USA IMO 2003

Function $f(n)=1$, for all $n\in \mathbb{N}$, is the only function satisfying the conditions of the problem.

Note that $$f(1)f(2n-1)=f(n^2)\text{and}f(3)f(2n-1)=f((n+1{)}^2)$$ for $n\ge 3$. Thus $$\frac{f(3)}{f(1)}=\frac{f((n+1{)}^2)}{f(n^2)}.$$ Setting $\frac{f(3)}{f(1)}=k$ yields $f(n^2)=k^{n-3}f(9)$ for $n\ge 3$. Similarly, for all $h\ge 1$, $$\frac{f(h+2)}{f(h)}=\frac{f((m+1{)}^2)}{f(m^2)}$$ for sufficiently large $m$ and is thus also $k$. Hence $f(2h)=k^{h-1}f(2)$ and $f(2h+1)=k^{h}f(1)$.

But $$\frac{f(25)}{f(9)}=\frac{f(25)}{f(23)}\cdots \frac{f(11)}{f(9)}=k^8$$ and $$\frac{f(25)}{f(9)}=\frac{f(25)}{f(16)}\cdot \frac{f(16)}{f(9)}=k^2,$$ so $k=1$ and $f(16)=f(9)$. This implies that $f(2h+1)=f(1)=f(2)=f(2j)$ for all $j,h$, so $f$ is constant. From the original functional equation it is then clear that $f(n)=1$ for all $n\in \mathbb{N}$.