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Let $n $ be even and define $$ Q(n)=\sum_{\substack{ p,q \ \textrm{ primes} \\p+q=n }}\left(\frac{p}{q} \right),$$ where $\left(\frac{p}{q} \right)$ is the quadratic Legendre symbol.

Has this sum been studied?

Here one would wish for
$$\lim_{n\to\infty}\frac{Q(n)}{\#\{ p,q \ \textrm{ primes}: p+q=n \} }=0 .$$ I do not have any connection to anything but I was wondering merely out of curiosity. If $n$ is a multiple of $4$ and $n>4$ one can show that $$\left(\frac{p}{q} \right)= \left(\frac{n}{p} \right)= \left(\frac{n}{q} \right) $$ for all terms in the sum, hence, perhaps the quadratic symbol may be slightly biased depending on $n$. The second moment $\sum_{n\leq x } Q(n)^2$ gives rise to sums of the form $$ \sum_{p,p',p'' \leq x} \left(\frac{p}{p'} \right)\left(\frac{p+p'}{p''} \right)$$ which I am sure ought to be $o(\pi(x)^3)$, thus possibly $Q(n)=o(n/\log n)$ for almost all $n$ a la binary Goldbach for almost all even $n$.

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    $\begingroup$ Just to be sure: The fractions here denote the Legendre symbol? $\endgroup$
    – HenrikRüping
    Commented Jul 7, 2022 at 13:39
  • $\begingroup$ I'm not sure what $R(n)$ is, but $Q(n)$ is obviously $O(n\log\log n/(\log n)^2)$, since $n$ has at most this many representations as a sum of two primes. $\endgroup$
    – Wojowu
    Commented Jul 7, 2022 at 15:12
  • $\begingroup$ @HenrikRüping. yes (edited) $\endgroup$
    – Dr. Pi
    Commented Jul 7, 2022 at 15:36
  • $\begingroup$ @Wojowu. Yes I clearly meant whether one can get a saving over the trivial upper bound where the quadratic symbol is replaced by $1$. Edited. $\endgroup$
    – Dr. Pi
    Commented Jul 7, 2022 at 15:37
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    $\begingroup$ Questions of this sort has been considered in the past. The "quadratic pattern" you mention looks a lot like the "spin" symbol introduced by Friedlander and Iwaniec in their landmark 1998 paper. There they proved that summing the spin over $x,y$ with $x^2 + y^4$ a prime is actually very small, getting a power-saving. $\endgroup$
    – Stanley Yao Xiao
    Commented Jul 7, 2022 at 23:42

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