# Sieve bound for the sum of two squares

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?

## 1 Answer

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.

• 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. – toshi Nov 27 '18 at 13:11