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This question is mainly inspired from a different problem I was working on.

Is there a value of $k$ such that, for each $n\in \mathbb N$, the equation $$\sum_{i=1}^{k}x_i^2=n$$ is solvable in $x_1,\dots ,x_k\in \mathbb P\cup \{0,1\}$? Here $\mathbb{P}$ is the set of prime numbers.

That is, can every nonnegative integer be written as a sum of squares of primes (together with $0$ and $1$), where the number of summands is absolutely bounded?

What about the same question but for sums of $n$th powers?

I want to know the research that has been done in this field.

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  • $\begingroup$ Can you use a prime more than once? I guess not, given the equation provided, and if you can it is trivially easy just using 1. And in that case the answer is no because it does not work for n = 2, 3, 6, 7, 8. I guess I am missing something here - I am not a mathematician - and I would be curious to know what. $\endgroup$
    – F2Andy
    Commented Nov 9, 2023 at 13:32
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    $\begingroup$ @F2Andy The point is that $k$ is independent of $n$. For example, it is known that every sufficiently large $n$ is a sum of $5$ prime squares. $\endgroup$
    – GH from MO
    Commented Nov 9, 2023 at 14:49
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    $\begingroup$ @GHfromMO There's some congruence condition, isn't there? Hua's result (Some results in the additive prime-number theory) is that every sufficiently large integer $n\equiv 5 \pmod{24}$ is the sum of five squares of primes. $\endgroup$
    – Timothy Chow
    Commented Nov 9, 2023 at 19:30
  • $\begingroup$ @TimothyChow Yes. Hua showed the 5 prime square result under this congruence condition. Sorry for being sloppy. $\endgroup$
    – GH from MO
    Commented Nov 9, 2023 at 21:06
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    $\begingroup$ See also my comment below the response of gnasher729. $\endgroup$
    – GH from MO
    Commented Nov 14, 2023 at 17:18

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The answer is yes, and this follows from known results concerning the Waring-Goldbach problem.

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You need at least three large primes to write every large s as the sum of three squares of primes; there are only about $n / \ln n$ primes below n, therefore only about $n^2 / (2 \cdot \ln^2 n)$ pairs of two primes, therefore much fewer than $n^2 / (2 \cdot \ln^2 n)$ sums of two squared primes less than $n^2$, we need at least $2 \cdot \ln^2 n$ other primes.

Every prime other than 2 or 3 is $6k ± 1$ and its square is $36 k^2 ± 12k + 1$. Using only primes, and each prime only once, $12k + 11$ could be the sum of 4 plus seven squares of larger primes, eight primes in total. Allowing 1 doesn't help. If we allow the same prime multiple then the worst case is $12k + 10$ which can be the sum of 4 and six squares of larger primes. ($12k + 11$ would not be the worst case because it can be 4 + 4 + 3 squares of large primes).

So 8 non-repeating squares and seven possibly repeating squares of primes may be enough to represent all large integers, 7 non-repeating or 6 repeating squares of primes are not.

The only way that $s = 12k + 0$ is the sum of eight or fewer squares of primes is s equal to 9 plus 3 large squares of primes; there are not that many combinations so we can expect quite large $s = 12k$ that are not the sum of four, and therefore not the sum of eight squares.

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    $\begingroup$ Hua proved in 1938 that every sufficiently large positive integer of the form $24k+5$ can be written as $x_1^2+\dotsb+x_5^2$ where each $x_j$ is a prime number. It follows that every sufficiently large positive integer can be written as $x_1^2+\dotsb+x_9^2$, where each $x_j$ is a prime number or $0$ or $1$. $\endgroup$
    – GH from MO
    Commented Nov 14, 2023 at 17:01

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