Suppose, $n\ge 24$ is a natual number. Then, $n$ can be written as a sum of squares of primes (the squares of the primes need not be distinct) because $n$ can always be written in the form $4a+9b$ with non-negative integers $a,b$. So, we can find primes (not necessarily distinct) such that $$(1)\ \ \ \ p_1^2+p_2^2+\cdots+p_k^2=n$$

But is there a positive integer $k$, such that every positive integer $n\ge16$ can be written as a sum of at most $k$ squares of a prime ? In other words, a number $k$ , such that $(1)$ has always a solution ?

  • $\begingroup$ According to my calculations so far, $k=7$ is not sufficient (take $n=32$), but $k=8$ might work. $\endgroup$
    – Peter
    Mar 11, 2017 at 23:21
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    $\begingroup$ This is not a trivial problem, it involves Weyl's and Vinogradov's bounds on exponential sums over squares or primes and Hardy's circle method. Due to arithmetic constraints $\pmod{8}$, my bet is that every sufficiently large positive integer can be written as the sum of $8$ prime squares, but probably "sufficiently large" means $n\geq N$ with $N\gg 16$. $\endgroup$ Mar 11, 2017 at 23:22
  • $\begingroup$ @JackD'Aurizio I currently search an $n$, for which $k=8$ is not sufficient. $\endgroup$
    – Peter
    Mar 11, 2017 at 23:23
  • $\begingroup$ It looks like a good idea to add some heuristics and classical methods in additive number theory for dealing with the $p^2$ terms, if there are any. $\endgroup$ Mar 11, 2017 at 23:26
  • $\begingroup$ what would happen if n is decomposed into 4 squares and then each of these squares is in turn decomposed into 4 squares...until we end up only with primes. $\endgroup$
    – user25406
    Mar 12, 2017 at 0:48

1 Answer 1


This is known as the "Waring-Goldbach problem". Waring's problem is of course your problem without the prime constraint, and Goldbach's problem is your problem for first powers. This article by Buttcane states that at least seven, and no more than nine, squares of primes are necessary to express every sufficiently large integer.

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    $\begingroup$ Just to clarify, I am not an expert on this. I just know how to use Google and had some time to kill. $\endgroup$ Mar 11, 2017 at 23:31

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