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Let $$S(n) = \sum_{p \le n} b(n-p),$$ where $b(a)=1$ is $a$ is a sum of two squares of positive integers and $b(a)=0$ otherwise. Trivially by PNT we have $$S(n) \ll \sum_{p \le n}1 \le \frac{n}{\log n}.$$ Could we do better or the above estimate is the best possible?

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The order of magnitude is correct. This paper by C. Hooley gives the asymptotic formula conditioned on GRH, with the order of magnitude $n/(\log n)$. There is also this unconditional result due to Linnik, where he gets a lower bound of the same order of magnitude. Your argument already gives an upper bound of the correct order of magnitude, so this is the exact order of magnitude.

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    $\begingroup$ But the theorem of Linnik and the result of Hooley count the number of solutions, no? Basically as I see Linnik gives asymptotic formula for $$\sum_{p \le n} r(n-p),$$ where $r(n)$ is the number of representations of $n$ as a sum of two squares, while I have $b(n-p)$ instead. $\endgroup$ – toshi Nov 27 '18 at 13:11

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