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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 !

GH from MO
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Stabilo
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  • For $k=2$, some positive integers are not the sum of two squares and some are in very few ways. – joro May 29 '16 at 12:38
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    This appears false for $k=4$ for primes. Check https://en.wikipedia.org/wiki/Jacobi%27s_four-square_theorem – joro May 29 '16 at 12:40
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    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. – Lucia May 29 '16 at 15:52
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    @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 $2$. – GH from MO May 29 '16 at 19:32
  • Jacobis identity (and modular forms and dimension counting) is not the only way to approach this problem. See the Hardy Littlewood circle method, which can be much more flexible in certain situations. – George Shakan May 29 '16 at 19:58
  • For $k\ge5$, the circle method actually gives an asymptotic formula for $r_k(n)$ which is even stronger than the desired lower bound. – Greg Martin May 30 '16 at 07:54
  • @joro: for $k=4$ and $n$ an odd prime, the formula you linked to gives $r_k(n) = 8(n+1) \gg n^{1-\varepsilon}$. – Greg Martin May 30 '16 at 07:55
  • Thanks a lot for your comments, I will check the Hardy-Littlewood circle method. @GregMartin, do you have some references for me? – Stabilo May 30 '16 at 09:06
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    @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 $k\geq 2^m+1$). – GH from MO May 30 '16 at 15:18

2 Answers2

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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 $r_k(n) \ge c(k) n^{k/2-1}$.

Douglas Zare
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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 $4$.)

Community
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GH from MO
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