数理解析研究所講究録 1319 巻 2003 年 125-130125

On sums of cubes of almost primes.
Koichi KAWADA (川田浩一)
Faculty of Education, Iwate University(岩手大学教育学部)

1. Introduction–The Waring-Goldbach problem for cubes.
Being based on Vinogradov’s method in his proof of the three prime
theorem, Hua started discussing representations of natural numbers assums of powers of prime numbers. This problem is often refered as theWaring-Goldbach problem. In 1938, Hua [4] proved, amongst others,
that every sufficiently large odd integer can be written as the sum of nine
cubes of primes. In fact, he established an asymptotic formula for thenumber of representations of large natural numbers as the sum of cubesof primes, for each exceeding 8.
The above result of Hua implies that for each exceeding 9, everylarge integer can be written as the sum of cubes of primes, because
for any integer , either $-(s-9)2^{3}$ or $n-(s-10)2^{3}-3^{3}$ is odd.
In this direction, therefore, the next target is the sum of eight cubes ofprimes. It is conjectured that every large even number can be written in
the latter manner. Although no one could succeed in proving it by now,some results were shown concerning this problem. First, Roth [6] provedthat every large integer can be written in the form
$s$$s$$s$$s$

$n$

$n$

$n$

$=p_{1}^{3}+\cdots+p_{7}^{3}+x^{3}$,

where ’s are primes and is anatural number. Briidern [2] improved
this result, by showing that when is even and large, one can restrictin the above representation. ( denotes anatural number having
toat most prime divisors, counted according to multiplicity.) By changingfor
the sieve procedure in Briidern’s proof, the author [5] substituted
in Briidern’s theorem. It is of course desired to replace further with
$p_{\dot{l}}$

$x$

$x$

$n$

$P_{f}$$P_{4}$$r$$P_{3}$$P_{4}$$P_{3}$

126, that is, aprime, but it seems to me that even the improvement tois way beyond the current state of technique.
We then proceed to the sum of seven cubes. Needless to say, the situation becomes harder than the case of eight cubes, but it is still possible
to show that every large integer can be written as the sum of seven cubesof almost primes. Indeed, Briidern [3] established that every large integer
can be written in the form
$P_{2}$$P_{1}$$n$$n=p^{3}+x_{1}^{3}+\cdots+x_{5}^{3}+y^{3}$,. The purpose of this short acwhere is aprime, :’s are , and iscount is to report several refinements on the latter result of Briidern. This
work was essentially done while Ivisited the University of Michigan atAnn Arbor through courtesy of Professor Trevor D. Wooley, and enjoyed
the benefits of aFellowship from the David and Lucile Packard Foundation, from April to June 1997. Iam disappointed and ashamed thateven now (as of March 2003) Icould not complete the paper of this workyet. Also, Iwould like to apologize to the organizer, Professor Noriko
Hirata-Kohno, for violating the deadline for this report (...as usual).
$p$

$x$

$P_{5}$$y$$P_{69}$

2. Problems on sums of seven cubes.
We are concerned with the conclusions of the following form:, where is . forevery large integer can be written as”
each .
As Briidern [3] mentioned, there are various combinations of ’s forwhich one can prove the latter statement. As regards this problem, personally Iam interested in three questions. Here they are:$x_{1}^{3}+\cdots+x_{7}^{3}$$x_{i}$$P_{r_{*}}$$i$$r_{\dot{l}}$(A) What is the possible least value of $\max\{r_{1}, \ldots, r_{7}\}$ ?

(B) What is the possible least value of $r_{1}+\cdots+r_{7}$ ?
(C) How many variables can we force to be primes? (In other words,
what is the possible largest number of with $r:=1?$ )$i’ \mathrm{s}$Probably one can expect that every sufficiently large natural number
can be written as the sum of seven cubes of primes, so we may conjecturethat the answers for these questions are 1for (A), and 7for both (B) and(C), in truth. Before writing down the answers that Iactually proved, we127should mention acongruence condition relating the sum of seven cubesof primes.
When we consider representaions of as the sum of seven cubes ofprimes, it is natural to impose the following condition on :For every
natural number , the congruence
$n$

