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Let $k\geq 4$. As usual, let $r_k(n)$ denote the number of ways to represent $n$ as the sum of $k$ squares. Is this true that for every $\varepsilon>0$, one has $r_k(n) \gg n^{\frac{k}{2}-1-\varepsilon}$ ? Is there an easy proof of this fact? What can we say about the cases $k=2,3$?

Many thanks !

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  • $\begingroup$ For $k=2$, some positive integers are not the sum of two squares and some are in very few ways. $\endgroup$
    – joro
    May 29, 2016 at 12:38
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    $\begingroup$ This appears false for $k=4$ for primes. Check en.wikipedia.org/wiki/Jacobi%27s_four-square_theorem $\endgroup$
    – joro
    May 29, 2016 at 12:40
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    $\begingroup$ For $k=4$ what you want follows from Jacobi; and then for all $k\ge 5$ follows easily by induction on $k$ (for example). For $k=3$ the number of representations of $n$ as a sum of three squares is a class number; and the inequality holds by Siegel's theorem. $\endgroup$
    – Lucia
    May 29, 2016 at 15:52
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    $\begingroup$ @Lucia: Of course for $k=4$ there is a $2$-adic obstruction. The lower bound is fine if $n$ is not divisible by a fixed power of $2$. $\endgroup$
    – GH from MO
    May 29, 2016 at 19:32
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    $\begingroup$ @Stabilo: For the Hardy-Littlewood circle method, the standard reference is Vaughan's book. You can find the asymptotic formula for $k\geq 5$ there (and also for the sum of $m$-th powers when $k\geq 2^m+1$). $\endgroup$
    – GH from MO
    May 30, 2016 at 15:18

2 Answers 2

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For $k=4$, your statement would be that $r_4(n) \gg n^{1-\epsilon}$. This is false. Jacobi's four-square theorem can be stated as that $r_4(n)/8$ is the sum of the divisors of $n$ that are not divisible by $4$. Let $n = 2^t m$ with $m$ odd and $t$ positive. $r_4(n) = 24 \sigma (m)$ independent of $t$. In particular, for $m=1$, you can express $n=2^t$ as a sum of $4$ squares in $24$ ways, scaled up versions of the $24$ ways to express $2$ and $4$ as sums of $4$ squares. $24$ doesn't grow with $n$.

For $k=4$, $n$ odd, the statement is trivially true (after Jacobi's four-square theorem) since the odd divisors of $n$ include $n$ itself.

For $k \gt 4$ this follows easily by reducing to the case of $k=4$, $n$ odd. Choose the first $k-4$ squares to be at most $\frac{n/2}{k-4}$ so that the remainder is odd. This can be done in about $\frac{1}{2} \left(\frac{n/2}{k-4}\right)^{(k-4)/2}$ ways. You can complete each of these to a representation of $n$ as a sum of $k$ squares in at least $n/2$ ways, so $r_k(n) \ge c(k) n^{k/2-1}$.

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  • $\begingroup$ That is what I was looking for, many thanks! $\endgroup$
    – Stabilo
    May 30, 2016 at 9:07
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Complementing Douglas Zare's answer, for $k=3$ see the responses here.

Regarding the case of $k=2$, the number of representations $r_2(n)$ as a sum of two squares is often zero (so there is no lower bound), but it behaves much like $d_2(n)$. Indeed, superficially the first quantity counts the number of representations $n=a^2+b^2$, while the second quantity counts the number of representations $n=ab$. At a deeper level, $r_2(n)$ equals $4\sum_{d\mid n}\chi(n)$, where $\chi$ is the nontrivial Dirichlet character modulo $4$, while $d_2(n)$ equals $\sum_{d\mid n} 1$ by definition. At any rate, the large values of $r_2(n)$ are much the same as those of $d_2(n)$, i.e. these functions have similar maximal order etc. (For $r_2(n)$ the largest values are produced by the products $n=q_1q_2\dots q_k$, where $q_1<q_2<\dots$ is the sequence of primes congruent to $1$ modulo $4$.)

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  • $\begingroup$ I like this point of view, thank you for sharing it! $\endgroup$
    – Stabilo
    May 30, 2016 at 9:09
  • $\begingroup$ @GH from Mo, on a slightly different note, would you happen to have a nice reference in mind for the computation of $r_2(n)$? $\endgroup$
    – user147650
    Jan 31, 2020 at 21:08
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    $\begingroup$ @user96343: See Theorem 278 in Hardy-Wright: An introduction to the theory of numbers. $\endgroup$
    – GH from MO
    Jan 31, 2020 at 22:43
  • $\begingroup$ Thank you very much! $\endgroup$
    – user147650
    Feb 1, 2020 at 12:55

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