Functions which satisfy $f(m^2+n^2)=f(m)^2+f(n)^2$

Question

Find all $f$: $f(m^2+n^2)$=$f(m)^2+f(n)^2$ for all $m,n\in \mathbb{N_0}$ where
$f(1)>0$,
$f\in\mathbb{N_0}$ for all inputs

I did $f(0)=2f(0)^2\implies f(0)=0$

$f(1)=f(1)^2\implies f(1)=1$

From this we can get $f(2)=2, f(4)=4$ etc. easily and also use some tricks like

$f(2^2+1^2)=f(5)=5, f(4)=4$

so $f(3)=\sqrt{5^2-4^2}=3$

I tried using the parametrization of Pythagorean triples but am unable to prove that $f(7)=7$

Please help

Answer 1

We see that $f(2n^2) = 2f^2(n)$. So $f(50)=2f^2(5)=2(25)=50$ So $f(50)=50$. But $50 = 1+49$. So $f(50) = 1 + f^2(7)$.

So $f(7)=7$