A Cubic Analogue of the Cauchy-Fermat Theorem

A Cubic Analogue of the Cauchy-Fermat TheoremAuthor(s): Alvin SugarSource: American Journal of Mathematics, Vol. 58, No. 4 (Oct., 1936), pp. 783-790Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2371249 .

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A CUBIC ANALOGUE OF THE CAUCHY-FERMAT THEOREM.'

By ALVIN SUGAR.

Introduction. In this paper we shall obtain an ideal universal WaringT, theorem for the polynomial

(1) P(x) = m(x3 - x)/6 + x, x integral and >O 0

where m is an integer > 16, i. e. we shall prove g(P) = m + 3 for m ? 16. In Part II of this paper we evaluate a constant C, = 1012in'10, which

maximizes the constants of papers of Dickson,2 Baker3 and Webber 4; this gives us the following theorem.

THEOREM 1. For m _ 7P every inZteger _ C, = 1012m10 is a sum of nzitne or ten values of (1) according ais thte congruence n =- 6 (mod 9) does not or does hold.

In Part I we develop a powerful ascension theorem and ascension methods, and by ascendilng beyond the constant C,, prove that every positive integer < C, is a sum of in + 3 values of P(x) for mn ? 16.

PART I. ASCENSION MIETHODS.

1. Ascension theorems. We shall ascend beyond the constant C, = 1012in'10

first for a fixed range and then for an arbitrary range of values of m. We write

F(a) =P(a+ 1) -P(a)

and apply a theorem of Dickson's 5 to our polynomial P(x).

THEOREMr 2. Let every integer n, c < n C g, be a sumrn of k 1 values of P(x), and let a be an integer ? 0 for which F(a) < g - c. Then every integer N, c < N < g + P(a + 1), is a sum of k 'values of P(x).

1 Presented to the Society, November 30, 1935. 2Tr-ansactions of the American Mathematical Society, vol. 36 (1934), pp. 1-12. 3Doctoral dissertation, Chicago, 1934. 4Tr-ansactions of the American Mathematical Society, vol. 36 (1934), pp. 493-510. 5Theorem 9 in the Bulletin of the American Mathematical Society, vol. 39 (1933),

p. 709. 783



784 ALVIN SUGAR.

Before going on to our next theorem, we note that in this series of theorems, g and c need not be integers.

THEOREM 3. Let every integer n, c < n ? g, be a sum of k 1 values of P(x), and let y be a real number ? 0 which satisfies the inequality F (y + 1) < g-c. Then every integer N,

(2) c ? N ? g + P(y + 1),

is a sum of k values of P(x).

By way of proof we observe that F([y] + 1) ? F(y + 1) and P(y + 1) < P( [y] + 2), since F(x) and P(x) are properly monotone increasing functions for x ? 1.

We are now in a position to introduce an important ascension theorem, which will enable us to breach a huge interval in one step. The inequality

(3) F(3t(3/2e + 1) < P(3te + 1)

holds for t ? 1. Let t be a real number ? 1 which satisfies the inequality F(3t + 1) < pm + q, and let every integer No, c < No ? c + pm + q, be a sum of ki values of P(x). Then from (2) we have that every integer N', c < N' ? c + pm + q + P(3t + 1), and hence every integer N1, c < N1 ? c + P(3t + 1), is a sum of k + 1 values. Similarly, since F(3t3/2 + 1) < P(3t + 1), by (3) with e = 1, then every integer N2, c < N2 _ c + P(3t3/2 + 1), is a sum of k + 2. And finally, every integer N2,

c < N2 _?c + P(3t(3/2)8-l + 1),

is a sum of k + s. The proof of this last statement is made by an induction on s. Since

(9/2)t2(3/2)8m < C + P(3t(3/2)8' + 1)

we may state the following theorem.

THEOREM 4. Let every integer n, c < n -< c + pm + q, be a sum of k values of P(x), and let t be a real number ? 1 which satisfies the inequality F(3t + 1) < pm + q. Then every integer N,

c < N ? (9/2)t2(3/2) m

is a sum of k + s values of P(x).

