get 5,17,27,37,47,51,57,61, and 92 as S^.
Hence 05, 757, and 59121 are S^. By adding 4l, 103132
and 134162 are S^. By adding 62, 124153 and I55I83 are
S5. By adding 95, we get 154, 157I86, and 188216 as S^.
By adding 105, 167196 and 198226 are S^. By adding 127,
189218 and 220248 are S^. By adding l6l, 223252 and
254282 are S^. Also 253 = 231+21+1 is an S^. Hence with
the exception of 6 and 58, 0282 are S^.
Apply the lemma of Dickson quoted in the introduc
tory part with h = 59, g = 282, and k = 6. The conditions
hold if
'Ibid.
9
D = 2k 2k
SJ+JL5iL
C1 =
~u 3
Then D and C will be integers for any Integral value
of k. Also we see that D(k) = D(kl), Cu(k) = C^(kl),
and C^ik) = Cu(kl) so that again we do not need to use
any k < 0. Thus we again get specific forms by assigning
nonnegative integral values to k.
VI. D = 4+
Ia
C = 12:
u
IF
C1 Wu
Setting 4+n2 = 0(mod g) gives n r 2(mod 4). Hence
let n = 4k+2. Then
D = l2k2k2
C = aUk.2+ZÂ£ r T r
u 3 C1 !cu
Then D and C will be Integers for any integral
value of k. We see that D(k) = D(kl), Cu(k) = C^(kl),
and C^ik) = Cu(kl) so that again we do not need to use
any k < 0. Thus we again get specific forms by assigning
nonnegative Integral values to k.
vil. D = S4ai
72
Then we must have g4+n2=0 whence n2^ 6(nod 9) which
is clearly impossible. Hence we get no results in this case.
15
(15) (2M1)2 < l+8(ghD).
The value of the right side is 2065 whence (15) holds if
M = 23. Hence 59 through 282+F(23) are S^. But 6 and 58
are Sg also. Thus six summands suffice from 0 through
gx = 282+F(23) = 1697.
Apply the lemma again with h = 0, g = g^ = 1697
and k = 7 The value of the right side of (15) is 13,857
whence (15) holds if M = 59. Hence seven summands suffice
to g2 = i697+f(59) = 3M^.
Apply the lemma again with h = 0, g = g2 = 3^,1^
and k = 8. The value of the right side of (15) is 272,665
whence (15) holds if M = 261. Hence eight summands suffice
to g^ = 34,l4g+F(26l) = 2,988,329.
Apply the lemma again with h = 0, g = g^ = 2,988,329,
and k = 9* The value of the right side of (15) is 23,906,913
whence (15) holds if M = 2400. Hence nine summands suffice
to g^ = 2,988,329+F(2^00) = 2,306,904,125.
The remaining three mild inequalities stated in the
theorem of Dickson quoted in the introduction are satisfied
trivially for this form. Hence by the theorem every inte
ger > 35*2 = 587,202,560 is the sum of nine values of
F(x). Hence every integer > 0 is the sum of nine values of
F(x) = i(x^x)35x+196 for integers x > t where l6 < t < 2.
6
12
D / 1 we have
(1*0 C2 >
 243
Using the values of D and Cu from III in ( l4) we
have 729k^+113^k^531k2+36k+4 < 0 which is impossible for
integral k > 0. Using the values of D and from III in
(l4) we have 729k^+172Â£k3+44lk215S4k96g <0. But since
k > 0 this is possible only for k = 0 which gives the form
(x^x)+l. But since this form arises in case I with n = 1,
6
case III gives no new usable forms.
Using the values of D and from V in (14) we
have 1296k^+1944k^+44lk2+36k+l < 0 which is clearly im
possible for nonnegative k. Using the values of D and
from V in (l4) we have 1296k^+324ok5+23g5k2+196k242 < 0
which is satisfied only by k = 0. But this gives the form
^(x^x)+l which arises in case I. Hence case V gives no
new usable forms.
Using the values of D and from VI in (l4) we
have 1296^+3240k^+3465k2+1764k+343 < 0 which is clearly
Impossible. Using the values of D and from VI in (l4)
we have 1296k^+1944k^+1521k2+630k+100 < 0 which is clearly
impossible also. Hence case VI gives no usable forms.
Using the values of D and Cu from VIII in (l4)
we have ll664k^+23976k^+l4265k2+73gk+10 < 0 which is also
32
D = 0, 2, 4, or 6(mod
4
2)
1
3
5
7 .
1
1,1
1,3
1.1
1,3
3
l.l
1.3
1.1
1.3
5
i.l
1.3
1,1
1.3
7
l.l
1,3
1,1
1.3
D= 3 or 7(od &)
1
3 .
5
7
1
1.3
1,1
1,1
1.1
3
1.3
1,1
1.1
1,1
5
1.3
1.1
1,1
1,1
7
1.3
1.1
1.1
1,1
'Table 21 The values of s, E, and are given
modulo 5.
D= 1(mod 5)
5]
 o
1
2
.3
4
i
2.3,4
1.3,*
1,2,3
1,2,4
1.3,4
2
1,2,3
1,2,4
1,2,3
2,3,4
1,2,4
3
2,3,4
1,2,4
1,2,3
1,3,4
1,2,3
4
1,3,4
1,2,3
2,3,4
1,3,4
1,2,3
16
Similarly with n = 7 in case II we get the form
7(x^x)29x+l45. With x > 4 we have F(x) = 0,1,7,9,23,
6
42,49,56,135, etc. Then 04, 712, 1425, 3034, 3670,
7276, and 7556 are S^. By adding 1, we see that 5,13,
29,35, and 71 are Sr. Adding 23, 5993 are S^. Adding
49, we have 55119 and 121125 as S^. Adding 56, we have
116120, 122156, and 164172 as S^. Adding 135, 149103
and 171205 are S^., Hence with the exception of 6, 0205
are S^.
Apply the lemma of Dickson quoted in the introduc
tory part with h = 7, g = 205, and k = 6. The conditions
of the lemma hold if (15) holds. The value of the right
side of (15) is 1517 whence (15) holds if M = 21. Hence
7 through 205+F(21) are S^, But 6 is also an so that
six summands suffice to g^ = 205+F(21) = 1254.
Apply the lemma again with h = 0, g = g^ = 1254,
and k = 7. The value of the right side of (15) is 10,505
whence (15) holds if M = 51 Hence seven summands suffice
to g2 = 1254+F(51) = 22,053.
Apply the lemma again with h = 0, g = g2 = 22,053,
and k = 5. The value of the right side of (15) is 176,657
whence (15) holds if M = 210. Hence eight summands suffice
to g^ = 22,053+F(210) = 1,559,576.
20
(23) PH h(mod 3k)
where h ~ 2(mod 3) and h Ed( mod k).
The product of (20) and (21) gives (E(6DE)jp) = 1.
Let N = E(6d~E). Then we wish to show that we can always
choose a prime p = h(mod 34) such that (N Â¡ p) = 1, N cannot
q n o
be a square since N = with K an integer gives 6ED = K+E
whence E2+K2Â£0(mod 3) and E2= K2(mod 3) But this is pos
sible only If ErK0(mod 3) contrary to (E,3) = 1. Hence
If N = SN' where S Is the largest square contained in N, we
see that N' ^ 1 while (NÂ¡p) = (N'p).
Let N' = pip2**pn wliere the P^ (i = 1,?,* ,n)
are distinct primes. Then since (E,3) = 1, we have (N,3) = 1
and (N',3) = 1. Hence p^ 3 for 1 = 1,2,' ,n. Since
(E,k) = 1 we have similarly that pi / 2 if k Is even. Hence
(N',3k) = 1 or 5 We thus have two cases to consider.
a) If p^ 5 (1 = 1, *>n), we write (N  p) =
P1.N11 P1.N11
(PIN1)(1) 2 2 = (1) 2 2 (pp,)(ppD)***(p!p ).
Thus we can choose a p which is a quadratic nonresidue of
p^ and a quadratic residue of p^ for 1 = 2,3* *,nlf
either p 1(mod 4) or N'SI (mod 4); while if p = N' = 3(mod 4)
we can choose a p which is a quadratic residue of all the
p^. This gives us p=J(mod N') and putting this with (23)
we have by the Chinese Remainder Theorem p=q(mod 34N')
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.
June, 1957
Dean, College of Arts
and Sciences
Dean, Graduate School
SUPERVISORY COMMITTEE
Chairman
49
21
where q = j(mod N) and. q = h(raod 3k). But this congruence
la satisfied by an Infinite number of primes.
b) Next suppose p^ = 5 When N'EOimod 5) then
N = E(6DE) E 0(mod 5) and either E E 0(mod 5) or E=D(mod 5)
First If N* = p_ = 5, then (N'p) = (51P) = (PI 5) If 5k,
then E^O(mod 5) since (E,k) = 1. Thus If 5k, then by
(22) vie have p=2 or (mo. 5) and (N  p) = (p5) = 1 with
p = h(mod 3^). If 5/k, then (N' ,k) = 1 so that If we choose
p=2 or 3(mod 5) we can put this with (23) and have by the
Chinese Hemalnder Theorem p = q(mod 15k) where q=h(mod 3k)
and q~2 or 3(nd 5)
P1.N11
If N' = 5P2P3*Pn then (N*  p) = (1) 2 2 (pjN')
P1 .N'1
= (1) 2 2 (pi 5) (pIp2> **(pPn) Then if either pEl
(mod 4) or N'El(mod 4) we choose p a quadratic residue
of p for 1 = 2,3,* *,n; while if pEN'E3(mod *0 we
choose p a quadratic nonresidue of p^ and a quadratic
residue of p^ for i *tn. This gives p = g
(mod N'/5) Thus if 5ik, then by (22) we have p= 2 or 3
(mod 5) and (pl5) = 1 so that we have (N'p) = 1 for
p = q(mod 3N'k/5) where qEh(mod 3k) and qEgimod N'/5)
Similarly if 5/k, then vie choose p = 2 or 3(md 5) and have
p = q(mod 3N'k) where q=h(raod 3k), q = 2 or 3(mod 5), and
q = g(mod N'/5). But this congruence is satisfied by an
BIOGRAPHICAL SKETCH
The author was born in Red Oak, Iowa, on June 15,
1931 He moved to Florida In 1944 and graduated from Palm
Beach High School In 1949. In June, 1952, he received the
Bachelor of Science In Education, Magna Cum Laude. from
Florida Southern College where he was valedictorian. In
June, 1954, he received the Master of Arts in Mathematics
from the University of Florida where he held a teaching
asslstantshlp for two years. He taught one year at Belle
Glade High School before returning to the University of
Florida in June, 1955 He again held a teaching assistant
ship for one year and has been the holder of a fellowship
from The Southern Fellowships Fund during the past year.
He is a member of Kappa Delta PI and the Mathematical
Association of America.
4g
PART II
SOME NECESSARY CONDITIONS FOR THE REPRESENTATION
OF ALL INTEGERS
In order that all positive Integers may be repre
sented as a sum of k values of the function
(1) F(x) = Â§(x^x)+Dx+C
6
the function should have nonnegative Integral values for
all Integers x > t and represent 0 and 1 for two such In
tegers, We wish to Investigate the necessary conditions
that must be placed on the coefficients E, D, and C for
these requirements to be fulfilled.
Assume that E is positive and odd. Let
(2) F(x^) = 0 and F(x2) 1.
Then E(x^x^) + 6Dxj+6C = 0 and E(x2^x2)+6Dx2+6C = 6
whence Ex^x^x^+x^J+Dx^x^) = 6 and
(3) (x2x1)[E(x22+x2x1+x121)+6d] = 6.
Let *2~xl = Then d6 and hence d = +1, +2, +3
or +6. Substituting
(^) x2 = x^+d
in (3) we have dÂ£(3xL2+3x^d+d2l)+6dD = 6 whence
3dEx^2+3d2Ex^+d^EdE+6dD6 = 0, Solving this quadratic
in x we have 3d2E+V72dE72d2DE+12d2E23dlfE2
Y. = 
1 6dE
3
modulo 2>.
1 or
\s
5(mod 5)
1
3
5 .
.7 .....
l
0,2,3,4,6
0,2,4,5,6
0,1,2,4,6
0,1,2,4,6
3
0,2,3,4,6
0,2,4,5,6
0,1,2,4,6
0,1,2,4,6
5
0,2,3,'4,6
0,2,4,5,6
0,1,2,4,6
0,1,2,4,6
7
0,2,3,4,6
0,2,4,5,6
0,1,?,4,6
0,1,2,4,6
D5 0. 2, 4, or 6(mod S)
^ V
....... 1
. 3
7
1
o.l,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5.6
0,1,2,4,5,6
3
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
5
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
7
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
2
Then every Integer which exceeds h+f(t) and Is < g+f(M)
is a sura of k or fewer numbers (A).
Theorem: Let F(x) >0 for all Integers x > t. Then
ph
every integer >35*2 is a sum of nine values of
F(x) for integers x > t provided that D, C, and t
satisfy the eight mild inequalities:
D < 927+36t if g < t < 0
D < 90g+36t if 16 < t < 9
D < ggl+36t if 24 < t < 17
3D < 27 if 1 < t < 4
2(6D1) < 216 if D > 0 and t < 0
Dt < i(22^+29) if D < 0 and t < 0
C < 6l*2S if 1 < t < 4
7D < 5211 if 1 < t < 4,
The symbol (a,b) will be used to denote the greatest
common divisor of a and b, while ajb denotes a divides b.
