Let $\tau$ be the divisor function, that is $$ \tau(n)=\sharp\{d \in \mathbb{N}, d|n\}. $$

I was wondering if anyone has ever proved an asymptotic estimate for the sum $$S(x):=\sum_{p,q\leq x}\tau(p^2+q^2),$$ where the summation is taken over pairs of primes.

One obviously expects $$S(x)\sim c\frac{x^2}{\log x}$$ as $x \to \infty,$ where $c$ is a positive constant which is an infinite product of Euler factors.

This is based on the heuristic that each of the $\pi(x)^2$ terms present in $S(x)$ is approximated by a constant multiple of $\log x$ on average.

Brun-Titchmarsch and Bombieri-Vinogradov can be used to prove the upper and the lower bound $$ c\frac{x^2}{\log x} (\frac{1}{2}+o(1)) \leq S(x) \leq c\frac{x^2}{\log x} (2+o(1)), $$ as $x\to \infty$ respectively.

But the question remains, $\textit{can we prove an asymptotic?}$

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    $\begingroup$ Actually, I have a different related question. By Fouvry and Iwaniec, we know that $a^2+p^2$ is infinitely often prime. Can we show the same for $p^2+q^2$, that it is infinitely often twice a prime? $\endgroup$ Aug 11, 2014 at 20:56
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    $\begingroup$ @Lucia: Can you be a bit more specific? $\endgroup$
    – GH from MO
    Aug 11, 2014 at 22:11
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    $\begingroup$ Getting ready for the Big Push, Darling? $\endgroup$
    – Asaf Karagila
    Aug 12, 2014 at 3:19
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    $\begingroup$ @Gerry: Computing asymptotics for the partial sums of $\tau(p+a)$, with $a$ fixed, is known as the Titchmarsh divisor problem. This was solved by Linnik, but nowadays can be done by the Brun--Titchmarsh and Bombieri--Vinogradov theorems. As for $\tau(x^2+y^2)$, it's maybe more natural to consider the sum extended over all pairs $x,y$ with $x^2+y^2 \le n$. That can be attacked by applying a mean-vale theorem of Wirsing to the function $n\mapsto \tau(n) r(n)$, where $r(n) = \frac{1}{4} \#\{(x,y):x^2+y^2=n\}$. (Note that $r$ is multiplicative.) $\endgroup$ Aug 12, 2014 at 5:33
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    $\begingroup$ @CaptainDarling: The asymptotic for $\sum_{n\leq x}\tau(n)r(n)$ is easy to get by Mellin inversion, because $\sum_{n=1}^\infty \tau(n)r(n)n^{-s}=\zeta(s)^2L(s,\chi_4)^2/L(2s,\chi_4)$. Perhaps you thought the same, but I found the wording "manifestly not untrivial" slightly confusing: did you mean "manifestly trivial"? $\endgroup$
    – GH from MO
    Aug 12, 2014 at 10:44

2 Answers 2


An answer regarding the use of large sieve suggested by Lucia (too long for a comment).

I guess her/his thought was along the following lines:

The divisors $d$ of $p^2+q^2$ are of order $x^2$ and the hyperbola trick reduces to estimating sums of the form $$\sum_{d\leq x}\sum_{\lambda^2=-1 (d)} \sum_{\substack{p,q \leq x \\ p= \lambda q (d)}} 1. $$ Inserting multiplicative characters in the sum over primes we will be left with an error term coming from the primitive characters, which looks like $$ \sum_{d\leq x}\frac{1}{\phi(d)} \sum^*_{\chi(d)} \sum_{\lambda^2=-1(d)}\overline{\chi(\lambda)} \ \Big| \sum_{p\leq x}\chi(p)\Big|^2 ,$$ which behaves as $$\frac{1}{x} \sum_{d\leq x}\tau(d)\frac{d}{\phi(d)} \sum^*_{\chi(d)} \ \Big| \sum_{p\leq x}\chi(p)\Big|^2.$$ Ideally we would like to show that this is $o(x^2/\log x)$. However the large sieve in the form of
[Th.4,p.g. 160, multiplicative Davenport] gives a bound for this quantity which is of order $O(x^2)$ and is therefore inadequate. Essentially this is the level of distribution problem.

What we described can show that the contribution of $d\leq x/(\log x)^A$ to the sums with primitive characters is indeed $o(x^2/\log x)$ for an appropriate value of $A>0$ but I am not sure whether the micro logarithmic savings coming from Hooley's-Delta function can be used to say something about the remaining range.

  • $\begingroup$ You're right; this is what I had in mind. But the range between $x/(\log x)^A$ and $x$ is problematic. But then in your question you state an upper bound for $S(x)$ which is not clear to me -- Brun-Titchmarsh doesn't seem to imply the bound, and one is off by a $\log \log x$ factor I think. (Because B-T would save only a factor of $\log (ex/d)$ and this is problematic for the large $d$.) $\endgroup$
    – Lucia
    Aug 12, 2014 at 14:36
  • $\begingroup$ Also, the large sieve argument would give the right asymptotic as a lower bound. (Note that you also have to use Siegel-Walfisz for small values of $d$; the large sieve works in the range $(\log x)^A \le d \le x/(\log x)^A$. $\endgroup$
    – Lucia
    Aug 12, 2014 at 15:21

When computing e.g. an asymptotic for $\sum_{p\leq x}d(p-1)$ you would like to estimate the number of primes $p$ such that $n$ divides $p-1$ by the prime number theorem as $\sim\frac{x}{\varphi(n)\log x}$. We do not know GRH, so we can't use this estimate for all $n<x^{1/2-\epsilon}$, but we do have Bombieri-Vinogradov, so we can use this estimate for almost all $n$ in the relevant range.

This approach does not work here, since for $\sum d(p^2+q^2)$ we have to take $n$ as large as $x^{1-\epsilon}$, however, the distribution of $p^2+q^2$ is a lot nicer than the distribution of $p$. So instead of a non-trivial bound for $\max_a\left|\pi(x, a, n) - \frac{x}{\varphi(n)\log x}\right|$ it suffices to bound $$ \sum_{a^2+b^2\equiv 0\pmod{n}} \pi(x,a,n)\pi(x, b, n) - \frac{x^2}{\varphi(n)^2\log^2 x}\sum_{a^2+b^2\equiv 0\pmod{n}} 1 $$ for almost all $n$. Since we are now averaging over $a, b$, we should be able to avoid Bombieri-Vinogradov but use the Barban-Davenport-Halberstam theorem, which is applicable to all $n$ up to $x/\log^A x$, which suffices.

Filling in the detail will probably be quite some work, but I guess that two or three pages of Cauchy-Schwarz and character calculations should do the trick.

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    $\begingroup$ Dear JCSP, this is indeed helpful. But what is to be done with the remaining moduli $x/\log^Ax<n<x$? $\endgroup$
    – Dr. Pi
    Aug 12, 2014 at 12:08
  • $\begingroup$ For this range an upper bound sieve should suffice. The real quantity is smaller than the main term by a factor $\frac{\log x}{\log\log x}$, so loosing a constant does not matter. Hence we can replace the set of primes e.g. by all integers having only prime factors $\geq x^{0.01}$. To do so by Selbergs sieve you need asymptotics for the number of solutions of $a^2+b^2\equiv 0\pmod{n}$, $a,b,\leq x$, $q|ab$, where $n$ and $q$ satisfy $n<x$, $q<$x^{0.02}$, which looks doable. $\endgroup$ Aug 13, 2014 at 14:03

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