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](https://en.wikipedia.org/wiki/Waring-Goldbach_problem) 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](https://doi.org/10.1134/S106456240904005X) paper of Chubarikov contains announcement of the complete proof, though I did not succeed in finding any reaction on it, even no MathSciNet-review.



Current status of Waring-Goldbach problem
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.