The symbol (np) is that of Jacobi in which p must be prime
to 2n in order for the symbol to have meaning. We shall use
maxia^, a2>* *, an) to signify the maximum of the n ele
ments ap a2, a^, and similarly mln(a1> a2>* *, an)
to signify the minimum of the n elements a^, a2>* *f a^
The function C^(m) as used in Part IV is Euler's
which denotes the number of positive Integers not exceeding
m which are relatively prime to ra. Finally the statement
"a is an S^" will mean that the Integer a can be represented
as a sum of 1 values of the particular function under dis
cussion
s
integral values to n.
III.
2n
C = 2n^+3n257n+5Â£
lS2
u
C1 = u
Setting 2nn2 = 0(mod IS) gives n =l(mod 3)* Hence
let n = 3k+l. Then
c 2k3+?k25K
C1 lcu
u o
Then D and C will .be integers for any integral value
of k. Also we see that D(k) = D(kl), Cu(k) = C^(kl),
and C^ik) = Cu(kl) so that we do not need to use any k < 0.
Thus we again get specific forms by assigning nonnegative
Integral values to k.
IV. d = 10+g+n2
Then we must have n2+n+l =0(mod 9) whence we have
(2n+l)2 =6(mod 9) which is an impossibility. Hence we get
no results in this case.
V. D = iia?
cu = iÂ£z
liosia
3
c. = lC
1 u
Setting 4n2E0(raod S) we get n=2(mod 4). Hence
let n = 4k+2. Then
41
h) Consider D = 3 and E = J( mod l6) and s = 2n. Let
1 = 2p2+l4pn+4. Then t=l(mod 16) gives M2= 5pz^+i(rao
while t = 3(mod l6) gives Spz^+2pz+i(mod l6). But since
i = g+gpn, this duplicates case f).
By remark 2) we have corresponding results for
E=9, 11, 13, or 15(mod 16) and D=3(mod 16).
i) Since in each of the preceding cases we have
5pz^+2(Vpz+VJ) (mod l6), excluded even values can arise
only for even z. Setting z = 2z' in remark 3) we see
that an integral amount =SE^pz'h(mod 16) is added to M2
when L is increased by an Integer 2h. With h = 2 this
added integral amount is EO(mod 16). Hence increasing
D by 4 gives duplicate results.
Thus for even s we have M2^O,4,g,12,l4(mod 16)
for D odd and either:
a) sS4 or 12(mod 16) or
b) s = 0(mod 16) and D = E+2(mod 4).
In either case we choose t=p = l(mod 16). Theorem 3 then
follows by means of Theorem 1.
Similarly, using other regular ternary forms, we
may obtain other representation theorems by making use of
Theorem 1.
By a similar proof based on the regular ternary
2 2 2
form x +2y +5w all of whose excluded progressions are of
37
M2^ 0,4,3,12,l4(mod 16).
Before considering even s = 2n, we shall make three
pertinent remarks.
1) For any D, E, and s = 2n we have with t = o(mod 16)
Mg= 2K+5pz^+llzp(mod l6) and with t = 2(mod l6)
2K+5pz^+7pz(mod 16) where K = E^pnE^Dpz+4E^EHp24. In
either case for any odd p we have M2E 0,2,4,6,3,10,12,14
(mod 16). Thus we can never show M2^0,4,3,12,l4(mod l6)
with t50 or 2(mod l6) and s even.
2) With any D if we Increase E by 2k (which we
must do since E is always odd), we have added
J = 2pk(3E2s+6Eks+4k2s6E2Dz12EkDz3k2Dz)+l6kD(3E2+6Ek+4k2)
to the previous value of M2. Since s = 2n, we have
J = 4pk(3E2n+6Ekn+4k2n3E2Dz6EkDz4k2Dz)+l6kD(3E2+6Ek+4k2).
If k = 4, we have JE0(raod 16). Hence increasing E by 3
gives duplicate results.
3) Increasing D by an even integer 2h adds
l6E^h4E^pzh to the previous M2# But this addition is
= 4E^pzh(mod l6).
Take DE2 and E E 1(mod l6) and s = 2n. Let
a = 2p2+2pn+12. Then with tEl(mod 16) we have ^pz^+6pz
+a(mod 16) and with tE3(mod 16) we have Ii2= 5pz^+l4pz+a
(mod 16). And in either case we have M2= a,a+p,a+3p,a+4p,
a+5p,a+7p,a+Sp,a+9p>a+llp,a+12p,a+13p,a+15p(mod l6) and
17
Apply the lemma again with h = 0, g = = 1,5591576
and k = 9. The value of the right side of (15) Is 12,476,341
whence (15) holds If M = 1766. Hence nine summands suffice
to g4 = 1,559,576+F(1766) = 919,462,065.
Again we have the mild inequalities trivially satis
fied so that by the theorem of Dickson stated in the intro
duction every integer > 35*2^ = 5^7,202,560 is the sura of
nine values of F(x). Hence every integer > 0 is the sum of
nine values of F(x) = Mx^x)29x+l4g for integers x > t
0
where 15 < t < 4.
Similarly with larger values of n we may get addi
tional forms for which it can be shown that all integers
are a sura of nine values of any one form.
ON WARINGS PROBLEM WITH
CUBIC FUNCTIONS
By
RICHARD LEE YATES
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
June, 1957
ACKNOWLEDGMENTS
The author wishes to acknowledge his indebtedness
to the members of his supervisory committee and in partic
ular to Dr. E. H. Hadlock for suggesting the topic of this
dissertation and for his continued interest and assistance
in its preparation. The contributions of others, direct
or indirect, were appreciated.
il
TABLE OP CONTENTS
Page
ACKNOWLEDGMENTS ii
PAST
I. INTSODUCriON 1
II.SOME NECESSARY CONDITIONS FOR THE
REPRESENTATION OF ALL INTEGERS 3
III, SPECIFIC RESULTS WITH E = 1 7
IV. A GENERALIZED REPRESENTATION THEOREM lg
V. SOME SPECIFIC RESULTS 2g
VI. SOME FORMS NOT USABLE IN THEOREM 1 ... 43
BIBLIOGRAPHY 47
BIOGRAPHICAL SKETCH 4g
CERTIFICATE OF APPROVAL 49
ill
PART I
INTRODUCTION
In 1770 2. Waring made the conjecture that every
positive integer is a sum of nine cubes, is a sum of nine
teen fourth powers, and is a sum of a limited number of n
powers. The study of the various portions of this conjecture
comprises part of what came to be known as Waring's Problem,
However Waring*s Problem has been generalized to Include the
representation of all, or almost all, or certain classes of
positive integers as sums of a finite number of values of a
given function. This paper will be concerned with the repre
sentation of positive Integers as the sum of a finite number
of cubic functions. In particular formulae will be derived
which give all such possible functions.
For this purpose we shall need in Part III a lemma"*"
p
and a theorem of L. E. Dickson.
Lemma: Let t be an integer and consider any set of
Integers > 0
(A) f(t), f(t+l), f(t+2), .
Make the hypothesis that every Integer 1 for which
h < 1 < g is a sura of k1 or fewer numbers (A),
allowing repetitions. Let f(j+l)f(j) < gh
(J = t, *, Ml), where the integer M exceeds t.
"^Leonard Eugene Dickson, Modern Elementary Theory
of Numbers.(Chicago.) 1939, p. l4o.
5lbX4., p. 131.
1
2
Then every Integer which exceeds h+f(t) and Is < g+f(M)
is a sura of k or fewer numbers (A).
Theorem: Let F(x) >0 for all Integers x > t. Then
ph
every integer >35*2 is a sum of nine values of
F(x) for integers x > t provided that D, C, and t
satisfy the eight mild inequalities:
D < 927+36t if g < t < 0
D < 90g+36t if 16 < t < 9
D < ggl+36t if 24 < t < 17
3D < 27 if 1 < t < 4
2(6D1) < 216 if D > 0 and t < 0
Dt < i(22^+29) if D < 0 and t < 0
C < 6l*2S if 1 < t < 4
7D < 5211 if 1 < t < 4,
The symbol (a,b) will be used to denote the greatest
common divisor of a and b, while ajb denotes a divides b.
The symbol (np) is that of Jacobi in which p must be prime
to 2n in order for the symbol to have meaning. We shall use
maxia^, a2>* *, an) to signify the maximum of the n ele
ments ap a2, a^, and similarly mln(a1> a2>* *, an)
to signify the minimum of the n elements a^, a2>* *f a^
The function C^(m) as used in Part IV is Euler's
which denotes the number of positive Integers not exceeding
m which are relatively prime to ra. Finally the statement
"a is an S^" will mean that the Integer a can be represented
as a sum of 1 values of the particular function under dis
cussion
PART II
SOME NECESSARY CONDITIONS FOR THE REPRESENTATION
OF ALL INTEGERS
In order that all positive Integers may be repre
sented as a sum of k values of the function
(1) F(x) = Â§(x^x)+Dx+C
6
the function should have nonnegative Integral values for
all Integers x > t and represent 0 and 1 for two such In
tegers, We wish to Investigate the necessary conditions
that must be placed on the coefficients E, D, and C for
these requirements to be fulfilled.
Assume that E is positive and odd. Let
(2) F(x^) = 0 and F(x2) 1.
Then E(x^x^) + 6Dxj+6C = 0 and E(x2^x2)+6Dx2+6C = 6
whence Ex^x^x^+x^J+Dx^x^) = 6 and
(3) (x2x1)[E(x22+x2x1+x121)+6d] = 6.
Let *2~xl = Then d6 and hence d = +1, +2, +3
or +6. Substituting
(^) x2 = x^+d
in (3) we have dÂ£(3xL2+3x^d+d2l)+6dD = 6 whence
3dEx^2+3d2Ex^+d^EdE+6dD6 = 0, Solving this quadratic
in x we have 3d2E+V72dE72d2DE+12d2E23dlfE2
Y. = 
1 6dE
3
4
whence
(O 3d2Ey where
X1 Sd
(6) y2 = 72dE72d2DE+12d2E23d4E2.
We then have two cases.
a) Assume d odd; then y must be odd and a multiple
of 3 Let
(7) y = 3(2n+D.
Then by (6)
(g) D = 72dE+l?d2E2^5d4E2y2
72d2E
whence substituting (7) we have
(9) D 24dE+4d2E2d4E212n2l?n3
24d2E
By (1), (2), and (5) w, )
1 bl *16* V
* $&) c o
(10) c = 64gd4E2DT2l6d2DEv+27d6E^T27d4E2y+qd2Ey2:ry3.logd4E3
136d2E2y^
1296d3E2
By (g) and (10)
(11)
64gd^E2:T2l6dEyT'd4E2y+i2y^
1296d^E2
C =
5
By (5) and (11)
(12) 12d^E2=F2dlE2n:P'dl4'E2TP4dEnTl2dE+gn^+12n2+6n+l
C = *
24d^E2
We then have four subcases.
I. Let d = 1. Then from (9) and (12)
D = gE+E24n24nl and
gE
12E2=T2E2n:PE2:F24En:ri2E+gn5+12n2+ 6n+l.
c = __^= = =
24e
II. Let d = 1. Then
D = E2gE4n24nl and
SE
12E2+E2+ 2E2nT24En;fl2ETgn^T12n2:r6nTl #
III.Let d = 3. Then
n 24E15E24n24nl and
v 72E
C =
3 24E2Tl62E2nT,glE2T7 PEn^ 6E+gn^jl 2n2+6n+l .
64gE2
IV. Let d = 3 Then
>2, ,.2
tv 24E+15E+4n+4n+l and
U ~ 72E
C
3 24E2+l62E2n+glE2;T72En:r3 6ETgn3:TL Pn2^^!
64gEc
b) Assume d even* then by (5) It follows that y
must be a multiple of 6. Let
6
(13) y = 6n.
Then D = 72dE+12d2E2^dl4E2y2 = 24dE+4d2E2d1E212n2
72d2E 24d2E
Prora (Â£>), (10), and (13) 6d^E2;Tl2dEn:Td^E2n+4n^
c =
12d^E
We then have four additional subcases.
V. Let d = 2. Then
n = ^En2 and
"E
12E2+6En+4E2n+n5
0 = p' *
24e2
VI. Let d = 2. Then
D = 811(1
12E2=r6En+4E2n*h3
C = ?
24e
VII. Let d = 6. Then
D 12E9oE2n2 and
72E
3 24E2:mgEnT3 24E2n+
C
64&ET
VIII. Let d *= 6. Then
D = 12E+96E2+n2 and
72E
3 24E2:FlSEn+ 3 24E2nTn^
G
64gE
PART III
SPECIFIC RESULTS WITH E = 1
We shall now Investigate particularly the cases
of the preceding part with E = 1 using the same numbering
for the eight cases.
I. D = 2nn2
2
Cu cj. iou
where C^ and C^ are the values of C using upper and lower
signs respectively.
We see that D(n) = D(nl), cu(n) = C^(nl), and
C^n) = Cu(nl). But If n < 0, then n1 > 0; hence we do
not need to use any n < 0. We see also that D and C will be
Integers for any integral value of n. Hence by assigning
nonnegative Integral values to n, we get specific forms
which may satisfy our original conditions.
11 d = Statn2.
2
cu = aaffi*3 cx = icu
We see that D(n) = D(nl), Cu(n) = C.^n1), and
C^(n) = Cu(nl) so that again we do not need to use any
n < 0. Also D and C will be integers for any integral value
of n. Again we get specific forms by assigning nonnegative
7
s
integral values to n.