$n$

$q$

$n$ $\equiv x_{1}^{3}+\cdots+x_{7}^{3}$$(\mathrm{m}\mathrm{o}\mathrm{d} q)$

are coprime to . By elementary
has asolution such that all theargument, one may easily confirm that the latter condition is eventually
For instance,equivalent to 2\dagger and 9{ . What happens if violates
suppose that $9|n$ . Then for $q=9^{*}$ , the above congruence has no solution
with $(x_{1}x_{2}\ldots x_{7},9)=1$ . Therefore, if such an is written as the sum ofseven cubes of primes, then at least one of the primes must be not coprime
to 9, namely, it must be 3. Thus is asum of seven cubes of primes, if andonly if $n-3^{3}$ is asum of six cubes of primes. Still may be written as thesum of seven cubes of primes, but the representation problem no longer
involves seven variables in practice. In this sense the above congruence
condition arises naturally, when one investigates the sum of seven cubesof primes. And as amodest form of the above conjecture, one may saythat every large satisfying 2\dagger and 9\dagger can be written as the sum ofseven cubes of primes.
$x_{i}’ \mathrm{s}$$q$$\mathrm{i}\mathrm{t}^{7}$

$n$

$n$

$n$

$n$

$n$

$n$

$n$

$n$

$n$

’It may be worth pointing out not only that 9looks similar to in shape, but. How
also that in Japanese, 9is pronounced exactly the same way as the letter$q$$” \mathrm{q}"$curious!

elementary deliberation on the above congruences may convince us that mostlikely this statement is true even in the cases where $2|n$ or $9|n$ . The hardest case will
. If such an is written as
be the integers satisfying $14|n$ andthe sum of seven cubes of primes, then three of the seven primes must be 2, 3and 7,
which means that $n-2^{3}-3^{3}-7^{3}$ must be the sum of four cubes of primes. Although
one may expect so if is large, one must face aproblem on four cubes which seems
very hard to solve.
In this context, moreover, it may be worth recording here that 7is presumably theleast value of for which every large integer can be written as the sum of cubes ofprimes. To see this, suppose that an odd natural number satisfies the congruences$n+1\equiv 0$ or, and that is written as the sum of six
andcubes of primes. Then elementary argument on congruences reveals that there mustsuch that $n-2^{3}-3^{3}-7^{3}=p_{1}^{3}+p_{2}^{3}+p_{3}^{3}$ . But if $n\leq X$ ,exist three primes , ,, so simply there are at most $O(X(\log X)^{-3})$ such numbers . Hencethenalmost all numbers satisfying the congruence conditions we are now assuming cannot
be written as the sum of six cubes of primes.
$\dagger \mathrm{A}\mathrm{n}$$n$$n\equiv \mathrm{O}\mathrm{o}\mathrm{r}-2(\mathrm{m}\mathrm{o}\mathrm{d} 9)$

$n$

$n$

$s$

$s$

$n$

$n\equiv\pm 1(\mathrm{m}\mathrm{o}\mathrm{d} 7)$$\pm 1(\mathrm{m}\mathrm{o}\mathrm{d} 9)$$\mathrm{p}\mathrm{i}$$p_{\dot{\iota}}\leq X^{1/3}$$p_{2}$$n$$p_{3}$$n$

128
3. Results.
In connection with the congruence condition observed in the previous
and as follows;
section, we introduce the numbers$b_{n}$$a_{n}$$a_{n}=\{_{2’}^{1}$$b_{n}=\{\begin{array}{l}\mathrm{l},\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}9\{n3,\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}9|n\end{array}$$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}n\mathrm{i}\mathrm{s}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}n\mathrm{i}\mathrm{s}\mathrm{o}\mathrm{d}\mathrm{d},$,,Then, as for the questions (A) and (B) above, the following conclusions

may be obtained.
Theorem 1Every sufficiently large integer
the following forms;
(i)

$n$

$=x_{1}^{3}+\cdots+x_{5}^{3}+(a_{n}y_{1})^{3}+(b_{n}y_{2})^{3}$are$P_{3}$.

$n$can be written as each of, where : ’s are$x$(ii) $n=p^{3}+x_{1}^{3}+\cdots+x_{4}^{3}+(a_{n}b_{n}x_{5})^{3}+y^{3}$ , where, and is .
$P_{3}$$y$$p$$P_{4}$, and$y_{1}$,$y_{2}$

is a prime, : ’s are$X$$P_{5}$The part (i) says that every large integer can be written as the sumof seven cubes of , and we may consequently answer 4to the question(A). By (ii), as regards odd integers with 9\dagger , we may answer 21 tothe question (B).
Theorem 1may be proved by adding two ingredients to the method ofBriidern [3]. One of them is concerned with the technique that is called,
for example, as “Vaughan’s iterative method restricted to minor arcs”
In [3], asimpler version of the latter method was adopted, but we may do
better at this point by appealing to the original argument of Vaughan [8].
Another point is about sieve methods. Briidern [3] used aweighted sieve,
but we apply the switching principle (or, the reversal role technique) withthe ordinary linear sieve. It seems that the switching principle may givestronger conclusions than weighted sieves, in most of the situations where
it can be applied.
We then proceed to the problem (C). Iconfess that all my efforts toanswer 4to this problem have ended in failure by now, while one caneasily answer 3to (C) by acouple of known results. In fact, it follows
immediately from the work of Vaughan [7] that when is alarge integer,
with primes
the number of positive integers of the form
$P_{4}$

