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Prove that, for any $N$, there exists an integer $n$ such that $r_2(n)>N$.

I don't know how to go about this problem, I only know the definition of $r_2(n)$, i.e. "the number of representations of $n$ as a sum of two squares of integers."

Example: $r_2(1)=4,\; r_2(5)=8$.

I've been using Page 9 of this document as a reference point, but I don't know how to go about a proof for this.

Jack D'Aurizio
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K.Garn
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    Maybe include the definition of r2(n). (I know it but it could make sense for certain readers.) You can [edit] the post. – quid Nov 02 '17 at 01:01
  • Also if $n = \prod_{j=1}^l p_j$ where the $p_j$ are distincts prime $\equiv 1 \bmod 4$ then $r_2(n) = 2^{l-1}$. This is so because $p_1p_2 = ((a_1+ib_1)(a_2+ib_2))((a_1-ib_1)(a_2-ib_2))$ $=((a_1+ib_1)(a_2-ib_2))((a_1-ib_1)(a_2+ib_2))$ so $p_1p_2$ is the sum of two squares in two different ways – reuns Nov 02 '17 at 02:11

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If you assume that $r_2(n)$ is bounded by some constant $M$ you have that the following sum

$$ \sum_{n=1}^{N}r_2(n), $$ counting the number of lattice points in the region $\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq N\}$, is bounded by $MN$. On the other hand, by Gauss circle problem a simple lower bound for such sum is given by $\pi N-2\sqrt{2N}$, so $M\geq 3$. Since $\mathbb{Z}[i]$ is a Euclidean domain it is a UFD too, and $r_2(n)$ turns out to be a multiple of a multiplicative function, namely $$ r_2(n) = 4\sum_{d\mid n}\chi_4(d),\qquad \chi_4(n)=\left\{\begin{array}{rcl}1 &\text{if}& n\equiv 1\pmod{4}\\ -1 &\text{if}& n\equiv 3\pmod{4}\\0 &&\text{otherwise}.\end{array}\right. $$ In particular $$ r_2(5^k) = 4k+4$$ $$ r_2(5\cdot 13\cdot 17\cdots p_k)=2^{k+2} $$ and $r_2$ is clearly unbounded.

Jack D'Aurizio
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    In case of interest: https://math.stackexchange.com/questions/2349068/which-numbers-less-than-5-billion-have-the-most-representations-as-the-sums-of-t/2349955#2349955 Nice Hat. – Will Jagy Nov 02 '17 at 01:52