2. The first ascension; 16 < m ? 1950. There follows a list of values of P(x).



A CUBIC ANALOGUE OF THE CAUCHY-FERMAT THEOREM. 785

0,1,a=mn+2, b=4m+3, c=lOm+4, d=20m+5, e=35m+6, f=56m+ 7, g9 84m+8, h=120m+9, i=165m+10.

We are also going to list a set of intervals such that an integer lying in anyone of these intervals will be a sum of in - 8 values of P(x). These intervals will overlap for m? 16. Therefore rn? - 8 values will suffice over the interval defined by the overlapping intervals.

We shall reconstruct a portion of the following list. We begin with 120m + 9. By adding n - 8 to this we obtain 121m + 1. It is evident that every integer from 120m + 9 to and not including 121m + 1 is a sum of m - 8 values. Now consider the integer 120m + 16 = a + e + g. By adding m - 10 to this we obtain 121m + 6. It is evident that every integer from 120m + 16 to 121m + 6 is a sum of m - 8 values, and continuing thus we come to the interval (120m + 24, 121m + 11) over which mi- 8 values will suffice. Since 121m + 11 = h + a, we can begin all over again as we did with 120m + 9 by adding n - 9 to 121m + 11 and repeating the above procedure. By inspection it may be verified that the following set of intervals overlap for m ? 16.

(h =120m + 9, 121m + 1), (a + e + g 120m + 16, 121m + 6), (2b + 2f = 120m + 20, 121m + 9), (2a + b + c + d + g = 120m + 24, 121m + 11), (a + h -- 121m + 11, 122m + 2), (2a + e + g - 121m + 18, 122m + 7), (c + d + e + f= 121m + 22, 122m. + 11), (3c + e + f

121im + 25, 122m + 13), (2a + h = 122m + 13, 123m + 3), (c + 2f 122m + 18,123m + 8), (a + c + d + e + f 122m + 24, 123m + 12), (a + 3c + e + f 122m + 27,123m + 14), (2a + c + 2d + 2e = 122m + 30, 123m + 16), (1 + 3a + h = 123m + 16, 124m + 4), (a + c + 2f 123m + 20,124rn + 9), (a + 2b + c + d + g 123m + 25, 124m + 12), (b + h = 124m + 12, 125m + 3), (a + b + e + g

124m + 19, 125m + 8), (4c + g 124m + 24, 125m + 12), (2a + 2b + c + d + g = 124m + 27, 125m + 13), (b + c + 2d + 2e = 124m + 29, 125m + 16), (2 + a + b + h = 125m + 16, 126m + 4), (a + 2d + g 125m + 20, 126m + 9), (b + c + d+ e+f 125m + 25, 126m + 13).

Hence if m ? 16, then every integer n, 120m + 8 < n ? 126m + 12, is a sum of m -8 values of P(x). Applying Theorem 2 we see that F(3) =6m + 1 < 6m + 4; then m- 7 values suffice from 120m + 8 to 120m + 8 + 16m + 8. Two more applications of this theorem give the result that m - 5 values will suffice from 120m + 8 to 120m + 8 + 216m.



786 ALVIN SUGAR.

The next ascent will be made in one step by employing Theorem 4. Since t = 19/3 satisfies the inequality F(3t + 1) < 216m, then every integer N,

120m + 8 < N ? c1m = (9/2) (19/3 ) 2 (3/2)8rM,

is a sum of m + 3 values. It is evident that the inequality 1012m10 < c1m holds for 16 < mn < 1950. Hence we have by employing Theorem 1 the following theorem.

THEOREM 5. Let m have the range 16 ? m ? 1950; then every positive integer > 120m + 8 is a sum of m + 3 values of P(x).

3. The second ascension. Again we construct a set of overlapping intervals. This time we begin with an arbitrary value P(A) = Rm + A, and we take mn - r as the number of values which will suffice over each interval. By adding m r (where r is a positive integer) to P(A), we obtain the first interval

(Em +A, (R+1)m+A- r),

We take r R - A - 10. The rest of the intervals can be written at once, as follows:

((R- 1 + t) m + 211-2 + 2t, (R + t)m +11 + t

((R + t) m + R + t-r, (E + 1 + t)rm +R-2r + t- 10)) (t ==1,~. . .,~ lo).

