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Let $n\ge 2$ an integer. Find the lowest integer $\kappa(n)$ such that every elements in $\mathbb{Z}/n\mathbb{Z}$ can be written as a sum of $\kappa(n)$ squares.

This statement can be found in the "smf 2017", a french contest. I was just wondering if it is a well-known result or are there references about it.

Thanks in advance !

Matthew Conroy
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Maman
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    Lagrange's four-square theorem suggests $\kappa(n) \leq 4$. Are nonzero squares required? Have you tried calculating $\kappa(n)$ for a few small $n$ and looking that up in the OEIS? – The Short One Jun 10 '17 at 19:23
  • May help: For $n=p$ a prime, $\kappa(p)=2$. (for fixed residue $m$, the sets of residues $x^2$ and $m-y^2$ each have $\frac {p+1}2$ elements hence must overlap). – lulu Jun 10 '17 at 19:26
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    Could you please add a link to this contest? Thanks! – Matthew Conroy Jun 10 '17 at 20:31
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    The answer is surprisingly not elegant. There are different cases according to whether : $8$ divides $n$, primes dividing $n$ that are $=3$ mod $4$ appear once or more in the decomposition of $n$, etc. If you want I can provide a full solution, but I think that's not what you want. If I understand correctly the rules of this contest, the questions aren't in the literature, and there should be no references to it besides the contest itself – Maxime Ramzi Jun 10 '17 at 20:41
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    @MatthewConroy http://smf.emath.fr/files/smf_junior_sujets.pdf problem 9 – Maman Jun 10 '17 at 20:47
  • @Max This is a theorem of Gauss, and it has appeared in the literature. Just reading the comments now. If I should delete my answer as this is a contest, let me know. – JavaMan Jun 10 '17 at 20:52
  • @Max it was just a curiosity, we see in class the case of two squares, four squares and for three squares it was more complicated. – Maman Jun 10 '17 at 20:56
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    @JavaMan : oh my bad, I guess the rules weren't applied then. You don't have to delete your answer, the contest is over now – Maxime Ramzi Jun 10 '17 at 21:00

2 Answers2

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In Charles Small's 1977 paper "Solutions of Waring's Problem $\bmod m$", Small called the following "Essentially a Theorem of Gauss":

$$ \kappa(n) \leq \left\{ \begin{array}{ccl} 1 & &n=2 \\ 2 & & p^2 \mid n \implies p \equiv 1 \pmod 4 \\ 3 & & 8 \nmid n \\ 4 & & \text{ all } n \end{array} \right. $$

JavaMan
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The statement is a straightforward consequence of Kneser's theorem.
Since $$\mathbb{Z}/(n\mathbb{Z})\simeq\mathbb{Z}/(2^{\alpha_0}\mathbb{Z})\times \mathbb{Z}/(p_1^{\alpha_1}\mathbb{Z}) \times \ldots \times \mathbb{Z}/(p_m^{\alpha_m}\mathbb{Z})$$ by the Chinese remainder theorem, the set $Q$ of quadratic residues in $\mathbb{Z}/(n\mathbb{Z})$ fulfills $$\left|Q\right| \geq \prod_{k=1}^{m}\frac{p_m^{\alpha_m}+1}{2} $$ and $\underbrace{Q+Q+\ldots+Q}_{\kappa(n)\text{ times}}\supseteq \mathbb{Z}/(n\mathbb{Z})$ for some $\kappa(n)\leq 2^{m+\alpha_0}$.

Jack D'Aurizio
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  • I don't follow this answer. Of course CRT and the bound on $|Q|$ is fine, but the bound $\kappa(n) < 2^{m+a_0}$ is quite suboptimal if $m+a_0 \geq 2$. Is there a deep connection I am missing here? For example, does this give an upper bound on the problem for cubes? – JavaMan Jun 15 '17 at 16:04
  • @JavaMan: I never said it was an optimal bound: it is very far from being optimal, indeed. I just wanted to show a straightforward technique for proving the claim. – Jack D'Aurizio Jun 15 '17 at 17:03