$n$

$n$

$n$

$n-p_{1}^{3}-p_{2}^{3}-p_{3}^{3}$$p_{i}$

129, for any fixed $>0$ . On the other hand, Briidern [1] showedisthat the number of positive integers less than that is not the sum of four, for any given
cubes is. Since $8/9>37/42$ , these results
together indicate the existence of an integer of the form
with primes
that is written as the sum of four cubes, whenever islarge. This tells that 3is apossible answer for (C).
Moreover, it is possible to refine the last conclusion by saying thatevery large can be written as $n=p_{1}^{3}+p_{2}^{3}+p_{3}^{3}+x_{1}^{3}+\cdots+x_{4}^{3}$ withprimes and almost primes . In this respect, Isuppose that the best
result may be given by the vector sieve of Briidern and Fouvry. Our
method of the proof of Theorem 1can also say something on this, due
to Wooley’s breaking classical convexity device. Via the latter way andsome numerical computation, Iconfirmed that in the last representation, for example. In, and toof , one can restrict ,andto
any case, rather than refining the quality of almost primes here, Iwould
like to make serious effort towards establishing 4as apossible answer for(C).
$>>n^{8/9+\epsilon}$$\epsilon$$n$$<<n^{37/42+\epsilon}$$\epsilon>0$$n-p_{1}^{3}-p_{2}^{3}-p_{3}^{3}$$n$$p_{i}$$n$$p_{i}$$x_{j}$$n$$P_{10}$$\mathrm{x}\mathrm{i}$$x_{2}$$x_{3}$$x_{4}$$P_{24}$

4. Sums of eight cubes of almost primes.
Finally Iwould like to discuss the questions on the sum of eight cubes
corresponding to (A), (B) and (C) above. As is written in the introduction, Roth [6] answered 7for the question corresponding to (C), and theimprovement on this seems far beyond our grasp at present. The resultof the author [5] shows that 10 is apossible answer for the question corresponding to (B). To beat it, one must work with the sum of seven cubesof primes and acube of , and it is again quite hard to do for now, Ithink. As regards (A), the best known answer is 3in view of [5], andthere is room for improvement on this. Actually, Idiscussed this problemwith Briidern, and obtained the following result recently. We now recallthe definition ofin the previous section.
$P_{2}$$a_{n}$

Theorem 2Every sufficiently large integer
$n$

where

$p_{\dot{l}}$$n$can be written as$=p_{1}^{3}+\cdots+p_{5}^{3}+(a_{n-1}p_{6})^{3}+x^{3}+y^{3}$

denotes primes,

$x$

and

$y$

are

$P_{2}$,.

Iconfess that Ispent quite alot of time on this problem, but at last it
could be proved within the techniques of Briidern [2] and Kawada [5]130References[1] J. Briidern, On Waring’s problem for cubes. Math. Proc. Cambridge
Philos. Soc. 109 (1991), 229-256.
[2] J. Briidern, A sieve approach to the Waring-Goldbach problem I:
Sums of four cubes. Ann. Scient. Ecole. Norm. Sup. (4) 28 (1995),461-476.
[3] J. Briidern, A sieve approach to the Waring-Goldbach problem, II..
On the seven cubes theorem. Acta Arith. 72 (1995), 211-227.
[4] L. K. Hua, Some results in additive prime number theory. Quart. J.
Math. Oxford 9(1938), 68-60.
[5] K. Kawada, Note on tie sum of cubes of primes and an almost prime.
Arch. Math. (Basel) 69 (1997), 13-19.
[6] K. F. Roth, On Waring’s problem for cubes. Proc. London Math.
Soc. (2) 53 (1951), 268-279.
[7] R. C. Vaughan, Sums of three cubes. Bull. London Math. Soc. 17(1985), 17-20.
[8] R. C. Vaughan, On Waring’s problem for cubes. J. Reine Angew.
Math. 365 (1986), 122-170