III.
2n
C = 2n^+3n257n+5Â£
lS2
u
C1 = u
Setting 2nn2 = 0(mod IS) gives n =l(mod 3)* Hence
let n = 3k+l. Then
c 2k3+?k25K
C1 lcu
u o
Then D and C will .be integers for any integral value
of k. Also we see that D(k) = D(kl), Cu(k) = C^(kl),
and C^ik) = Cu(kl) so that we do not need to use any k < 0.
Thus we again get specific forms by assigning nonnegative
Integral values to k.
IV. d = 10+g+n2
Then we must have n2+n+l =0(mod 9) whence we have
(2n+l)2 =6(mod 9) which is an impossibility. Hence we get
no results in this case.
V. D = iia?
cu = iÂ£z
liosia
3
c. = lC
1 u
Setting 4n2E0(raod S) we get n=2(mod 4). Hence
let n = 4k+2. Then
9
D = 2k 2k
SJ+JL5iL
C1 =
~u 3
Then D and C will be integers for any Integral value
of k. Also we see that D(k) = D(kl), Cu(k) = C^(kl),
and C^ik) = Cu(kl) so that again we do not need to use
any k < 0. Thus we again get specific forms by assigning
nonnegative integral values to k.
VI. D = 4+
Ia
C = 12:
u
IF
C1 Wu
Setting 4+n2 = 0(mod g) gives n r 2(mod 4). Hence
let n = 4k+2. Then
D = l2k2k2
C = aUk.2+ZÂ£ r T r
u 3 C1 !cu
Then D and C will be Integers for any integral
value of k. We see that D(k) = D(kl), Cu(k) = C^(kl),
and C^ik) = Cu(kl) so that again we do not need to use
any k < 0. Thus we again get specific forms by assigning
nonnegative Integral values to k.
vil. D = S4ai
72
Then we must have g4+n2=0 whence n2^ 6(nod 9) which
is clearly impossible. Hence we get no results in this case.
VIII. p 108Ht2
72
ou =1 U
Setting 10S+n2=0(mod 72) gives n = 6(mod 12).
Hence let n = 12k+6. Then
D = 22k2k2
cu = Cl = 1CU
Then D and C will be integers for any integral value
of k. Also we see that D(k) = D(kl), Cu(k) = C^(kl),
and C1(k) = Cu(kl) so that again we do not need to use
any k < 0, Thus we again get specific forms by assigning
nonnegative Integral values to k.
We shall nov* show that in every case we have D < 1
and that D = 1 only in case I with n = 0. But by (1) the
two forms that this gives are ^(x^x)+x and i
o 6
both of which have been covered by Dickson.^ Thus we shall
assume D < 0 in seeking further forms not covered by Dickson
2 2
In case I we have D = 1 But > o. Thus
D < 1 and D = 1 when n = 0.
2 2
In case II D = 1 But 21IH > o. Thus D < 1
2 2
In case III D = But > 0 so that D < 0.
1Ibld.. p. 141,
11
In case V we have D = (2k+2k2) and since 2k+2k2 > 0,
we see that D < 0.
In case VI D = l(2k+2k2) and since 2k+2k2 > 0, we
have D < 1,
Finally In case VIII we have D = 2(2k+2k2). Then
since 2k+2k2 > 0 we have D < 2.
Note? Since D < 1, the first five of the eight mild
Inequalities of the theorem of Dickson quoted in Part I are
always satisfied.
Since F'(x) = ^x2 ^ +D and Fw(x) = x, we see that
2 o
if D j 1 so that 1 > 6d the function F(x) has a maximum at
Xq = and a minimum at x^ = Also F(x)
is an Increasing function for x < Xq and x > x^ and a de
creasing function for Xq < x < x^.
If d / 1 and x2 > xi > x^> then from (4) we have
d > 1. Hence there exists an integer x]_ < x4 < x2 for which
0 = F(x^) < F(x^) < F(Xp) = 1, contrary to F(x) being an
integer for all integral values of x. If d < 0 and
X1 > x2 >x3 then 0 = F(xx) < F(x2) =1 contrary to F(x)
being an increasing function for all x > x^.
He;qce since we must have F(x) >0 for every
x > mlnCx^, x2), we must have F(x^) > 0. This gives
16dUVTPh5T +c o whenoe c J. U=fla^g=^t ,,4 wlth
12
D / 1 we have
(1*0 C2 >
 243
Using the values of D and Cu from III in ( l4) we
have 729k^+113^k^531k2+36k+4 < 0 which is impossible for
integral k > 0. Using the values of D and from III in
(l4) we have 729k^+172Â£k3+44lk215S4k96g <0. But since
k > 0 this is possible only for k = 0 which gives the form
(x^x)+l. But since this form arises in case I with n = 1,
6
case III gives no new usable forms.
Using the values of D and from V in (14) we
have 1296k^+1944k^+44lk2+36k+l < 0 which is clearly im
possible for nonnegative k. Using the values of D and
from V in (l4) we have 1296k^+324ok5+23g5k2+196k242 < 0
which is satisfied only by k = 0. But this gives the form
^(x^x)+l which arises in case I. Hence case V gives no
new usable forms.
Using the values of D and from VI in (l4) we
have 1296^+3240k^+3465k2+1764k+343 < 0 which is clearly
Impossible. Using the values of D and from VI in (l4)
we have 1296k^+1944k^+1521k2+630k+100 < 0 which is clearly
impossible also. Hence case VI gives no usable forms.
Using the values of D and Cu from VIII in (l4)
we have ll664k^+23976k^+l4265k2+73gk+10 < 0 which is also
13
impossible. Using the values of D and from VIII in
(14) we have Il664k^+226g0k3+12321k2+2520k+1225 < 0 which
is obviously impossible also. Hence case VIII also gives
no usable forms.
Using the values of D and from II in (l4) we
have gln^+4g6n^+1197n2+1764n+1372 < 0 which is clearly
Impossible. Thus in case II we can get usable results
only for D and C^.
Consider now case I where d = 1. Here we have
X, = V.3.l.Ol. and X, = 3V^172D. if c < 0, then since
3 3 6
D < 0 we have F(x^) = ( +c < 0. Hence we
must have x^ > x^. This gives 3+V8172D > 2V316D)
which is obviously impossible for the lower sign. We thus
have (using the upper sign) 3+ V372D > which
gives D > 11/9, i. e. D = 0 or D = 1, But D = 1 does
not arise while D = 0 for n = 1 from which we have the two
forms g(x^x) and ^Kx^x)+l both of which have been covered
by Dickson.1 But C1 < 0 for all values of n except n = 0
and n = 1 both of which we have already considered. Hence
with these exceptions we get no usable forms for in
Thus except for the two forms 4(x^x)+l and
6
case I
14
^(x^x)+x+l which already have been considered, we get
6
distinct usable forms only from D and In case I and
from D and In case II. For 0 < n < 7 I*1 case I and
0 < n < 6 In case II we duplicate the results of Dickson,^
However for larger values of n we get new usable forms.
For instance, letting n = 8 in case I, we get the
form x^x)35x+196 for which we shall show that all ln
6
tegers are the sum of nine values.
With x > 2 we have F(x) = 0,1,7,11,21,31,41,62,66,
95,105,127,161,231, etc. Then 04, 716, 1826, 2836,
3846, 4850, 5256, 5960, 6291, and 93121 are S^. By
adding 1, we get 5,17,27,37,47,51,57,61, and 92 as S^.
Hence 05, 757, and 59121 are S^. By adding 4l, 103132
and 134162 are S^. By adding 62, 124153 and I55I83 are
S5. By adding 95, we get 154, 157I86, and 188216 as S^.
By adding 105, 167196 and 198226 are S^. By adding 127,
189218 and 220248 are S^. By adding l6l, 223252 and
254282 are S^. Also 253 = 231+21+1 is an S^. Hence with
the exception of 6 and 58, 0282 are S^.
Apply the lemma of Dickson quoted in the introduc
tory part with h = 59, g = 282, and k = 6. The conditions
hold if
'Ibid.
15
(15) (2M1)2 < l+8(ghD).
The value of the right side is 2065 whence (15) holds if
M = 23. Hence 59 through 282+F(23) are S^. But 6 and 58
are Sg also. Thus six summands suffice from 0 through
gx = 282+F(23) = 1697.
Apply the lemma again with h = 0, g = g^ = 1697
and k = 7 The value of the right side of (15) is 13,857
whence (15) holds if M = 59. Hence seven summands suffice
to g2 = i697+f(59) = 3M^.
Apply the lemma again with h = 0, g = g2 = 3^,1^
and k = 8. The value of the right side of (15) is 272,665
whence (15) holds if M = 261. Hence eight summands suffice
to g^ = 34,l4g+F(26l) = 2,988,329.
Apply the lemma again with h = 0, g = g^ = 2,988,329,
and k = 9* The value of the right side of (15) is 23,906,913
whence (15) holds if M = 2400. Hence nine summands suffice
to g^ = 2,988,329+F(2^00) = 2,306,904,125.
The remaining three mild inequalities stated in the
theorem of Dickson quoted in the introduction are satisfied
trivially for this form. Hence by the theorem every inte
ger > 35*2 = 587,202,560 is the sum of nine values of
F(x). Hence every integer > 0 is the sum of nine values of
F(x) = i(x^x)35x+196 for integers x > t where l6 < t < 2.
6
16
Similarly with n = 7 in case II we get the form
7(x^x)29x+l45. With x > 4 we have F(x) = 0,1,7,9,23,
6
42,49,56,135, etc. Then 04, 712, 1425, 3034, 3670,
7276, and 7556 are S^. By adding 1, we see that 5,13,
29,35, and 71 are Sr. Adding 23, 5993 are S^. Adding
49, we have 55119 and 121125 as S^. Adding 56, we have
116120, 122156, and 164172 as S^. Adding 135, 149103
and 171205 are S^., Hence with the exception of 6, 0205
are S^.
Apply the lemma of Dickson quoted in the introduc
tory part with h = 7, g = 205, and k = 6. The conditions
of the lemma hold if (15) holds. The value of the right
side of (15) is 1517 whence (15) holds if M = 21. Hence
7 through 205+F(21) are S^, But 6 is also an so that
six summands suffice to g^ = 205+F(21) = 1254.
Apply the lemma again with h = 0, g = g^ = 1254,
and k = 7. The value of the right side of (15) is 10,505
whence (15) holds if M = 51 Hence seven summands suffice
to g2 = 1254+F(51) = 22,053.
Apply the lemma again with h = 0, g = g2 = 22,053,
and k = 5. The value of the right side of (15) is 176,657
whence (15) holds if M = 210. Hence eight summands suffice
to g^ = 22,053+F(210) = 1,559,576.
17
Apply the lemma again with h = 0, g = = 1,5591576
and k = 9. The value of the right side of (15) Is 12,476,341
whence (15) holds If M = 1766. Hence nine summands suffice
to g4 = 1,559,576+F(1766) = 919,462,065.
Again we have the mild inequalities trivially satis
fied so that by the theorem of Dickson stated in the intro
duction every integer > 35*2^ = 5^7,202,560 is the sura of
nine values of F(x). Hence every integer > 0 is the sum of
nine values of F(x) = Mx^x)29x+l4g for integers x > t
0
where 15 < t < 4.
Similarly with larger values of n we may get addi
tional forms for which it can be shown that all integers
are a sura of nine values of any one form.
PART IV
A GENERALIZED REPRESENTATION THEOREM
We shall now consider the representation of certain
sufficiently large Integers as sums of eight values of the
function f(x) = Dx+^kx^x) with x > 0 and with E > 0 and
6
D > 0. Let
(16) G = A+B+C and H = A^+B^+C^
with A, B, and C arbitrary positive Integers and let s be
any positive Integer.
With every ternary quadratic form there is associated
a set of progressions1 given by 2k(Sm+a), piki(p1m+bi) where
the Pj^ (i = 1,2, ,n) are odd prime divisors of the deter
minant and (p^jb^) = 1. Then it is possible to find a least
n
k < g^n^p^ suh that the set of progressions kra+rj with
(J = 1,2,* *,t) contains the progressions associated with
the form.
We first define a condition relating to a regular
ternary form Ax2+By2+Cw2 and the function f(,x) = Dx+(x^x)
6
Condition V: Given the regular ternary form
p 2 2
Ax +3y +Cw whose excluded progressions are contained in
the progressions km+rj for j = 1,2,* *,t and (k,3) = 1
Burton W. Jones, MA New Definition of Genus for
Ternary Quadratic Forms," American Mathematical Society
Transactions. XXXIII (193l), p. 99.
19
and not contained In any set of progressions with a smaller
value of k. This is equivalent to saying that the excluded
progressions are contained in the number classes which are
= r^Tg,* *,rt(mod k) with (kf3) = 1 where k is the small
est possible value for the modulus. Then it is possible to
prove that, fpr any z and the proper choices of d, r, and a
prime p, that M2^rl,r2'* *,rt(mod k) where
By examination of the 102 regular ternary forms,
we see that excluding those for which k would not be prime
to 3 gives a k dividing l600 in every case. Hence the dis
tinct prime factors of k can be only 2 and 5
Lemma 1: With k as defined in Condition V and
(E,3k) = 1, there exist primes p with the properties:
(18)
(19)
(20)
(21)
(p,3kE) = 1
(p,EoD) = 1
(3 Ip) i
(3E(e6d)p) = i
(22) p = d(mod k) with d = 2 or 3(mod 5) if E=D(mod 5)
and 5k, otherwise d arbitrary.
