The function given here is $f(x)f(y)=f\left(\sqrt{x^2+y^2}\right)$ where $x,y\in\mathbb{R}$ and $f:\mathbb{R}\mapsto\mathbb{R}$
A trivial solution is $\forall x\in\mathbb{R},f(x)=0$.
So, let's assume $\exists x\in\mathbb{R}$ such that $f(x)\neq0$. Let's assume, $f(a)\neq0$ where $a\in\mathbb{R}$.
$\therefore f(a)f(x)=f(a)f(-x)=f\left(\sqrt{a^2+x^2}\right)\Rightarrow f(x)=f(-x)$.
Let's plug in $x=y\geq0$ in the defining equation of $f$,
$f(\sqrt2x)=f(x)^2=f(-\sqrt2x)\geq0$.
So, $f$ is an even function and $\forall x,f(x)\geq0$.
Let's plug in $x=y=0$ in the defining equation.
$f(0)=f(0)^2\Rightarrow f(0)\in\{0,1\}$
If $f(0)=0$, we can say, $0=f(0)f(a)=f\left(\sqrt{0^2+a^2}\right)=f(a)$ which contradicts our assumption that $f(a)\neq0$.
So, the other case must be true which is, $f(0)=1$.
By a few case checking, we can show that this is also a valid solution:
$$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$$
So, let's assume $\exists a\in\mathbb{R}$ such that $a>0$ and $f(a)>0$.
But $f(a)>0\Rightarrow f(a)^2>0\Rightarrow f\left(\sqrt2a\right)>0$.
By proceeding this way, we can find arbitrarily large positive real number $x$ for which $f(x)>0$
Now assume, $\exists b$ such that $f(b)=0$.
$\therefore \forall x\geq0, 0=f(b)f(x)=f\left(\sqrt{b^2+x^2}\right)$
That means $\forall y\geq b, f(y)=0$ since we can always find $x\geq0$ such that $y=\sqrt{b^2+x^2}$.
But this arises a contradiction. the existence of $a$ implies that we can find arbitrarily large real number $x$ for which $f(x)>0$, in the other hand, the existence of $b$ implies that $\forall x\geq b, f(x)=0$. Both statement can't be true simultaneously.
Now, existence of $b$ gives us the piecewise function we got earlier.
And existence of $a$ implies that $\forall x\in\mathbb{R}, f(x)>0$.
Now we are assuming that this $a$ exists. We get the following from this assumption
Now, let's assume $x,y\in\mathbb{R}_{\geq0}$
We got, $f\left(\sqrt{x}\right)f\left(\sqrt{y}\right)=f\left(\sqrt{x+y}\right)$
$\Leftrightarrow \ln{f\left(\sqrt{x}\right)}+\ln{f\left(\sqrt{y}\right)}=\ln{f\left(\sqrt{x+y}\right)}$
Let's define $\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$ to be the function $\phi(x)=\ln{f\left(\sqrt{x}\right)}$
Since $\forall x,y\in\mathbb{R}_{\geq0}, \phi(x)+\phi(y)=\phi(x+y)$, The function $\phi$ satisfies the
Cauchy's functional equation,
$\phi$ is a solution to Cauchy's functional equation.
The most trivial such function is $\forall x\in\mathbb{R}_{\geq0},\phi(x)=kx$ where $k\in\mathbb{R}$.
For the other non trivial solutions, there is no closed formula. Let's denote all of them by just $\phi$.
So, the solutions to the given function are:
1. $\forall x, f(x)=0$
2. $$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$$
3. $f(x)=e^{\phi(x^2)}$ where $\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$ and $\phi$ satisfies Cauchy's functional equation.
tanmoy bhaiya ektu confirm den problem to hoiche kina)
