Celebrating 2020 with a functional equation.

Making $y=0$ in the equation $f(x)f(y)=g(\sqrt{x^2+y^2}),$ we obtain that $2020f(x)=g(\sqrt{x^2}).\quad\quad \quad (*)$

Then we can rewrite the given functional equation as $f(x)f(y)=2020f(\sqrt{x^2+y^2})$ . Now by making $h(x)=\frac{1}{2020}f(x),$ and substitution into the previous equation and multiplying both side of the equation by a convenient number, we get the equation $h(x)h(y)=h(\sqrt{x^2+y^2}). \quad\quad (**)$ As we have done it before, replacing the $y$ by 0 in the last equation, we obtain that $h(x)=h(\sqrt{x^2})$ and therefore there is a continuous function $m$ defined on the interval $[0, \infty]$ with positive values, such that $h(x)= m(x^2).$ Then the equation $(**)$ written in terms of $m$ turns to be $m(x^2)m(y^2)= m(x^2+y^2),$ Making $t=x^2$ and $s=y^2,$ it follows that $m(t)m(s)=m(t+s).$ It is known that if $m$ is defined and continuous on $[0, \infty)$ with positive values and satisfies the previous functional equation , then $m(t)=e^{kt},$ where $k$ is a real constant. Then using backsubstitution, we obtain that $f(x)=2020h(x)=2020m(x^2)=2020e^{kx^2}$

Now using the fact that $f(1)= \frac{2020}{e},$ we obtain the equation $e^k=\frac{1}{e},$ and, therefore, $k=-1.$ Then the answer is $\left\lfloor 10000 f(\pi)\right\rfloor=\left\lfloor 10000 *2020 e^{-1\pi^2 }\right\rfloor =\boxed{1044}.$

If the function $f(x)$ is continuous, then the function $g(\sqrt{x^2+y^2})$ is also continuous. This allows us to proceed as follows. COnsider:

$f(x)f(y) = g(\sqrt{x^2+y^2})$

Partially differentiating the above with respect to x gives:

$f'(x)f(y) = \frac{x g'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}} \ \dots (1)$

Partially differentiating the above with respect to y gives:

$f(x)f'(y) = \frac{y g'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}} \ \dots (2)$

Dividing (1) and (2) and rearranging gives:

$\frac{f'(x)}{x \ f(x)} = \frac{f'(y)}{y \ f(y)}$

Replacing $y$ by any constant reduces the right-hand side to a constant (say $A$ ) thus leading to:

$\frac{f'(x)}{f(x)}=Ax$

Integrating both sides and simplifying leads to:

$f(x) = B\mathrm{e}^{Ax^2}$

Where $B$ is an arbitrary constant. Now $f(0) = 2020$ , thus $B = 2020$ .

$f(x) = 2020\mathrm{e}^{Ax^2}$

Now $f(1) = 2020/\mathrm{e}$ and this leads to $A = -1$ . Therefore, the solution is:

$f(x) = 2020\mathrm{e}^{-x^2}$

From here, the required answer follows.

$f(\pi) \approx 1044.8$