Jalil_Huseynov wrote: Let $P(x,y)=(x^2+y^2)\cdot Q(x,y)$. ... You need first to prove that $x^2+y^2$ divides $P(x,y)$ before writing this (except if you dont assume that $Q(x,y)$ is a polynomial but then your link would be useless)

I definitely have seen this in older literature. With $x=\frac{a+b}{2}$, $y=\frac{a-b}{2}$ we get $P(a,b) = 2P(\frac{a+b}{2},\frac{a-b}{2})$ for any $a$ and $b$. For $x=\frac{a}{2}$, $y=\frac{b}{2}$ we get $P(\frac{a+b}{2},\frac{a-b}{2}) = 2P(\frac{a}{2},\frac{b}{2})$. Therefore $P(x,y) = 4P(\frac{x}{2},\frac{y}{2})$. In particular, for the coefficient $a_{mn}$ at $x^my^n$ we get $a_{mn} = 4a_{mn}\frac{1}{2^m}\frac{1}{2^n}$, i.e. $a_{mn} = 0$ or $m+n=2$. So we must have $P(x,y) = bx^2 + cxy + dy^2$ for some constants $b$, $c$, $d$. Direct verification shows that all solutions are $P(x,y) = (c+d)x^2 + cxy + dy^2$ for arbitrary real constants $c$ and $d$.

See the Iran MO (2nd round ) 2006