Proof.Property (20) will be satisfied if we take
p = 2(mod 3). Putting this with (22) we have by the Chinese
Remainder Theorem
20
(23) PH h(mod 3k)
where h ~ 2(mod 3) and h Ed( mod k).
The product of (20) and (21) gives (E(6DE)jp) = 1.
Let N = E(6d~E). Then we wish to show that we can always
choose a prime p = h(mod 34) such that (N Â¡ p) = 1, N cannot
q n o
be a square since N = with K an integer gives 6ED = K+E
whence E2+K2Â£0(mod 3) and E2= K2(mod 3) But this is pos
sible only If ErK0(mod 3) contrary to (E,3) = 1. Hence
If N = SN' where S Is the largest square contained in N, we
see that N' ^ 1 while (NÂ¡p) = (N'p).
Let N' = pip2**pn wliere the P^ (i = 1,?,* ,n)
are distinct primes. Then since (E,3) = 1, we have (N,3) = 1
and (N',3) = 1. Hence p^ 3 for 1 = 1,2,' ,n. Since
(E,k) = 1 we have similarly that pi / 2 if k Is even. Hence
(N',3k) = 1 or 5 We thus have two cases to consider.
a) If p^ 5 (1 = 1, *>n), we write (N  p) =
P1.N11 P1.N11
(PIN1)(1) 2 2 = (1) 2 2 (pp,)(ppD)***(p!p ).
Thus we can choose a p which is a quadratic nonresidue of
p^ and a quadratic residue of p^ for 1 = 2,3* *,nlf
either p 1(mod 4) or N'SI (mod 4); while if p = N' = 3(mod 4)
we can choose a p which is a quadratic residue of all the
p^. This gives us p=J(mod N') and putting this with (23)
we have by the Chinese Remainder Theorem p=q(mod 34N')
21
where q = j(mod N) and. q = h(raod 3k). But this congruence
la satisfied by an Infinite number of primes.
b) Next suppose p^ = 5 When N'EOimod 5) then
N = E(6DE) E 0(mod 5) and either E E 0(mod 5) or E=D(mod 5)
First If N* = p_ = 5, then (N'p) = (51P) = (PI 5) If 5k,
then E^O(mod 5) since (E,k) = 1. Thus If 5k, then by
(22) vie have p=2 or (mo. 5) and (N  p) = (p5) = 1 with
p = h(mod 3^). If 5/k, then (N' ,k) = 1 so that If we choose
p=2 or 3(mod 5) we can put this with (23) and have by the
Chinese Hemalnder Theorem p = q(mod 15k) where q=h(mod 3k)
and q~2 or 3(nd 5)
P1.N11
If N' = 5P2P3*Pn then (N*  p) = (1) 2 2 (pjN')
P1 .N'1
= (1) 2 2 (pi 5) (pIp2> **(pPn) Then if either pEl
(mod 4) or N'El(mod 4) we choose p a quadratic residue
of p for 1 = 2,3,* *,n; while if pEN'E3(mod *0 we
choose p a quadratic nonresidue of p^ and a quadratic
residue of p^ for i *tn. This gives p = g
(mod N'/5) Thus if 5ik, then by (22) we have p= 2 or 3
(mod 5) and (pl5) = 1 so that we have (N'p) = 1 for
p = q(mod 3N'k/5) where qEh(mod 3k) and qEgimod N'/5)
Similarly if 5/k, then vie choose p = 2 or 3(md 5) and have
p = q(mod 3N'k) where q=h(raod 3k), q = 2 or 3(mod 5), and
q = g(mod N'/5). But this congruence is satisfied by an
22
infinite number of primes.
Thus (20), (21), and (22) are satisfied simulta
neously by an infinite number of primes chosen as previ
ously described. But since JkE and E6D are constants,
any prime p > max(3kE, E6D) must be prime to both. Thus
all five conditions are satisfied simultaneously by an
infinite number of primes.
Lemma 2: Let p be a prime having the properties of
Lemma 1, s a given integer prime to p, d as in Lemma 1, and
r an arbitrary integer. Then there exist positive integers
t and z such that
(24)
(25)
(26)
(27)
Proof.Since (p,3E) = 1, there exists an integer
K such that 3EK = l(mod p) Hence (3EKp) = 1. Then from
(21) we have 1 = (3EKp)(3E(E6D)p) = (K(ED)p). Hence
there exists an Integer t^ such that t^E K(E6D) (mod p)
whence = 3EK(E6D)c E6d(mod p) and (24) is satisfied
by t^. Then by the Chinese Remainder Theorem (since (k,p)
= 1) there exists an Integer tg with 0 < t2 < kp such that
t2(mod p) and t^r r(mod k). Hence congruences (24),
(25), and (26)^ are satisfied simultaneously by t = t2
23
Next If (z,p) = 1, then 2Dz+y(z^+3t2zz) is also
prime to p since (24) holds. Suppose that also (u,p) = 1
and 2Dz+j( z^+3t2zz) = 2Du+j(u^+3t2uu) (mod p^) with
(25) kp < z < p^+kp; kp < u < p^+kp.
Then (z^u^)E+(zu)(3Et2+6DE)= O(mod p^) whence
(zu)[E(z2+zu+u2)+3Et2(E6D)]=O(mod p^). If the second
factor Is a multiple of p, then by (15) and (24) we have
z2+zu+u2=O(mod p) whence (2z+u)= 3u2(mod p). But this
is impossible since (u,p) = 1 and (31p) = 1. Therefore
z=u(mod p3) whence zu = Mp^ and (25) ^ plus the negative
of (25) 2 gives kp(p^+kp) < zu < p^+kpkp or p^ < zu < p^.
But this says that M = 0 and hence that z = u. We have
thus shown that the
(29) 2Dz+j( z.^+3t2zz) with kp < z < p^+kp
are prime to p and further that no two of them are congruent
to each other modulo p^. Hence every Integer s which is
prime to p is congruent modulo p^ to one of the integers
(29). Therefore congruence (26) has a solution whioh
satisfies (27)2
Lemma : When s is sufficiently large there exists
at least one prime p having the properties (15)(22) of
Lemma 1 which does not divide s and which satisfies:
(30) 2H+2ep9 < s < 2h+5ep9
24
(32) 33Ep9 I(p3+kp)3(p5+kp)(2D+Ek2p2)2DGp3+fkp hp9
b 3 3 3
+Lgp3 > o
3
where G and H are given by (l6).
Proof.Assume that there exists a number class of
primes p=d(mod k) with (d,k) = 1 tfhich satisfy conditions
(IS)(22) of Lemma 1. It is known that the number of primes
p=d(mod k) with (d,k) = 1 which satisfy the inequalities
x < p < x+ax Increases Indefinitely with x.1 Let
x =
2H+ si 9 Then for s sufficiently
2H+3J
large there are at least ten primes satisfying the condi
tions (lS)(22) of Lemma 1 which satisfy
(33)
r 6s 19 < p < [6s 19
[E(2K+5)J P Â£ (_E(2H+3)J .
10
Their product is greater than f 6s 1 9 =
[E(2H+f;)J [
6s
E(2H+5)
3'
> S
for s sufficiently large. Hence for s sufficiently large
atleast one of the ten primes does not divide s for if all
ten distinct primes divide s, then their product must divide
Edmund Landau, "liber elne Anwendung der Primzahl
theorie auf das Waringsche Problem in der elementaren Zahlen
theorie," Hathematlsche Annalen. LXVI (1909), pp. 103104,
25
s contrary to their product being greater than s. Hence if
s > for some Integer C^, there exists at least one prime
which does not divide s and which satisfies (33) and the
conditions (12$)(22) of Lemma 1. But (33) is equivalent
to (30).
Again (3D Is equivalent to Ep9 > 2p^G(E6D) which
is satisfied by any p sufficiently large. But we can in
crease p by increasing s. Hence if s > for some integer
C2, then all primes satisfy (3D.
Similarly (32) is equivalent to
Ep9 > 6Ekp7+12Ek2p5+2p3[E(^k3G)+6D(G+l)]+2kp(6DE) which
is satisfied by any p sufficiently large. But since we can
increase p by Increasing s, if s > for some Integer C^,
then all primes satisfy (32).
Hence for s > raax(C^, C2> C^) there is at least one
prime which satisfies the conditions of Lemma 3
Theorem 1: Let s be a given Integer and Ax^+By^+Cw2
a regular ternary form satisfying Condition V. If one of
the sets of values of z and p determined in Lemmas 1, 2,
and 3 satisfies the congruence
(3*0 s =2Dz+2DGp3(mod E) ,
then s can be represented as the sum of eight values of
f(x) = Dx+^(x^x) with x > 0.
Proof .We first determine p by means of Lemmas 1
2 6
and 3. By Lemma 2 there exist integers t and z which
satisfy congruences (25) and (26). Thus we have
s = 2Dz+ z^+3t2zz) (mod p3) whence
(35) s = 2Dz+ j( z^+3t2zz)+p^M.
Let PL be determined by
(36) K = 2DG+&P6 So+K1#
Then vre have
(37) s = 2Dz+y(z3+3t2zz)+2DQp3+5Hp9 ^GpVp^K^.
Since (E,p) = 1 and by (3*0 and (37) we see that 0
(mod E), Let
(3^) M EM2.
Prom inequalities (27) and E > 0 we obtain jz > ^kp,
2Dz > 2D(p3+kp) Sz3 > y(p3+kp)3 and finally
Et2z > E(kp)2(p3+kp) whence
(39) 2Dz j(z^+3t2zz) > j(p3+kp)3(p^+kp) (2IHEk2p2)+kp.
But also since E>0, D > 0, z > 0, and z3z > 0, we have
(40) 2Dz ~(z^+3t2zz) <0.
Hence from (30), (39) and (40) we obtain
(41) ^Ep9 (p3+kp)3.(p3+kp) (2EHEk2p2)+^kp < s2Dz
 j( z^+3t2zz) < ^Ep^
whence by (37) and (41) we have
(42) 2H+3.Ep9_ (p3+kp)3_(p3+kp)(2I>fEk2p2)+4cp_2DQp3_ gp9
27
+ap3 < p^K1 < 2^^Ep92DGp^. Hp9+op3.
Then by inequalities (3D* (32), and (42), when p
is sufficiently large, we have 0 < p^M^ < Ep^ whence
(43) o < m2 < p6.
Since Condition V holds, there exist Integers X,
Y, and Z such that
(44) M2 = AX2+BY2+CZ2.
We thus obtain from (35), (36)* (3&), and (44)
s = 2Dz+~(z^+3t2zz)+p^M
= f(z+t)+f(zt)+p^(2DO+Hp6 laJ+p^ML
= f(z+t)+f(zt)+p^(2DO+Hp6 o)+p^E(AX2+BY2+CZ2)
= f(z+t)+f(zt)+f(Ap^+X)+f(Ap^X)+f(Bp^+Y)+f(Bp^Y)
+f(Cp5+Z)+f(Cp^Z).
From (43) and (44) it follows that 0 < AX2+BY2+CZ2
< p^ and hence that Â¡Xj, y, and z are each less than
p^. From (27) it follows that C < t < z. Hence z+t, ztf
Ap^+X, Ap^X, Bp^+Y, Bp^Y, Cp^+Z, and Cp^Z are all posi
tive and thus that s is the sum of eight values of the cubic
function f(x) = Dx^(x^x) with x > 0.
PART V
SOME SPECIFIC RESULTS
Theorem 2: Let s be a sufficiently large given
Integer. If one of the sets of values of z and p deter
mined in Lemmas 1, 2, and 3 satisfies the congruence
s =2Dz+26Dp3(mod E), then s can be represented as the sum
of eight values of f(x) = Dx+Â£(x^x) with (E,30) = 1 and
o
x > 0.
We shall show that by using the regular ternary
form x2+2y2+10w2 Condition V is satisfied. The excluded
progressions are all of the form 25k(25n+5) and 3n+7 and
are thus (with k = 40) included in the progressions 40n,
40n+5, 40n+7, 40n+10, 40n+15, 40n+20, 40n+23, 40n+25,
40n+30, 40n+31, 40n+35i and 40n+39 Hence we need to shovr
M2Â£ 0,5,7,10,15,20,23,25,30,31,35,39(mod 4o). This we do
by showing 7(od 6) and by (38) that = EM2jÂ£ 0(mod 5)
and by using the fact that E is prime to 5 Table 2 shows
0(mod 5) while Table 3 shows that M2^= 7(mod &) The
choices of d and r are given in Table 1. (To save space
Table 1 has been written in two separate parts modulo 5
and g. For example r=l6(mod 4o) is given as r = l(mod 5)
in Table la and r = 0(mod g) in Table lb.)
As an example of the use of Table la, consider the
case Dr EJ1 and s = 0(nod 5). Table la with (22) and (25)
2S
29
then tells us to take p E d = 2(mod 5) and t = r= 1(mod 5)
Also with the form x2+2y2+10w2 we have G = 13 and H = 1009.
