<s>[INST] 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 $x=0$.
 [/INST] let $x=y=0 \to f(0)=f^2(0) \to f(0)=0$ or $ f(0)=1  $,
case I:$f(0)=0$
let $x-y=0  \to f(2x^2)=f(2x)*f(0)=0 \to f(x)=0$ when $x \ge 0 $, 
let $x+y=u,x-y=v \to f(\dfrac{u^2+v^2}{2})=f(u)f(v), $if $v=c <0 ,f(c) \not=0$, $ f(u)=\dfrac{f(\dfrac{u^2+c^2}{2})}{f(c)}=f(-u) \to f(x)=0 $ when $x<0$, contradict. so $f(x)=0$ for all $x$.
EDIT: following is 3rd version and I think it is OK now.
case II: $f(0)=1 \to f(x^2)=(f(x))^2$ and $f(x)=f(\dfrac{x^2}{2})$ and $f(x)=f(-x)$
for any $x>\sqrt{2}$  we can have $ f(x)=(f(\sqrt{x}))^2=(f(x^{\frac{1}{2n}}))^{2n} $, when n is big enough, we have $x^{\frac{1}{2n}} < \sqrt{2}$
for any $x<\sqrt{2}$ we have $f(x)= f(\dfrac{x^2}{2})$, so $\dfrac{x^2}{2}<1$
for any $x<1 \to f(x)=f(\dfrac{x^2}{2})=f(\dfrac{x^{2n}}{2^{2n+1}})$, when n is big enough ,we have $f(x)=C$ when $x \to 0$
that gives for any $x>0, f(x)=C$, and $C=C^2 \to C=0 $ or $ C=1$
if $f(0)$ is not continue with $f(x)$ when $x=0$, we have $f(x)=0$ when $x\not=0$,or $f(x)=1$
so it proof that the OP's guess. QED
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