Is the following statement proved?

For any positive integer $k$ there exists positive integer $n$ such that all sufficiently large integers may be represented as $p_1^k+p_2^k+\dots+p_n^k$ for primes $p_1,\dots,p_n$.

This Wiki article claims that some progress is made only for $k$ up to 7, on the other hand, it refers to Hua Lo Keng's monograph, in which this statement is proved, if we may take zero terms instead some prime powers. This zero does not seem to be so essential on the first glance... And on the third hand, this paper of Chubarikov contains announcement of the complete proof, though I did not succeed in finding any reaction on it, even no MathSciNet-review.

  • $\begingroup$ I'm pretty sure this is known for all k but I'm afraid I don't know a reference. $\endgroup$ – gowers Jul 14 '11 at 20:03
  • $\begingroup$ Sorry, I take that back -- I was thinking of the result with zeros. But I think the zeros are needed for congruence reasons and it's not clear that one couldn't use some trick. E.g. if k=6 then each kth power is 1 mod 7, except that we can throw in a few $p_i=7$s to deal with that. $\endgroup$ – gowers Jul 14 '11 at 20:22
  • $\begingroup$ wiki says n is an absolute constant for all k. Is this the case? $\endgroup$ – user16007 Aug 12 '11 at 8:45
  • $\begingroup$ @unknown(yahoo): definitely no! For fixed $n$ and large $k$ result is not true by simple counting argument (density of representable numbers is 0), even if you take all positive integers instead primes. $\endgroup$ – Fedor Petrov Aug 14 '11 at 6:51

This is a corrected version of my original response, incorporating a nice argument by Fedor Petrov.

Hua in his book (cf. review of MR0124306) proved that there are integers $s,K,N>0$ such that every $n>N$ with $n\equiv s\pmod{K}$ is a sum of $s$ $k$-th powers of primes. For any $t>0$ let $M(t)$ denote the set of residues modulo $K$ which can be represented by a sum of $t$ $k$-th powers of primes. Clearly $M(t+1)$ contains $M(t)+p^k$ for any prime $p$, hence $|M(t+1)|> |M(t)|$ unless $M(t+1)$ equals $M(t)+p^k$ for any prime $p$. In this case $M(t)$ is invariant under the shift of $p^k-q^k$ for any two distinct primes $p$ and $q$. The shifts are coprime (e.g. $p^k-q^k$ is coprime with $q$), hence $M(t)=M(t)+1$, and $M(t)$ contains all residues modulo $K$. This argument shows that $|M(K)|=K$, i.e. modulo $K$ every residue class is a sum of $K$ $k$-th powers of primes. If $p$ denotes the largest of the $K^2$ primes used in the latter representation, and $M$ equals $N+Kp^k$, then we have the following. For every $m>M$ there is a sum of $K$ $k$-th powers of primes, denote it by $m'$, such that $m-m'\equiv s\pmod{K}$ and $m-m'>N$. By Hua's theorem, $m-m'$ is a sum of $s$ $k$-th powers of primes, hence in fact every $m>M$ is a sum of $s+K$ $k$-th powers of primes.

To summarize: the statement in Fedor Petrov's original question follows from Hua's theorem.

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    $\begingroup$ Well, if Hua's theorem is that, then indeed the rest is easy. I do not take your argument literary, by the way (Dirichlet's theorem requires some coprimeness etc), but it looks that we may avoid even Dirichlet's theorem for finishing the proof. Indeed, if $M(n)$ denote the set of residues of $p_1^k+\dots+p_n^k$ modulo $K$, then either $|M(n+1)|\geq \min(M(n)+1,K)$. $\endgroup$ – Fedor Petrov Jul 14 '11 at 21:30
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    $\begingroup$ (continuation) Indeed, $M(n+1)$ contains $M(n)+p^k$ for each prime $p$, so if $|M(n+1)|=|M(n)|$, then $M(n+1)=M(n)+p^k$, so $M(n)$ is invariant under shift to $p_1^k-p_2^k$ for any two primes $p_1$, $p_2$. But such shifts are coprime for obvious reasons (if they all are divisible by prime $q$, then choose $p_1=q$ and $p_2\neq q$). $\endgroup$ – Fedor Petrov Jul 14 '11 at 21:30
  • $\begingroup$ Dear Fedor, thanks for the correction. I realized the problem with coprimeness shortly after my response, but fortunately your argument saved the day. I updated my response accordingly. $\endgroup$ – GH from MO Jul 14 '11 at 22:24

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