By (17) of Condition V we have
= P3EM2= s2Dz y(z3+3t2zz)2DGp3 tp9+Gp3(mod 5)*
Multiplying through by 6 we get
p3tf = s+3Dz+3E(z3+3t2zz)+3DGp3+3EHp9+2EGp3(mod 5). Re
placing D, E, s, p, t, G, and H by their above values, we
have 3M1= 3z3+4z+4(mod 5) whence Mj_= z3+3z+3(mod 5) Then
taking z = 0,l,2,3,4(mod 5) we have 2,3,4(mod 5) so that
M^j O(mod 5) Similarly Table lb is used to show that we
have S),
Table la: (Note that d= 2 or 3(mo
have EED(mod 5).) The values of E, s, r, and d are given
modulo 5
D = 1(mod 5)
0
1
2
3
4
1
1,2
2,3
1,2
2,2
2,2
2
0,2
2,2
2,1
2.1
2,3
3
1,1
1,3
0,3
0,1
1,1
4
0,1
1,1
1,1
1,2
30
D = 2(mod 5)
0
1
2
3
4
1
1.1
0,1
1.3
0,3
o,3
2
1.2
1.3
2,3
2,2
1.2
3
0,1
1.2
1.1
1.*
1.1
4
0,2
2,1
2,2
2,3
2,2
D 3(mod 5)
\s
0
. 1
2
3
4
1
0,2
2,1
2,3
2,2
2,1
2
0,1
1.1
M
1.1
1,2
3
1.2
1.2
2,2
2,3
1,3
4
1,1
o,3
1,1
1.3
0,1
D E 4(mod 5)
0
1
2
3
4
1
0,1
iA
1,2
1,2
1,1
2
1,1
1,1
0,1
0,4
1,3
3
0,2
2,3
2,1
2,1
2,2
4
1,2
2,2
1,3
1,2
2,3
31
Table lb: Suppose s Is even, that is s = 2n. Then
(17) becomes after multiplying by 9
(45) pEKpE s2Dz3E(z5+3t2z2)2Dp4Ep(mod g),
Take pEl and t = 0(mod g), Then (45) becomes
M2= 2n2Dz3Ez^+3Ez2D4E = 2(nDzD2E)3E(z^z)(mod g)
and since z^z is even we have g). The case with
s odd is covered by the following table and Table 3 The
values of E, s, r, and d are given modulo g,
DE 1 or 5(mod g)
E^O
1
1
. 7
1
1,1
1,1
1,3
1,1
3
1,1
1,1
1,3
1,1
5
1,1
1.1
1,3
1,1
7
1,1
1,1
1,3
1,1
32
D = 0, 2, 4, or 6(mod
4
2)
1
3
5
7 .
1
1,1
1,3
1.1
1,3
3
l.l
1.3
1.1
1.3
5
i.l
1.3
1,1
1.3
7
l.l
1,3
1,1
1.3
D= 3 or 7(od &)
1
3 .
5
7
1
1.3
1,1
1,1
1.1
3
1.3
1,1
1.1
1,1
5
1.3
1.1
1,1
1,1
7
1.3
1.1
1.1
1,1
'Table 21 The values of s, E, and are given
modulo 5.
D= 1(mod 5)
5]
 o
1
2
.3
4
i
2.3,4
1.3,*
1,2,3
1,2,4
1.3,4
2
1,2,3
1,2,4
1,2,3
2,3,4
1,2,4
3
2,3,4
1,2,4
1,2,3
1,3,4
1,2,3
4
1,3,4
1,2,3
2,3,4
1,3,4
1,2,3
33
D 2 2(mod 5)
A
... o
1
2
3
4
1
1,3,4
1,2,3
2,3,4
1,3,4
1,2,4
2
1.3,4
1,2,4
1,2,3
1,2,3
1,2,4
3
1,2.3
1,2,3
1,2,4
1.2,4
1,3,4
4
1,2,4
1,3.4
2,3,*
2,3,4
1,2,3
3(mod 5)
0
1
2
3
4
i
1.3,4
1,3,4
1,2,3
1,2,3
1,2,*
2
2,3,4
1,2,4
1,3,4
1,3,4
2,3.4
3
1,2,4
1,3,4
2,3,4
2.3,4
1,3.4
4
1.2,4
1,3,4
1,3,*
1,2,3
2,3,4
4(mod 5)
*4
0
1
2
3
4
i
1,2,4
2,3,4
1,2,4
1,3,4
2,3,4
2
1,2,3
2,3,4
1,2,4
1,2,*
1,3,4
3
2,3.4
1,3,4
1,2,3
2,3,4
1,3,4
4
1,2,3
1,2,4
2,3,*
2,3,4
1,2,4
modulo 2>.
1 or
\s
5(mod 5)
1
3
5 .
.7 .....
l
0,2,3,4,6
0,2,4,5,6
0,1,2,4,6
0,1,2,4,6
3
0,2,3,4,6
0,2,4,5,6
0,1,2,4,6
0,1,2,4,6
5
0,2,3,'4,6
0,2,4,5,6
0,1,2,4,6
0,1,2,4,6
7
0,2,3,4,6
0,2,4,5,6
0,1,?,4,6
0,1,2,4,6
D5 0. 2, 4, or 6(mod S)
^ V
....... 1
. 3
7
1
o.l,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5.6
0,1,2,4,5,6
3
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
5
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
7
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
0,1,2,4,5,6
35
D 5 3 or 7(od S)
A
l
3
5 .
7. ..
1
0,1,2,4,6
0,1,2,4,6
0,2,3,4,6
0,2,4,5,6
3
0,1,2,4,6
0,1,2,4,6
0,2,3,4,6
0,2,4,5,6
5
0,1,2,4,6
0,1,2,4,6
0,2,3,4,6
0,2,4,5,6
7
0,1,2,4,6
0,1,2,4,6
0,2,3,4,6
0,2,4,5,6
Theorem 2 then follows by an application of Theorem 1.
Theorem 3: Let s be a sufficiently large given
integer satisfying one of the conditions:
i) s is odd
il) D is odd and s = 4 or 12(raod l6)
ill) D is odd, s = 0(mod l6) and D=E+2(mod 4).
If one of the sets of values of z and p determined in
Lemmas 1, 2, and 3 satisfies the congruence s = 2Dz+SDp^
(mod E), then s can be represented as the sura of eight
values of the function f(x) = Dx+^ix^x) with (E,6) = 1
and x > 0.
We shall show that by using the regular ternary
form x2+y2+2w2 Condition V is satisfied. The excluded
progressions are all of the form 4k(l6n+l4). Hence we
need to show 0,4,7?,12,l4(raod l6) Since p must be
large, we can assume p ^ 2, Then we have
= s2Dz y( z^+3t2zz)3Dp3 ^Ep^+^Ep^ whence
36
EM^E ps2Dpz+5pE(z^+3t2zz)+8EH2Ep24E(mod l6) and since
E is odd we obtain
M^= E^ps2DE^pz+5p(z^+3t2zz)+5DE^+2p24(mod l6).
Take DH l(raod l6) Then we have
Mg= E^ps2E^pz+5p( z^+3t2zz)+SE^+2p24(rood l6). Then take
ESI, sil, ti 0, and p = 1( mod l6). Then we have
m2= 5z3+9z+7= l,3,5,7,9,ll,l3(mod 16).
Increasing s by an even Integer 2k will add 2pk to
which will still give a set of seven odd values and hence
will not contain any of the excluded values. Thus any odd
s is satisfactory for E=DEl(mod l6).
Since E is odd, any Increase must be a multiple of
2, say 2n. But this adds 6E2psn+12En^ps+2n^ps12E^npz
24En"pzl6n^pz+4<3E2n+96En2+64n^
= 2pn(3E2s+6Ens+4n2s6E2z12EnzSn2z)+l6n(3E2+6En+4n2) to
the value of Mg. Since this value is even, it will again
give a set of seven odd values which are thus distinct from
the excluded values. Thus with DIE 1( mod l6) for any odd E
and odd s we choose t=0 and pil(mod 16) and have
M2/ 0,4,3,12,l4(mod l6) .
But if we Increase D by one we add BE^2E^pz
= 2E^(4pz) to Mg. But this is even and hence will again
give a set of seven odd values. Thus for any D, any odd
E, and any odd s we choose t = 0 and pEl(mod l6) and have
37
M2^ 0,4,3,12,l4(mod 16).
Before considering even s = 2n, we shall make three
pertinent remarks.
1) For any D, E, and s = 2n we have with t = o(mod 16)
Mg= 2K+5pz^+llzp(mod l6) and with t = 2(mod l6)
2K+5pz^+7pz(mod 16) where K = E^pnE^Dpz+4E^EHp24. In
either case for any odd p we have M2E 0,2,4,6,3,10,12,14
(mod 16). Thus we can never show M2^0,4,3,12,l4(mod l6)
with t50 or 2(mod l6) and s even.
2) With any D if we Increase E by 2k (which we
must do since E is always odd), we have added
J = 2pk(3E2s+6Eks+4k2s6E2Dz12EkDz3k2Dz)+l6kD(3E2+6Ek+4k2)
to the previous value of M2. Since s = 2n, we have
J = 4pk(3E2n+6Ekn+4k2n3E2Dz6EkDz4k2Dz)+l6kD(3E2+6Ek+4k2).
If k = 4, we have JE0(raod 16). Hence increasing E by 3
gives duplicate results.
3) Increasing D by an even integer 2h adds
l6E^h4E^pzh to the previous M2# But this addition is
= 4E^pzh(mod l6).
Take DE2 and E E 1(mod l6) and s = 2n. Let
a = 2p2+2pn+12. Then with tEl(mod 16) we have ^pz^+6pz
+a(mod 16) and with tE3(mod 16) we have Ii2= 5pz^+l4pz+a
(mod 16). And in either case we have M2= a,a+p,a+3p,a+4p,
a+5p,a+7p,a+Sp,a+9p>a+llp,a+12p,a+13p,a+15p(mod l6) and
3*3
since a is even we always get either M2= 0,1,3,4,5,789.
11,12,13,15(mod l6) or K2= 1,2,3,5,6,7,9.10,11,13,14,15
(mod 16). Hence at least one of the excluded values always
arises. Thus in this case there are no values of p and t
such that we obtain MgÂ£0,4,3,12,l4(nod l6), With D=2
and E = 3, 5. or 7(n
remark 2) we see that the results will also be duplicated
for E = 9, 11, 13, or 15(mod l6).
But by remark 3) we see that Increasing D by 2h
adds an Integer = ^E^pzhimod l6) to the previous Mg. This
does not affect the even values we are getting. Consequently
if D and s are both even, there are no values of p and t
such that 0,4,3,12,l4(mod l6),
a) Consider D = E E 1(mod 16) and s = 2n. Let
b = 2p2+2pn+4. Then we have for t = l(mod l6) NgE 5pz^+3pz+b
(mod 16) and for t= 3(md l6) 5pz^+b(mod l6) which
gives in either case MgE b,b+p,b+3p,b+5p,b+7p,b+3p,b+9p,
b+llp,b+13p,b+15p(mod 16), And with nE2 or 6 and p E 1
(mod 16) we have Mg= l,2,3,5,7,9,10,H,13,15(mod l6) Thus
with sE4 or 12(mod l6) we take t = p = l(mod 16),
b) Consider DEI and E E3(mod l6) and s = 2n, Let
c = 2p^+6pn+4. Then for t = l(mod 16) we have Mg= 5Pz^+4pz+c
(mod 16) and for t=3(nod l6) we have Mg= 5pz^+12pz+c
(nod 16), In either case we obtain MgE c,c+p,c+3p,c+5p,
39
c+7p,c+9p,c+llp,c+13p,c+15p(mod l6). Then with n30(mod l6)
we get lf35,6f7,9,H,13.15(mod 16) with pHl(mcd l6).
With n = 2 and p = l(mod l6) we get Mr,E 1,2,3,5,79H>13>15
(mod l6). With n = 6 and p=l(mod l6) we get
M2H l,3,5,7,9,10,ll,13.15(mod l6). Hence with s = 0, 4, or 12
(mod 16) we take t=pEl(mod l6).
c) Consider D r 1 and E = 5(niod l6) and s = 2n. Let
d = 2p^+10pn+4, Then for t=l(nod l6) we have MgS 5Pz';+d
(mod 16) and for t=3(md 16) we have MgS 5Pz^+*pz+d(mod l6)
But since d = b+gpn, this duplicates the results of case a).
Hence with s = 4 or 12(mod 16) we take t = pEl(mod l6) .
d) Consider DEI and EE7(od l6) and s = 2n. Let
e = 2p2+l4pn+4, Then t=l(mod l6) gives 5Pz^+12pz+e
(mod 16) while t = 3(mod 16) gives M2= 5Pz^+4pz+e(raod 16),
In either case we obtain M^E e,e+p,e+3Pe+5Pe+7pe+9p*
e+llp,e+13p,e+15p(mod 16). Then with n=0 and p21(nod 16)
we have M r l,3,56,79ll1315(mod 16). With n = 2 and
p = l(mod 16) we have K2 = l,2,3,5,79,11,13*15(md l6).
With n=6 and p=l(raod 16) we have Kg = 1,3,5,7,9,10,11,13,15
(mod 16). Hence with srO, 4, or 12(mod 16) we take
t = p = l(mod 16) Then by remark 2) we have corresponding
results for DEI and EE 9, 11, 13, or 15(mod 16).
e) Consider D = 3 and EEKmod l6) and s = 2n. Let
f = 2p^+2pn+4. Then t=l(mod 16) gives 5Pz^+4pz+f
4o
(mod 16) while t = 3(mod l6) gives 5pz3+12pz+f (mod l6).
