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Nov 11, 2019 at 10:01 vote accept asad
Jan 14, 2018 at 13:19 comment added GH from MO @asad: $A(x)$ equals the integral part of $\sqrt{x-1}$, which explain their choice $X=\sqrt{x}$. The number of solutions of the congruence $m^2+1\equiv 0\pmod{p}$ equals $1$ when $p=2$, equals $2$ when $p\equiv 1\pmod{4}$, and equals $0$ when $p\equiv 3\pmod{4}$. Therefore $A_p(x)/\sqrt{x}$ tends to $1/2$ when $p=2$, tends to $2/p$ when $p\equiv 1\pmod{4}$, and tends to $0$ when $p\equiv 3\pmod{4}$. This explains their choices of $g(p)$, since the idea is that $A_p(x)\approx g(p)X$.
Jan 14, 2018 at 4:52 comment added asad thanks, but I did not get it yet, for instance why they chose $g(2)=1/2$ and $g(p)=2/p$, and $X=\sqrt{x}$
Jan 13, 2018 at 20:30 history answered GH from MO CC BY-SA 3.0