I want to know number of ordered pairs $(x,y)$ satisfying $x^2 +y^2 =a$.
Let $n(a)$ be the number of ordered pairs $(x,y)$ satisfying $x^2 +y^2 =a$.
If $x^2 +y^2 =a$ and $z^2 +w^2 =b$, then $(xz+yw)^2 +(xw-yz)^2 =ab$.
So, I guess that if $a,b$ is coprime, $n(ab)=n(a)n(b)$.
To prove it, I need to prove if $x^2 +y^2 =a$ and $z^2 +w^2 =ab$, then $\frac{xz+yw}{a}$ is integer.
How can I prove it?
edit: It can't hold...
The number $n(a)$ is equal to zero for all $a$ tabulated at OEIS A022544. So we have no solutions for $$ a=3,6,7,11,12,14,15,19,21,22,23,\cdots $$ In fact, a positive integer $a$ is the sum of two squares if and only if each prime factor $p$ of $a$ such that $p \equiv 3 \bmod 4$ occurs to an even power in the prime factorization of $a$. Also for such integers, the number $n(a)$ can be computed. In general we have the formula $$ n(a)=r_2(a)=4(d_1(a)-d_3(n)), $$ where $d_j(a):=\sum_{d\mid c,d\equiv j(4)}1$.