let $y=0$,then$f(x^2)=0$ let$x=x+1,y=x$ $f(2x+1)=f(x)f(x+1)$hence$f(2k+1)=0$; let$x=x+2,y=x,f(4x+4)=f(x+2)f(x)$ hence$f(4k)=0$ as for $4k+2$,if two of them x,y,$f(x)f(y)\ne 0$,hence$f(x^2-y^2)\ne 0$contradiction. hence all solutions are: $f(x)=0,x\ne 4k+2$; $c,x=4k+2$ where $k,c\ge 0$ are given.

littletush wrote: let $y=0$,then$f(x^2)=0$ let$x=x+1,y=x$ $f(2x+1)=f(x)f(x+1)$hence$f(2k+1)=0$; @littletush Can you explain more in this step ?

Your solution is not much obvious. We can also write it as \[f(ab)=f\left(\frac{a+b}2\right)f\left(\frac{a-b}2\right)\ \ \ \ (\dagger)\] for $a, b$ of same parity. Set $y=x-1$ since $y<x$. We have $f(2x-1)=f(x)f(x-1)$. Setting $x=1,f(1)=$ and by induction $f(2k-1)=0$ for all $k\ge 1$ because $f(2x+1)=f(x)f(x+1)\implies f(x-1)f(2x+1)=f(x+1)f(2x-1)$. Again, \[f\left((a+1)^2-(2a)^2 \right)=f(a+1)f(2a)=f((a-1)^2)=0\] Set $a$ odd, then $f(a+1)=0$ or $f(2a)=0$.

let f(x)=g(x)-x so FE <=>g(x²-y²)+y²-x²='(g(x)-x)(g(y)-y) f(0)=0 => g(0)=0 if y=0 => g(x²)=x² hence f(x)=0

ryuzaki wrote: let f(x)=g(x)-x so FE <=>g(x²-y²)+y²-x²='(g(x)-x)(g(y)-y) f(0)=0 => g(0)=0 if y=0 => g(x²)=x² hence f(x)=0 Not sure if this is entirely correct, since not all naturals are squares.