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Let $r_2(n)$ denote the number of ways in which a positive integer $n$ is expressed as the sum of two squares (integers). I would like to know if there is a result that gives the exact behavior of (as $X\to\infty$) $$ \sum_{n\leq X}r_2(n^2+d^2) $$ for a fixed $d\in\mathbb{N}$. I think that it is expected to have the behaviour $A(d)X \log X + B(d) X + ERROR$. I am interested in the constants $A(d)$ and $B(d)$. Do you know if some paper establishes this result?

Thank you very much

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    $\begingroup$ There is not much difference between $r_2(n^2+d^2)$ and $\tau(n^2+d^2)$ with $\tau(n)$ being the divisor function. The divisor function on quadratic polynomials is well known -- see for example Hooley's beautiful paper projecteuclid.org/euclid.acta/1485889355 . (You should also be able to evaluate $r_2$ on general quadratic polynomials -- I don't know a reference offhand.) $\endgroup$ Commented Sep 14, 2020 at 1:50

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Just to expand a bit on Lucia's comment: It is well known that $$r_2(n) = 4(\tau_1(n)-\tau_3(n)) = 4\tau(n) - 8\tau_3(n)$$ Where $d\tau_1(n)$ is the number of divisors of $n$ that are $1\mod4$, $\tau_3(n)$ likewise for divisors that are $3\mod4$, and $\tau(n)$ is the ordinary divisor function.

The paper referenced in Lucia's comment shows that $$\sum_{n\leq X}{\tau(n^2+d^2)} = A(d)X\log X + B(d)X + O(x^{8/9}\log^3(x))$$ So it will suffice to consider the sum $$\sum_{n\leq X}{\tau_3(n^2+d^2)}$$ However, notice that $\tau_3(n^2+d^2) = \tau_3(\gcd(n,d)^2)$, because $\tau_3$ is multiplicative and no prime number $q=3\mod4$ divides a sum of two coprime squares. So $\tau_3(n^2+d^2)$ is periodic$\mod n$, and therefore our sum is approximately linear (with an error of $O(1)$) as requested.

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  • $\begingroup$ The second equality in your first equation seems to assume $n$ odd (e.g. we have $\tau_1(2)=1, \tau_3(2) =0, \tau (2) =2$). I don't think this causes a big problem though as one can also handle the even divisors separately. $\endgroup$ Commented Sep 14, 2020 at 23:44
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    $\begingroup$ It is not true that $\tau_3(n^2+d^2) = \tau_3(\gcd(n,d)^2)$. For $(n,d)=(6,3)$, the LHS is $3+15=18$, while the RHS is $3$. $\endgroup$ Commented Sep 15, 2020 at 0:41
  • $\begingroup$ Oh dear I shouldn't post when I'm tired. Fixing it soon. $\endgroup$ Commented Sep 15, 2020 at 13:31
  • $\begingroup$ Thank you very much for helping me. $\endgroup$ Commented Nov 4, 2020 at 0:57

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