, 2003

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

Note that
$$f(1)f(2n-1) = f(n^2) \quad \text{and} \quad f(3)f(2n-1) = f((n+1)^2)$$
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 $n \in \mathbb{N}$.