In either case we obtain K2= f ,f+p,f+3p,f'f5Pf+7Pf+9P
f+llp,f+13p,f+15p(nod l6). With n=0 and p=l(mod l6)
M2= l,3,5,6f7,9,H,13,15(mod 16). With n = 2 and p = l
(mod 16) M2= l,3,5,7,9,10,ll,13,15(mod 16). with n = 6 and
p=l(mod l6) we have 1,2,3,5,7,9 ,11,13,15(mod l6).
Thus with s = 0, 4, or 12(nod l6) we choose t=p=l(mod 16).
f) Consider Dr EE 3(md 16) and s = 2n. Let
g = 2p2+6pn+4. With t=l(mod l6) we have M25 5pz^+Â£pz+g
(mod l6) and with t=3(mod 16) we have I^E 5P2^+g(mcd 16) .
In either case we have M2 = g,g+p,g+3P,g+5P,g+7P,g+Sp,g+9p,
g+llp,g+13p,g+15p(mod 16). With n=2 or 6 and pE l(mod 16)
we obtain M2 = 1,2,3,5,7,9, 1C,11,13,15(niod 16). Hence with
s = 4 or 12(mod 16) we choose t = p = 1( mod 16).
g) Consider D= 3 and EE 5(rood 16) and s = 2n. Let
h = 2p^+10pn+4. With t=l(mod 16) we have M2S 5pz^+12pz+h
(mod l6) and with t = 3(md l6) 5pz^+4pz+h(mod l6). In
either case we have h,h+p,h+3p,h+5p,h+7p,h+9p,h+llp,
h+13p,h+15p(mod l6) With n50 and pEKmod l6) we have
M2= l,356,7,9,ll,13,15(^od 16). With n = 2 and p= 1
(mod 16) we have Mg= 1,3,5,7,9,10,11,13,15(mod l6). With
n = 6 and p = l(mod l6) we have M2= 1,2,3,5,7,9,11,13,15
(mod 16). Thus with s = 0, 4, or 12(mod 16) we choose
t = p = l(mod l6) .
41
h) Consider D = 3 and E = J( mod l6) and s = 2n. Let
1 = 2p2+l4pn+4. Then t=l(mod 16) gives M2= 5pz^+i(rao
while t = 3(mod l6) gives Spz^+2pz+i(mod l6). But since
i = g+gpn, this duplicates case f).
By remark 2) we have corresponding results for
E=9, 11, 13, or 15(mod 16) and D=3(mod 16).
i) Since in each of the preceding cases we have
5pz^+2(Vpz+VJ) (mod l6), excluded even values can arise
only for even z. Setting z = 2z' in remark 3) we see
that an integral amount =SE^pz'h(mod 16) is added to M2
when L is increased by an Integer 2h. With h = 2 this
added integral amount is EO(mod 16). Hence increasing
D by 4 gives duplicate results.
Thus for even s we have M2^O,4,g,12,l4(mod 16)
for D odd and either:
a) sS4 or 12(mod 16) or
b) s = 0(mod 16) and D = E+2(mod 4).
In either case we choose t=p = l(mod 16). Theorem 3 then
follows by means of Theorem 1.
Similarly, using other regular ternary forms, we
may obtain other representation theorems by making use of
Theorem 1.
By a similar proof based on the regular ternary
2 2 2
form x +2y +5w all of whose excluded progressions are of
42
the form 5rai but without the general Theorem 1, L. K. Hua^
proved a representation theorem stated as follows:
Theorem: Let s be a given Integer. If one of the
sets of values of z and p determined in Lemmas 1,
2, and 3 satisfies the congruence s=2Dz+l6Dp^
(mod E), then s can be represented as the sum of
eight values of f(x) = DxfÂ§(x^x) where E is prime
to 15. b
However we can show that this and Theorem 2, al
though they may overlap for some sets of s, D, and E, do
not duplicate each other in every case. For example, Hua's
theorem will not hold if s2Dzl6Dp^ O(raod E) for all
pairs of p and z determined in Lemmas 1, 2, and 3 But
the two conditions s2Dzl6Dp^j O(mod E) and s2Dz26Dp^ = 0
(mod E) can hold simultaneously if 10Dp^Â£ O(mod E) which
is always the case if E and D are relatively prime since
(E,30) = 1 and (E#p) = 1. Hence there can exist sets of
s, D, and E for which Hua's theorem will not hold while
Theorem 2 does hold.
By similar arguments we can show that each of the
Theorems 1 and 2 and Hua's theorem covers cases excluded
by the other two.
^Lookeng Hua, "On Waring Theorems with Cubic Poly
nomial Summands," Mathematlsche Annalen. CXI (1935), p. 624.
PART VI
SOME FORMS NOT USABLE IN THEOREM 1
In contrast to the theorems of the preceding two
parts it is not possible to prove a representation theorem
from Theorem 1 using certain of the 102 regular ternary
forms. Specifically we have the following three theorems.
Theorem 4: No regular ternary form which has the
progressions 4k(gn+7) in its set of excluded progressions
can be used to prove a representation theorem by means of
Theorem 1.
First we show that a theorem cannot be proved using
the regular ternary form x2+y2+w^. For to do so we would
have to show M^J O,4,7(mod g) With x2+y2+w^ we have
p^EM2 = s2Dz (z^+3t2zz)6Dp3Ep9+Ep^ whence as sinning
p / 2 (which we can do since p must be a large prime) and
thus that p is odd, we have EK2= ps+6pDz+5Ep(z^+3t2zz)+2D
(mod g) from which by assuming E odd (which it must be
since (E,k) = 1 and k = g) we have
(46) Mg= Eps+6EpDz+5p(z^+3t2zz)+2ED(mod g).
Taking D = E =1(mod g) and s = 0(mod g) we have
K2= 5pz^+pz+7pt2z+2(mod g). For distinct results we must
have t = 0, 1, or 2(mod g). With t = 0(mod g) we have
^2 5Pz^+Pz+2 = 2,2p+2,4p+2,6p+2(mod g) and for any choice
^3
44
of p we have Mg2 0,2,4,6(mod g) With t = l(mod g) we have
M2= 5pz^+2=2,p+2,3p+2,5p+2,7p+2(mod g) and for any choice
of p we get 1,2,3*57(mo^ *$) With tS2(mod g) we have
= 5pz^+5pz+2 = 2,2p+2,4p+2,6p+2(mod g) and for any choice
of p we have M^S 0,2,4,6(raod g). Hence this choice of D,
E, and s fails.
But from (46) increasing s by an integer n gives
a new mJ = Mg+Enp and we see that for any value of t the
addition of a multiple of p still gives one of the sets
of values containing excluded number classes. Hence we
are unsuccessful for E ED = 1( mod g) and any value of s.
But Increasing E by an even Integer 2n (since E
must remain odd) gives a new mJ = M2+2nps+12npz+4n = Mg+2m
where m *= nps+6npz+2n. But adding an even integer to M2
leaves the sets of four values with at least one excluded
value unchanged. Hence we are unsuccessful for DEl(mod g)
and any s and odd E,
But by (46) increasing D by an Integer n gives a
new M2* M2+6Enpz+2En = Mg+2m where m = 3E&PZ+E*1, Thus
again we get the same sets of four values including at
least one excluded value. Hence we fall for any combina
? 2 2
tion of D, s, and odd E and cannot use the form x +y +w
to prove a representation theorem by means of Theorem 1,
2 2 2
Next let Ax +By +Cw be any regular ternary form
45
which does not represent 4k(gn+7) In order to prove a
representation theorem using this form we must be able to
prove 0,4,7( mod g) in addition to any other values
arising from an excluded progression.
We have (again letting 0 = A+B+C and H = A^+B^+C^)
p^EM2 = s2Dz (z3+3t2zz)2DGp3 Hp94Qp3. Then by
assuming p / 2 and E odd, we have
(47) M2= Eps+bDEpz+5p(z3+3t2zz)+6DEQ+5H+3G(mod g).
2 2 2
But v.e showed in connection with the form x +y +w
that (with M signifying the Mp of the form x2+y2+w2)
Eps+6DEpz+5p(z^+3t2zz)+2ED(mod g) gives in all cases
either MÂ£50,2,4,6 or Ea,1,3,57(mod g) where a is even.
But from (47) we have (with g = 5H+3G) MÂ£+2ED(3Gl)+g
(mod g) whence M2 = (0,2,4,6)+g or M2= (6,l,3,5,7)+g(mod g)
where 6 = a+2ED(3Gl) is even. Hence whether g is even or
odd we get a set of values Including at least one excluded
value and hence we cannot prove a theorem using this form.
This theorem excludes 20 of the 102 regular ternary forms.
Theorem o: No regular ternary form which has the
progressions 4k(gn+3) in its set of excluded progressions
can be used to prove a representation theorem by means of
Theorem 1.
In order to prove a theorem using such a form, we
would have to be able to show M2^O,3,4(mod g). The remain
46
der of the proof Is Identical with that above since we get
the same sets of four values including at least one of the
excluded values in every case. This theorem excludes six
additional forms.
Similarly we can prove;
Theorem 6: No regular ternary form which has either
4k(Sn+5) or 4k(gn+l) in its set of excluded progressions
can be used to prove a representation theorem by means of
Theorem 1.
These exclude ten and three additional regular ter
nary forms respectively.
These three theorems together with the exclusion of
those regular ternary forms for which k would not be prime
to 3 leave exactly fifteen regular ternary forms for possible
use in proving specific representation theorems from the
general Theorem 1,
BIBLIOGRAPHY
Dickson, Leonard Eugene. Modern Elementar:/ Theory of Numbers.
Chicago: The University of Chicago Press, 1939
. "A New Method for Waring Theorems with Polynomial
Summands ". American Mathematical Society Transactions,
xxxvi (193V). pp. 731^s.
. "Waring1s Problem for Cubic Functions". American
Mathematical. Society Transactions. XXXVI (193^). PP.
112.
Hua, Lookeng, "On Waring Theorems with Cubic Polynomial
Summands". Hathematlsche Annalen. CXI (1935). PP# 622
62g.
James, R. D. "The Representation of Integers as Sums of
Values of Cubic Polynomials". American Journal of
Mathematics. LVI (193*0. PP. 303315.
. "The Representation of Integers as Sums of Values
of Cubic Polynomials. II". American Journal of Kathe
g&SJLfig, UX (1937). PP. 3933937
Jones, Burton W. A New Definition of Genus for Ternary
Quadratio Forms". American Mathematical Society Trans
actions XXXIII (1931). PP. 92110.
Landau, Edmund. "Uber eine Anwendung der Primzahltheorie
auf das Waringsche Problem In der elementaren Zahlen
theorie". Mathematlsche Armale^. LXVI (1909). pp. 102
105.
Webber, Cuthbert G. "Waring's Problem for Cubic Functions".
Mericafl MatftamtjcaA s
(193*0. pp. 793510.
**7
BIOGRAPHICAL SKETCH
The author was born in Red Oak, Iowa, on June 15,
1931 He moved to Florida In 1944 and graduated from Palm
Beach High School In 1949. In June, 1952, he received the
Bachelor of Science In Education, Magna Cum Laude. from
Florida Southern College where he was valedictorian. In
June, 1954, he received the Master of Arts in Mathematics
from the University of Florida where he held a teaching
asslstantshlp for two years. He taught one year at Belle
Glade High School before returning to the University of
Florida in June, 1955 He again held a teaching assistant
ship for one year and has been the holder of a fellowship
from The Southern Fellowships Fund during the past year.
He is a member of Kappa Delta PI and the Mathematical
Association of America.
4g
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.
June, 1957
Dean, College of Arts
and Sciences
Dean, Graduate School
SUPERVISORY COMMITTEE
Chairman
49
6
(13) y = 6n.
Then D = 72dE+12d2E2^dl4E2y2 = 24dE+4d2E2d1E212n2
72d2E 24d2E
Prora (Â£>), (10), and (13) 6d^E2;Tl2dEn:Td^E2n+4n^
c =
12d^E
We then have four additional subcases.
V. Let d = 2. Then
n = ^En2 and
"E
12E2+6En+4E2n+n5
0 = p' *
24e2
VI. Let d = 2. Then
D = 811(1
12E2=r6En+4E2n*h3
C = ?
24e
VII. Let d = 6. Then
D 12E9oE2n2 and
72E
3 24E2:mgEnT3 24E2n+
C
64&ET
VIII. Let d *= 6. Then
D = 12E+96E2+n2 and
72E
3 24E2:FlSEn+ 3 24E2nTn^
G
64gE
45
which does not represent 4k(gn+7) In order to prove a
representation theorem using this form we must be able to
prove 0,4,7( mod g) in addition to any other values
arising from an excluded progression.
We have (again letting 0 = A+B+C and H = A^+B^+C^)
p^EM2 = s2Dz (z3+3t2zz)2DGp3 Hp94Qp3. Then by
assuming p / 2 and E odd, we have
(47) M2= Eps+bDEpz+5p(z3+3t2zz)+6DEQ+5H+3G(mod g).
2 2 2
But v.e showed in connection with the form x +y +w
that (with M signifying the Mp of the form x2+y2+w2)
Eps+6DEpz+5p(z^+3t2zz)+2ED(mod g) gives in all cases
either MÂ£50,2,4,6 or Ea,1,3,57(mod g) where a is even.