We observe that (R -1 + t)m + 2R -2 + 2t (R1-1 + t)a and that for this range of t the integer 6 (R + t) m, + R + t- r==P(A) + ta + 10 - t is a sum of 11 values of P(x). By inspection it is evident that these intervals will overlap for m? > Q(A) =3R 2 - 1 (A3-5A)/2 --1. We also see that r = (A3 - 7A)/6 -10.

LEMAIA 1. For- m ? Q(A), A _ 5, every integer n,

-Rm + A ? n?< (R + 10)m + A + 20,

is a sum of in- r values of P(x).

In the following discussion we shall prove statenments (Si) and (S,). We begin with P(A) and show (S,) that for Q(A) < m < Q(A + 1) every

6 As a matter of fact the value assigned to r was obtained by requiring that r be the greatest initeger for which the inequality (R + t) m + R + t - i-r P (A) + ta, (t = 1 ,. . ., 10) holds.



A CUBIC ANALOGUE OF THE CAUCHY-FERMrAT THEOREMI. 787

integer > P(A) is a sum of m't + 3 values, provided A ? 10. We also show (S2) thact for m _ Q (A), in + 3 values will suffice from P (A) to P (A + 1) inclusive, when A ? 10. Since Q (A) is an increasing function, then, by (S2) and an induction on A, we conclude (S3) that for mn _ Q (A), every integer n, P(10) ? n < P(A + 1), will be a sum of m + 3 values. Hence from (S1) and (S3) we have the following theorem.

THEOREM 6. Every integer ? P(10) is a sum of m + 3 values of P(x) for mn> Q(10) = 474.

This and Theorem 5 give us the next theorem.

THEOREM 7. Every integer ? P(10) is a sum of m + 3 values of P(x) for oi 16.

There remains yet to be proved, the Statements (9S,) and (S2). In establishing these statements we make use of a pair of inequalities which are derived from the expansion of kx, into the power series

~'~==i -xiog'c x2 log2lk x 3log3 k k.$ =- I + x log k + p - !- - - -

Since this series converges for all x, we have the following inequalities holding for a positive x.

x2 log2 kc x3 log3 k (4) k 2 > r 2

k 6 > 6

To the results of Lemma 1 we apply Theorems 2 and 4, and we find that every integer N,

(5) PE(A) `N _ c2m (9/2) (10)2(3/2)rm,

is a sum of in + 3 values of P(x) for m ? Q(A). We know that

1012m10 _ c2Cm, when Q(A) ? m < 1 = (1O-12c2)1/9.

Let y = A3/10; it is then evident that 7

MI > ( 10) (2/9) (3/2)V-12/9;

for, X=(A3 -7A)/6- 10 > A3/1O.

From (4,), we have

2 (3)8 12 > y2log2 15 12 > 1O-4A6 92 9> 9 9

7 For the remainder of this discussion we shall take A - 10.



788 ALVIN SUGAR.

Write z = 10-4A6; then Ml > 1Oz. Employing (42), we obtain

lOz > 1O-12A18 > A3 > (A3 + 3A2 2A - 6)/2 = Q(A + 1).

Therefore 11 > Q (A + 1), and we have proved (S1). It is evident that

c2m> Mm >m Q(A) > P(A + 1).

This result and (5) prove (S2).

4. The positive integers < P(10). We shall prove another lemma.

LEMMA 2. Every positive integer _ c lOim + 4 is a sum of m + 3 values of P(x) for m ? 4.

It is evident that every integer < 3m + 6 = 3a is a sum of m + 3 values. Adding m + 1 to 3m + 6, we see that every integer < 4m + 7 = 4 + b is a sum of m + 3 values. Adding m - 1 to this, we see that every integer < 5m + 6 =1 + a + b is a sum of m + 3 values. Repetition of this argument gives the following list:

6m+7= 2a+b, 7m+8=8 6a-+-m-4, 7m+1O=-=1+3a+b, 8m+9=3+2b, 9m+8=-a-+ 2b, 1Om + 4 ==c.

