Find all function $f:\mathbb{R}\mapsto\mathbb{R}$ such that $f(x^2+y^2)=f(x+y)f(x-y)$.

Question

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 $x=0$.

Answer 1

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.

Answer 2

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