But from (47) we have (with g = 5H+3G) MÂ£+2ED(3Gl)+g
(mod g) whence M2 = (0,2,4,6)+g or M2= (6,l,3,5,7)+g(mod g)
where 6 = a+2ED(3Gl) is even. Hence whether g is even or
odd we get a set of values Including at least one excluded
value and hence we cannot prove a theorem using this form.
This theorem excludes 20 of the 102 regular ternary forms.
Theorem o: No regular ternary form which has the
progressions 4k(gn+3) in its set of excluded progressions
can be used to prove a representation theorem by means of
Theorem 1.
In order to prove a theorem using such a form, we
would have to be able to show M2^O,3,4(mod g). The remain
25
s contrary to their product being greater than s. Hence if
s > for some Integer C^, there exists at least one prime
which does not divide s and which satisfies (33) and the
conditions (12$)(22) of Lemma 1. But (33) is equivalent
to (30).
Again (3D Is equivalent to Ep9 > 2p^G(E6D) which
is satisfied by any p sufficiently large. But we can in
crease p by increasing s. Hence if s > for some integer
C2, then all primes satisfy (3D.
Similarly (32) is equivalent to
Ep9 > 6Ekp7+12Ek2p5+2p3[E(^k3G)+6D(G+l)]+2kp(6DE) which
is satisfied by any p sufficiently large. But since we can
increase p by Increasing s, if s > for some Integer C^,
then all primes satisfy (32).
Hence for s > raax(C^, C2> C^) there is at least one
prime which satisfies the conditions of Lemma 3
Theorem 1: Let s be a given Integer and Ax^+By^+Cw2
a regular ternary form satisfying Condition V. If one of
the sets of values of z and p determined in Lemmas 1, 2,
and 3 satisfies the congruence
(3*0 s =2Dz+2DGp3(mod E) ,
then s can be represented as the sum of eight values of
f(x) = Dx+^(x^x) with x > 0.
Proof .We first determine p by means of Lemmas 1
39
c+7p,c+9p,c+llp,c+13p,c+15p(mod l6). Then with n30(mod l6)
we get lf35,6f7,9,H,13.15(mod 16) with pHl(mcd l6).
With n = 2 and p = l(mod l6) we get Mr,E 1,2,3,5,79H>13>15
(mod l6). With n = 6 and p=l(mod l6) we get
M2H l,3,5,7,9,10,ll,13.15(mod l6). Hence with s = 0, 4, or 12
(mod 16) we take t=pEl(mod l6).
c) Consider D r 1 and E = 5(niod l6) and s = 2n. Let
d = 2p^+10pn+4, Then for t=l(nod l6) we have MgS 5Pz';+d
(mod 16) and for t=3(md 16) we have MgS 5Pz^+*pz+d(mod l6)
But since d = b+gpn, this duplicates the results of case a).
Hence with s = 4 or 12(mod 16) we take t = pEl(mod l6) .
d) Consider DEI and EE7(od l6) and s = 2n. Let
e = 2p2+l4pn+4, Then t=l(mod l6) gives 5Pz^+12pz+e
(mod 16) while t = 3(mod 16) gives M2= 5Pz^+4pz+e(raod 16),
In either case we obtain M^E e,e+p,e+3Pe+5Pe+7pe+9p*
e+llp,e+13p,e+15p(mod 16). Then with n=0 and p21(nod 16)
we have M r l,3,56,79ll1315(mod 16). With n = 2 and
p = l(mod 16) we have K2 = l,2,3,5,79,11,13*15(md l6).
With n=6 and p=l(raod 16) we have Kg = 1,3,5,7,9,10,11,13,15
(mod 16). Hence with srO, 4, or 12(mod 16) we take
t = p = l(mod 16) Then by remark 2) we have corresponding
results for DEI and EE 9, 11, 13, or 15(mod 16).
e) Consider D = 3 and EEKmod l6) and s = 2n. Let
f = 2p^+2pn+4. Then t=l(mod 16) gives 5Pz^+4pz+f
4o
(mod 16) while t = 3(mod l6) gives 5pz3+12pz+f (mod l6).
In either case we obtain K2= f ,f+p,f+3p,f'f5Pf+7Pf+9P
f+llp,f+13p,f+15p(nod l6). With n=0 and p=l(mod l6)
M2= l,3,5,6f7,9,H,13,15(mod 16). With n = 2 and p = l
(mod 16) M2= l,3,5,7,9,10,ll,13,15(mod 16). with n = 6 and
p=l(mod l6) we have 1,2,3,5,7,9 ,11,13,15(mod l6).
Thus with s = 0, 4, or 12(nod l6) we choose t=p=l(mod 16).
f) Consider Dr EE 3(md 16) and s = 2n. Let
g = 2p2+6pn+4. With t=l(mod l6) we have M25 5pz^+Â£pz+g
(mod l6) and with t=3(mod 16) we have I^E 5P2^+g(mcd 16) .
In either case we have M2 = g,g+p,g+3P,g+5P,g+7P,g+Sp,g+9p,
g+llp,g+13p,g+15p(mod 16). With n=2 or 6 and pE l(mod 16)
we obtain M2 = 1,2,3,5,7,9, 1C,11,13,15(niod 16). Hence with
s = 4 or 12(mod 16) we choose t = p = 1( mod 16).
g) Consider D= 3 and EE 5(rood 16) and s = 2n. Let
h = 2p^+10pn+4. With t=l(mod 16) we have M2S 5pz^+12pz+h
(mod l6) and with t = 3(md l6) 5pz^+4pz+h(mod l6). In
either case we have h,h+p,h+3p,h+5p,h+7p,h+9p,h+llp,
h+13p,h+15p(mod l6) With n50 and pEKmod l6) we have
M2= l,356,7,9,ll,13,15(^od 16). With n = 2 and p= 1
(mod 16) we have Mg= 1,3,5,7,9,10,11,13,15(mod l6). With
n = 6 and p = l(mod l6) we have M2= 1,2,3,5,7,9,11,13,15
(mod 16). Thus with s = 0, 4, or 12(mod 16) we choose
t = p = l(mod l6) .
2 6
and 3. By Lemma 2 there exist integers t and z which
satisfy congruences (25) and (26). Thus we have
s = 2Dz+ z^+3t2zz) (mod p3) whence
(35) s = 2Dz+ j( z^+3t2zz)+p^M.
Let PL be determined by
(36) K = 2DG+&P6 So+K1#
Then vre have
(37) s = 2Dz+y(z3+3t2zz)+2DQp3+5Hp9 ^GpVp^K^.
Since (E,p) = 1 and by (3*0 and (37) we see that 0
(mod E), Let
(3^) M EM2.
Prom inequalities (27) and E > 0 we obtain jz > ^kp,
2Dz > 2D(p3+kp) Sz3 > y(p3+kp)3 and finally
Et2z > E(kp)2(p3+kp) whence
(39) 2Dz j(z^+3t2zz) > j(p3+kp)3(p^+kp) (2IHEk2p2)+kp.
But also since E>0, D > 0, z > 0, and z3z > 0, we have
(40) 2Dz ~(z^+3t2zz) <0.
Hence from (30), (39) and (40) we obtain
(41) ^Ep9 (p3+kp)3.(p3+kp) (2EHEk2p2)+^kp < s2Dz
 j( z^+3t2zz) < ^Ep^
whence by (37) and (41) we have
(42) 2H+3.Ep9_ (p3+kp)3_(p3+kp)(2I>fEk2p2)+4cp_2DQp3_ gp9
PART VI
SOME FORMS NOT USABLE IN THEOREM 1
In contrast to the theorems of the preceding two
parts it is not possible to prove a representation theorem
from Theorem 1 using certain of the 102 regular ternary
forms. Specifically we have the following three theorems.
Theorem 4: No regular ternary form which has the
progressions 4k(gn+7) in its set of excluded progressions
can be used to prove a representation theorem by means of
Theorem 1.
First we show that a theorem cannot be proved using
the regular ternary form x2+y2+w^. For to do so we would
have to show M^J O,4,7(mod g) With x2+y2+w^ we have
p^EM2 = s2Dz (z^+3t2zz)6Dp3Ep9+Ep^ whence as sinning
p / 2 (which we can do since p must be a large prime) and
thus that p is odd, we have EK2= ps+6pDz+5Ep(z^+3t2zz)+2D
(mod g) from which by assuming E odd (which it must be
since (E,k) = 1 and k = g) we have
(46) Mg= Eps+6EpDz+5p(z^+3t2zz)+2ED(mod g).
Taking D = E =1(mod g) and s = 0(mod g) we have
K2= 5pz^+pz+7pt2z+2(mod g). For distinct results we must
have t = 0, 1, or 2(mod g). With t = 0(mod g) we have
^2 5Pz^+Pz+2 = 2,2p+2,4p+2,6p+2(mod g) and for any choice
^3
ON WARINGS PROBLEM WITH
CUBIC FUNCTIONS
By
RICHARD LEE YATES
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
June, 1957
35
D 5 3 or 7(od S)
A
l
3
5 .
7. ..
1
0,1,2,4,6
0,1,2,4,6
0,2,3,4,6
0,2,4,5,6
3
0,1,2,4,6
0,1,2,4,6
0,2,3,4,6
0,2,4,5,6
5
0,1,2,4,6
0,1,2,4,6
0,2,3,4,6
0,2,4,5,6
7
0,1,2,4,6
0,1,2,4,6
0,2,3,4,6
0,2,4,5,6
Theorem 2 then follows by an application of Theorem 1.
Theorem 3: Let s be a sufficiently large given
integer satisfying one of the conditions:
i) s is odd
il) D is odd and s = 4 or 12(raod l6)
ill) D is odd, s = 0(mod l6) and D=E+2(mod 4).
If one of the sets of values of z and p determined in
Lemmas 1, 2, and 3 satisfies the congruence s = 2Dz+SDp^
(mod E), then s can be represented as the sura of eight
values of the function f(x) = Dx+^ix^x) with (E,6) = 1
and x > 0.
We shall show that by using the regular ternary
form x2+y2+2w2 Condition V is satisfied. The excluded
progressions are all of the form 4k(l6n+l4). Hence we
need to show 0,4,7?,12,l4(raod l6) Since p must be
large, we can assume p ^ 2, Then we have
= s2Dz y( z^+3t2zz)3Dp3 ^Ep^+^Ep^ whence
ACKNOWLEDGMENTS
The author wishes to acknowledge his indebtedness
to the members of his supervisory committee and in partic
ular to Dr. E. H. Hadlock for suggesting the topic of this
dissertation and for his continued interest and assistance
in its preparation. The contributions of others, direct
or indirect, were appreciated.
il
VIII. p 108Ht2
72
ou =1 U
Setting 10S+n2=0(mod 72) gives n = 6(mod 12).
Hence let n = 12k+6. Then
D = 22k2k2
cu = Cl = 1CU
Then D and C will be integers for any integral value
of k. Also we see that D(k) = D(kl), Cu(k) = C^(kl),
and C1(k) = Cu(kl) so that again we do not need to use
any k < 0, Thus we again get specific forms by assigning
nonnegative Integral values to k.
We shall nov* show that in every case we have D < 1
and that D = 1 only in case I with n = 0. But by (1) the
two forms that this gives are ^(x^x)+x and i
o 6
both of which have been covered by Dickson.^ Thus we shall
assume D < 0 in seeking further forms not covered by Dickson
2 2
In case I we have D = 1 But > o. Thus
D < 1 and D = 1 when n = 0.
2 2
In case II D = 1 But 21IH > o. Thus D < 1
2 2
In case III D = But > 0 so that D < 0.
1Ibld.. p. 141,
22
infinite number of primes.
Thus (20), (21), and (22) are satisfied simulta
neously by an infinite number of primes chosen as previ
ously described. But since JkE and E6D are constants,
any prime p > max(3kE, E6D) must be prime to both. Thus
all five conditions are satisfied simultaneously by an
infinite number of primes.
Lemma 2: Let p be a prime having the properties of
Lemma 1, s a given integer prime to p, d as in Lemma 1, and
r an arbitrary integer. Then there exist positive integers
t and z such that
(24)
(25)
(26)
(27)
Proof.Since (p,3E) = 1, there exists an integer
K such that 3EK = l(mod p) Hence (3EKp) = 1. Then from
(21) we have 1 = (3EKp)(3E(E6D)p) = (K(ED)p). Hence
there exists an Integer t^ such that t^E K(E6D) (mod p)
whence = 3EK(E6D)c E6d(mod p) and (24) is satisfied
by t^. Then by the Chinese Remainder Theorem (since (k,p)
= 1) there exists an Integer tg with 0 < t2 < kp such that
t2(mod p) and t^r r(mod k). Hence congruences (24),
(25), and (26)^ are satisfied simultaneously by t = t2
42
the form 5rai but without the general Theorem 1, L. K. Hua^
proved a representation theorem stated as follows:
Theorem: Let s be a given Integer. If one of the
sets of values of z and p determined in Lemmas 1,
2, and 3 satisfies the congruence s=2Dz+l6Dp^
(mod E), then s can be represented as the sum of
eight values of f(x) = DxfÂ§(x^x) where E is prime
to 15. b
However we can show that this and Theorem 2, al
though they may overlap for some sets of s, D, and E, do
not duplicate each other in every case. For example, Hua's
theorem will not hold if s2Dzl6Dp^ O(raod E) for all
pairs of p and z determined in Lemmas 1, 2, and 3 But
the two conditions s2Dzl6Dp^j O(mod E) and s2Dz26Dp^ = 0
(mod E) can hold simultaneously if 10Dp^Â£ O(mod E) which
is always the case if E and D are relatively prime since
(E,30) = 1 and (E#p) = 1. Hence there can exist sets of
s, D, and E for which Hua's theorem will not hold while
Theorem 2 does hold.