This completes the proof of the lemma. The following set of intervals, which overlap for m > 16, give rise to the

conclusion that every integer n,

(6) Om + 3 < n? 13m + 9,

is a sum of m-1 values of P(x).

(c =Om + 4, llm + 3), (6a + b =- Om + 15, lm + 8), (2 + a + c - rm + 8, 12m + 4), (7a + b = rm + 17, 12m + 9), (1 + 2a + c 12m + 9, 13m + 5), (8a + b - 12m + 19, 13m + 10).

Applying Theorem 2 to (6) four timnes, we obtain the following result.

LEMMA 3. For m 16, every integer n, 1Oim + 3 < n ? 217m + 32, is a sum of m + 3 values of P(x).

This result along with Lemma 2 and Theorem 7 completes the proof of the Principal Theorem.



A CUBIC ANALOGUE OF THE CAUCHY-FERMAT THEOREM. 789

PRINCIPAL THEOREM. Every positive integer is a sum of m + 3 values of mn(X3 - X)/6 + x for non-negative integers x, where m ? 16.

PART II. EVALUATION OF THE CONSTANT.

The proof of Theorem 2 of Dickson's and of similar theorems of Baker's and Webber's 8 depends upon the existence of an integer C lying in each interval of a triple of intervals of the form

(7) f (m,~ b ) -- 3 i-'C _ F (m, b j) (i 1~, 2, 3)

where the bi are suitably chosen positive odd integers. For m w 6 (mod 9), (7) takes on the form (8). For m - 6 (mod 9), (7) becomes (9).

(8) 2m + y + (9/8)mb3< _ 3t-1C ? (3/2)mbi3 + vrnj3 (i 1, 2, 3), J (9m4 + 1)/2 ((m, 3)=1)

(M |m4 + 81) /6 ((m, 3) 1);

125 + 26 + 3 125 mb3 + = (9) mb=Y- i1 125 mib=1, 24 2~~5 18 ' 3

where y has some value similar in form to those of (8).9

An inspection will verify that for m> 7,

(10) b, < 7m (i 1,2,3)

for all the 10 bi 6f papers A. Replacing bi by 7m in the right hand side of (8), we get

(11) ~~~~3'-1C < 520M4 (i 1, 2, 3).

Whence C < 60m4. We also seek a value for 32n which satisfies

32n

b? 3A)Y_ 8

and hence which satisfies

(12) 34n _-[384bi ( - m) -432bi4 32n + m (m-6-3bi) +l92i2.

8 Op. cit. These papers shall henceforth be referred to as papers A. 9f and F originally contained terms of the form fi divided by a power p = p (n)

of 3, but for n = v these terms become negligible. 10 In his paper, Webber did not list the bi corresponding to the case m = 18e + 12,

. e. m = 3a, a even and _ I (mod 3). They are here supplied by the author: b =c 20e+ 11, b2==28e + 17, b3== 40e + 23.

11 Dickson, op. cit., p. 7, (28).

9



790 ALVIN SUGAR.

For convenience we state the following lemma.

LEMAIA 4. For a positive a, the inequtality X2 > ax + ,8 iS satisfied by x>1M= max(c,,8) +1.

For each of three cases, ii negative, /B > a, aid ,8 < a, the proof may be made by substituting Ml for x.

If we write (12) in the form X2 > ax + / - 8, where /3 is the sum of all the positivTe terms free of 32n, we knowv that this inequality is satisfied for x > max(a,f/) + 1, by Lemma 4. But by (10) and (11) we have

5,000,000 m4 > 384bi3'C/m > max(a, ,) + 1.

Hence we havTe for m > 77

C33V < 1012m10 = C1.

A similar argument for (9) produces a smaller constant than C1.

THE UNIVERSITY OF CALIFORNIA.

BERKELEY, CALIFORNIA.



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