Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy $f\left(a^{2}+b^{2}\right)=f(a) f(b)$ and $f\left(a^{2}\right)=f(a)^{2}$ for all positive integers $a$ and $b$.
The answer is $f \equiv 1$ only, which works. We prove it's the only one.
The bulk of the problem is:
Claim - If $f(a)=f(b)=1$ and $a>b$, then $f\left(a^{2}-b^{2}\right)=f(2 a b)=1$.

Proof. Write
$$
\begin{aligned}
1=f(a) f(b) & =f\left(a^{2}+b^{2}\right)=\sqrt{f\left(\left(a^{2}+b^{2}\right)^{2}\right)} \\
& =\sqrt{f\left(\left(a^{2}-b^{2}\right)^{2}+(2 a b)^{2}\right)} \\
& =\sqrt{f\left(a^{2}-b^{2}\right) f(2 a b)} .
\end{aligned}
$$

By setting $a=b=1$ in the given statement we get $f(1)=f(2)=1$. Now a simple induction on $n$ shows $f(n)=1$ :
- If $n=2 k$ take $(u, v)=(k, 1)$ hence $2 u v=n$.
- If $n=2 k+1$ take $(u, v)=(k+1, k)$ hence $u^{2}-v^{2}=n$.