If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$ As we know, for the $$f(x)f(y)=f(x+y)$$  $f(x)=\mathrm e^{\alpha x}$ is a solution.
What about 
$f(x)f(y)=f(\sqrt{x^2+y^2})$? 
Does anybody know about the solution of the function equation?
I tried to find $f(x)$.
See my attempts below to find $f(x)$.
$$f(x)=a_0+a_1x+\frac{a_2x^2}{2!}+\frac{a_3x^3}{3!}+\cdots$$
$$f(y)=a_0+a_1y+\frac{a_2y^2}{2!}+\frac{a_3y^3}{3!}+\cdots$$
$$f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
$$f(\sqrt{x^2+y^2})=a_0+a_1\sqrt{x^2+y^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3(x^2+y^2)^{3/2}}{3!}+\cdots=$$
$$f(\sqrt{x^2+y^2})=a_0+a_1y\sqrt{1+(x/y)^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3y^2(1+(x/y)^2)^{3/2}}{3!}+\cdots=f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
if we use binom expansion for $(1+(x/y)^2)^{m}$
$$(1+(x/y)^2)^{m}=1+\frac{mx^2}{y^2}+\frac{m(m-1)x^4}{2!y^4}+\frac{m(m-1)(m-2)x^6}{3!y^6}+\cdots$$
Let's put the expansion to the equation $f(\sqrt{x^2+y^2})$
$$
\begin{align}
& f(\sqrt{x^2+y^2}) =a_0 + a_1 y \left( 1 + \frac{(1/2)x^2}{y^2} + \frac{(1/2)((1/2)-1)x^4}{2!y^4} \right. \\  \\
& \left. {} + \frac{(1/2)((1/2)-1)((1/2)-2)x^6}{3!y^6} + \cdots\right) + \frac{ a_2 (x^2+y^2)}{2!} \\  \\
&  + \frac{a_3y^2 \left(1+\frac{(3/2)x^2}{y^2}+\frac{(3/2)((3/2)-1)x^4}{2!y^4}+\frac{(3/2)((3/2)-1)((3/2)-2)x^6}{3!y^6}+\cdots\right)}{3!} +\cdots \\  \\
& = a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots
\end{align}
$$
If we equal for all $x^n$ terms in both sides
we can see $a_{2n-1}=0$, but to find  $a_{2n}$  seems hard for me.
Any idea to find $a_{2n}$
Thanks in advice.
 A: Change variable, $g(u) = f(\sqrt{u})$.  You need to decide what you want for negative $u$.  Then this functional equation becomes $g(u+v)=g(u)g(v)$.
A: If you set $$g(x) := f(\sqrt{x})$$ for $x \in [0, \infty)$ then you get $$g(x)g(y) = f(\sqrt{x})f(\sqrt{y}) = f(\sqrt{x+y}) = g(x+y)$$
You see that $g(x) \geq 0$, and if $g(x) = 0$ for some $x > 0$ then $g \equiv 0$. Thus, you can look at $$h(x) := \log(g(x))$$ It satisfies $$h(x) + h(y) = \log(g(x)g(y)) = \log(g(x+y)) = h(x+y)$$ If you impose any reqularity condition on $f$ you can think of, you will get $h(x) = \alpha x$, and consequently $$f(x) = \exp(\alpha x^2)$$ for $x > 0$. You can generalise this result to $x < 0$ using the fact that from the initial equation it follows that $f$ is even.  
A: Let' assume you want solution to $f:\mathbb{R}\mapsto\mathbb{R}$ that satisfies this equation $$\forall x,y\in\mathbb{R},f(x)f(y)=f\left(\sqrt{x^2+y^2}\right)\tag{1}$$
Let's plug in $(x,y)=(0,0)$ in the equation $(1)$,
$$f(0)^2=f(0)\Rightarrow f(0)\in\{0,1\}$$
Case-1 : $f(0)=0$
For $x>0, f(x)=f\left(\sqrt{x^2+0^2}\right)=f(x)f(0)=0$.
For $x<0, f(x)^2=f(x)f(x)=f\left(\sqrt{x^2+x^2}\right)=f\left(\sqrt{2}|x|\right)=0\Rightarrow f(x)=0$.
So, if $f(0)=0$, then $\forall x\in\mathbb{R}, f(x)=0$
Case-2 : $f(0)=1$
In this case, we claim that this function must be even.
Proof :
$f(x)=f(x)f(0)=f\left(\sqrt{x^2+0^2}\right)=f\left(\sqrt{(-x)^2+0^2}\right)=f(-x)f(0)=f(-x)\tag{2}$
Now, we will consider two assumptions under this case.
Assumption-A :
$\exists a>0, f(a)=0$.
Then, $\forall x\geq a$, $$f(x)=f\left(\sqrt{a^2+\left(\sqrt{x^2-a^2}\right)^2}\right)=f(a)f\left(\sqrt{x^2-a^2}\right)=0$$
Assumption-B :
$\exists b>0, f(b)>0$
Then, $$f\left(\sqrt2b\right)=f\left(\sqrt{b^2+b^2}\right)=f(b)^2>0$$
Proceeding this way, we can find arbitrarily large real number $x$ for which $f(x)>0$. Which contradicts with Assumption-A that says after reaching a bound, we can't find and real number $x$ for which $f(x)>0$.
So, we conclude that $a$ and $b$ can't exist simultaneously. If we assume the existence of $a$, We get the solution : $$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$$
We can check and indeed, this is a valid solution.
Now we assume the existence of $b$, that is, $\exists b>0, f(b)>0$. Since this assumption excludes the possibility of the existence of real number $a$ for which $f(a)=0$, we only have two options : either $f(x)>0$ or $f(x)<0$ for any real number $x$. We claim that $\forall x, f(x)>0$.
Proof :
$$f(x)=f\left(|x|\right)=f\left(\sqrt{\left(\frac{x}{\sqrt2}\right)^2+\left(\frac{x}{\sqrt2}\right)^2}\right)=\left(f\left(\frac{x}{\sqrt2}\right)\right)^2>0$$
So, $\forall x, f(x)>0$.
Now, we will plug in $(\sqrt{x},\sqrt{y})$ where $x,y\geq0$ in equation $(1)$,
$$f\left(\sqrt{x}\right)f\left(\sqrt{y}\right)=f\left(\sqrt{x+y}\right)\\\Leftrightarrow \ln\left(f\left(\sqrt{x}\right)\right)+\ln\left(f\left(\sqrt{y}\right)\right)=\ln\left(f\left(\sqrt{x+y}\right)\right)$$
Let's define $\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$ such that $\phi(x)=\ln\left(f\left(\sqrt{x}\right)\right)$. Note that $\phi(x)+\phi(y)=\phi(x+y)$ which satisfies the Cauchy's functional equation. So, $\phi$ is a solution to Cauchy's functional equation. The most trivial solution is $\forall x\geq0,\phi(x)=kx$ where $k\in\mathbb{R}$. There are non trivial solutions which are highly pathological functions.
So, analyzing all the cases we got this solutions :

*

*$\forall x, f(x)=0$

*$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$

*$\forall x, f(x)=e^{\phi(x^2)}$ where $\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$ and $\phi$ satisfies Cauchy's functional equation.

A: The answer to this question is a well known result called Maxwell's theorem, after James Clerk Maxwell.  This earlier question deals with it:
very elementary proof of Maxwell's theorem
