Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$. Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$.
Some solutions I found are $f\equiv0,f\equiv1$, $f(x)=0$ if $x\neq0$ and $f(x)=1$ if $x=0$.

Perform the substitution $x=\frac{a+b}{2}, y=\frac{a-b}{2}$ to get $f(\frac{a^2+b^2}{2})=f(a)f(b)$. We shall denote this statement with $P(a, b)$.
$P(a, a)$: $f(a^2)=f(a)^2$. In particular, when $a=0$, we have $f(0)=f(0)^2$ so $f(0)=0$ or $1$.
If $f(0)=0$, then $P(a, 0)$: $f(\frac{a^2}{2})=0$, so $f(x)=0 \, \forall x \geq 0$. Thus $f(x)^2=f(x^2)=0 \, \forall x \in \mathbb{R}$, so $f(x)=0 \, \forall x \in \mathbb{R}$.
Otherwise $f(0)=1$. Then $P(a, 0)$: $f(\frac{a^2}{2})=f(a)$. Thus $f(a^2)=f(a)^2=f(\frac{a^2}{2})^2$, so $f(2x)=f(x)^2 \, \forall x \geq 0$. Thus $f(x^2)=f(x)^2=f(2x)=f(\frac{(2x)^2}{2})=f(2x^2) \, \forall x \geq 0$, so $f(2x)=f(x) \, \forall x \geq 0$, so $f(x)=f(2x)=f(x)^2 \forall x \geq 0$. Thus $f(x)=0$ or $1 \, \forall x \geq 0$.
Note that $f(-a)=f(\frac{(-a)^2}{2})=f(\frac{a^2}{2})=f(a)$. If $f(c)=0$ for some $c>0$, then $P(a, c)$: $f(\frac{a^2+c^2}{2})=0$, so $f(x)=0$ for $x \geq \frac{c^2}{2}$. Consider an arbitrary $x>0$. Using $f(2x)=f(x)$, it is easy to prove by induction that $f(2^nx)=f(x) \, \forall n \in \mathbb{Z}^+$. Clearly there exists a positive integer $N_x$ s.t. $2^{N_x}x \geq \frac{c^2}{2}$, so $f(x)=f(2^{N_x}x)=0$. Therefore $f(x)=0 \, \forall x>0$, so $f(x)=0 \, \forall x \not =0, f(0)=1$. 
Otherwise $f(x)=1 \, \forall x \geq 0$, so since $f$ is even, $f(x)=1 \, \forall x \in \mathbb{R}$
To conclude, we have 3 solutions: $f(x)=0 \, \forall x \in \mathbb{R}$, $f(x)=1 \, \forall x \in \mathbb{R}$, and $f(x)=0 \, \forall x \not =0, f(0)=1$. It is easily checked that these are all solutions.