By similar arguments we can show that each of the
Theorems 1 and 2 and Hua's theorem covers cases excluded
by the other two.
^Lookeng Hua, "On Waring Theorems with Cubic Poly
nomial Summands," Mathematlsche Annalen. CXI (1935), p. 624.
23
Next If (z,p) = 1, then 2Dz+y(z^+3t2zz) is also
prime to p since (24) holds. Suppose that also (u,p) = 1
and 2Dz+j( z^+3t2zz) = 2Du+j(u^+3t2uu) (mod p^) with
(25) kp < z < p^+kp; kp < u < p^+kp.
Then (z^u^)E+(zu)(3Et2+6DE)= O(mod p^) whence
(zu)[E(z2+zu+u2)+3Et2(E6D)]=O(mod p^). If the second
factor Is a multiple of p, then by (15) and (24) we have
z2+zu+u2=O(mod p) whence (2z+u)= 3u2(mod p). But this
is impossible since (u,p) = 1 and (31p) = 1. Therefore
z=u(mod p3) whence zu = Mp^ and (25) ^ plus the negative
of (25) 2 gives kp(p^+kp) < zu < p^+kpkp or p^ < zu < p^.
But this says that M = 0 and hence that z = u. We have
thus shown that the
(29) 2Dz+j( z.^+3t2zz) with kp < z < p^+kp
are prime to p and further that no two of them are congruent
to each other modulo p^. Hence every Integer s which is
prime to p is congruent modulo p^ to one of the integers
(29). Therefore congruence (26) has a solution whioh
satisfies (27)2
Lemma : When s is sufficiently large there exists
at least one prime p having the properties (15)(22) of
Lemma 1 which does not divide s and which satisfies:
(30) 2H+2ep9 < s < 2h+5ep9
44
of p we have Mg2 0,2,4,6(mod g) With t = l(mod g) we have
M2= 5pz^+2=2,p+2,3p+2,5p+2,7p+2(mod g) and for any choice
of p we get 1,2,3*57(mo^ *$) With tS2(mod g) we have
= 5pz^+5pz+2 = 2,2p+2,4p+2,6p+2(mod g) and for any choice
of p we have M^S 0,2,4,6(raod g). Hence this choice of D,
E, and s fails.
But from (46) increasing s by an integer n gives
a new mJ = Mg+Enp and we see that for any value of t the
addition of a multiple of p still gives one of the sets
of values containing excluded number classes. Hence we
are unsuccessful for E ED = 1( mod g) and any value of s.
But Increasing E by an even Integer 2n (since E
must remain odd) gives a new mJ = M2+2nps+12npz+4n = Mg+2m
where m *= nps+6npz+2n. But adding an even integer to M2
leaves the sets of four values with at least one excluded
value unchanged. Hence we are unsuccessful for DEl(mod g)
and any s and odd E,
But by (46) increasing D by an Integer n gives a
new M2* M2+6Enpz+2En = Mg+2m where m = 3E&PZ+E*1, Thus
again we get the same sets of four values including at
least one excluded value. Hence we fall for any combina
? 2 2
tion of D, s, and odd E and cannot use the form x +y +w
to prove a representation theorem by means of Theorem 1,
2 2 2
Next let Ax +By +Cw be any regular ternary form
PART V
SOME SPECIFIC RESULTS
Theorem 2: Let s be a sufficiently large given
Integer. If one of the sets of values of z and p deter
mined in Lemmas 1, 2, and 3 satisfies the congruence
s =2Dz+26Dp3(mod E), then s can be represented as the sum
of eight values of f(x) = Dx+Â£(x^x) with (E,30) = 1 and
o
x > 0.
We shall show that by using the regular ternary
form x2+2y2+10w2 Condition V is satisfied. The excluded
progressions are all of the form 25k(25n+5) and 3n+7 and
are thus (with k = 40) included in the progressions 40n,
40n+5, 40n+7, 40n+10, 40n+15, 40n+20, 40n+23, 40n+25,
40n+30, 40n+31, 40n+35i and 40n+39 Hence we need to shovr
M2Â£ 0,5,7,10,15,20,23,25,30,31,35,39(mod 4o). This we do
by showing 7(od 6) and by (38) that = EM2jÂ£ 0(mod 5)
and by using the fact that E is prime to 5 Table 2 shows
0(mod 5) while Table 3 shows that M2^= 7(mod &) The
choices of d and r are given in Table 1. (To save space
Table 1 has been written in two separate parts modulo 5
and g. For example r=l6(mod 4o) is given as r = l(mod 5)
in Table la and r = 0(mod g) in Table lb.)
As an example of the use of Table la, consider the
case Dr EJ1 and s = 0(nod 5). Table la with (22) and (25)
2S
27
+ap3 < p^K1 < 2^^Ep92DGp^. Hp9+op3.
Then by inequalities (3D* (32), and (42), when p
is sufficiently large, we have 0 < p^M^ < Ep^ whence
(43) o < m2 < p6.
Since Condition V holds, there exist Integers X,
Y, and Z such that
(44) M2 = AX2+BY2+CZ2.
We thus obtain from (35), (36)* (3&), and (44)
s = 2Dz+~(z^+3t2zz)+p^M
= f(z+t)+f(zt)+p^(2DO+Hp6 laJ+p^ML
= f(z+t)+f(zt)+p^(2DO+Hp6 o)+p^E(AX2+BY2+CZ2)
= f(z+t)+f(zt)+f(Ap^+X)+f(Ap^X)+f(Bp^+Y)+f(Bp^Y)
+f(Cp5+Z)+f(Cp^Z).
From (43) and (44) it follows that 0 < AX2+BY2+CZ2
< p^ and hence that Â¡Xj, y, and z are each less than
p^. From (27) it follows that C < t < z. Hence z+t, ztf
Ap^+X, Ap^X, Bp^+Y, Bp^Y, Cp^+Z, and Cp^Z are all posi
tive and thus that s is the sum of eight values of the cubic
function f(x) = Dx^(x^x) with x > 0.
11
In case V we have D = (2k+2k2) and since 2k+2k2 > 0,
we see that D < 0.
In case VI D = l(2k+2k2) and since 2k+2k2 > 0, we
have D < 1,
Finally In case VIII we have D = 2(2k+2k2). Then
since 2k+2k2 > 0 we have D < 2.
Note? Since D < 1, the first five of the eight mild
Inequalities of the theorem of Dickson quoted in Part I are
always satisfied.
Since F'(x) = ^x2 ^ +D and Fw(x) = x, we see that
2 o
if D j 1 so that 1 > 6d the function F(x) has a maximum at
Xq = and a minimum at x^ = Also F(x)
is an Increasing function for x < Xq and x > x^ and a de
creasing function for Xq < x < x^.
If d / 1 and x2 > xi > x^> then from (4) we have
d > 1. Hence there exists an integer x]_ < x4 < x2 for which
0 = F(x^) < F(x^) < F(Xp) = 1, contrary to F(x) being an
integer for all integral values of x. If d < 0 and
X1 > x2 >x3 then 0 = F(xx) < F(x2) =1 contrary to F(x)
being an increasing function for all x > x^.
He;qce since we must have F(x) >0 for every
x > mlnCx^, x2), we must have F(x^) > 0. This gives
16dUVTPh5T +c o whenoe c J. U=fla^g=^t ,,4 wlth
TABLE OP CONTENTS
Page
ACKNOWLEDGMENTS ii
PAST
I. INTSODUCriON 1
II.SOME NECESSARY CONDITIONS FOR THE
REPRESENTATION OF ALL INTEGERS 3
III, SPECIFIC RESULTS WITH E = 1 7
IV. A GENERALIZED REPRESENTATION THEOREM lg
V. SOME SPECIFIC RESULTS 2g
VI. SOME FORMS NOT USABLE IN THEOREM 1 ... 43
BIBLIOGRAPHY 47
BIOGRAPHICAL SKETCH 4g
CERTIFICATE OF APPROVAL 49
ill
24
(32) 33Ep9 I(p3+kp)3(p5+kp)(2D+Ek2p2)2DGp3+fkp hp9
b 3 3 3
+Lgp3 > o
3
where G and H are given by (l6).
Proof.Assume that there exists a number class of
primes p=d(mod k) with (d,k) = 1 tfhich satisfy conditions
(IS)(22) of Lemma 1. It is known that the number of primes
p=d(mod k) with (d,k) = 1 which satisfy the inequalities
x < p < x+ax Increases Indefinitely with x.1 Let
x =
2H+ si 9 Then for s sufficiently
2H+3J
large there are at least ten primes satisfying the condi
tions (lS)(22) of Lemma 1 which satisfy
(33)
r 6s 19 < p < [6s 19
[E(2K+5)J P Â£ (_E(2H+3)J .
10
Their product is greater than f 6s 1 9 =
[E(2H+f;)J [
6s
E(2H+5)
3'
> S
for s sufficiently large. Hence for s sufficiently large
atleast one of the ten primes does not divide s for if all
ten distinct primes divide s, then their product must divide
Edmund Landau, "liber elne Anwendung der Primzahl
theorie auf das Waringsche Problem in der elementaren Zahlen
theorie," Hathematlsche Annalen. LXVI (1909), pp. 103104,
33
D 2 2(mod 5)
A
... o
1
2
3
4
1
1,3,4
1,2,3
2,3,4
1,3,4
1,2,4
2
1.3,4
1,2,4
1,2,3
1,2,3
1,2,4
3
1,2.3
1,2,3
1,2,4
1.2,4
1,3,4
4
1,2,4
1,3.4
2,3,*
2,3,4
1,2,3
3(mod 5)
0
1
2
3
4
i
1.3,4
1,3,4
1,2,3
1,2,3
1,2,*
2
2,3,4
1,2,4
1,3,4
1,3,4
2,3.4
3
1,2,4
1,3,4
2,3,4
2.3,4
1,3.4
4
1.2,4
1,3,4
1,3,*
1,2,3
2,3,4
4(mod 5)
*4
0
1
2
3
4
i
1,2,4
2,3,4
1,2,4
1,3,4
2,3,4
2
1,2,3
2,3,4
1,2,4
1,2,*
1,3,4
3
2,3.4
1,3,4
1,2,3
2,3,4
1,3,4
4
1,2,3
1,2,4
2,3,*
2,3,4
1,2,4
3*3
since a is even we always get either M2= 0,1,3,4,5,789.
11,12,13,15(mod l6) or K2= 1,2,3,5,6,7,9.10,11,13,14,15
(mod 16). Hence at least one of the excluded values always
arises. Thus in this case there are no values of p and t
such that we obtain MgÂ£0,4,3,12,l4(nod l6), With D=2
and E = 3, 5. or 7(n
remark 2) we see that the results will also be duplicated
for E = 9, 11, 13, or 15(mod l6).
But by remark 3) we see that Increasing D by 2h
adds an Integer = ^E^pzhimod l6) to the previous Mg. This
does not affect the even values we are getting. Consequently
if D and s are both even, there are no values of p and t
such that 0,4,3,12,l4(mod l6),
a) Consider D = E E 1(mod 16) and s = 2n. Let
b = 2p2+2pn+4. Then we have for t = l(mod l6) NgE 5pz^+3pz+b
(mod 16) and for t= 3(md l6) 5pz^+b(mod l6) which
gives in either case MgE b,b+p,b+3p,b+5p,b+7p,b+3p,b+9p,
b+llp,b+13p,b+15p(mod 16), And with nE2 or 6 and p E 1
(mod 16) we have Mg= l,2,3,5,7,9,10,H,13,15(mod l6) Thus
with sE4 or 12(mod l6) we take t = p = l(mod 16),
b) Consider DEI and E E3(mod l6) and s = 2n, Let
c = 2p^+6pn+4. Then for t = l(mod 16) we have Mg= 5Pz^+4pz+c
(mod 16) and for t=3(nod l6) we have Mg= 5pz^+12pz+c
(nod 16), In either case we obtain MgE c,c+p,c+3p,c+5p,
PART IV
A GENERALIZED REPRESENTATION THEOREM
We shall now consider the representation of certain
sufficiently large Integers as sums of eight values of the
function f(x) = Dx+^kx^x) with x > 0 and with E > 0 and
6
D > 0. Let
(16) G = A+B+C and H = A^+B^+C^
with A, B, and C arbitrary positive Integers and let s be
any positive Integer.
With every ternary quadratic form there is associated
a set of progressions1 given by 2k(Sm+a), piki(p1m+bi) where
the Pj^ (i = 1,2, ,n) are odd prime divisors of the deter
minant and (p^jb^) = 1. Then it is possible to find a least
n
k < g^n^p^ suh that the set of progressions kra+rj with
(J = 1,2,* *,t) contains the progressions associated with
the form.
We first define a condition relating to a regular
ternary form Ax2+By2+Cw2 and the function f(,x) = Dx+(x^x)
6
Condition V: Given the regular ternary form
p 2 2
Ax +3y +Cw whose excluded progressions are contained in
the progressions km+rj for j = 1,2,* *,t and (k,3) = 1
Burton W. Jones, MA New Definition of Genus for
Ternary Quadratic Forms," American Mathematical Society
Transactions. XXXIII (193l), p. 99.

