## Task 1.

Determine all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ such that

$$
f\left(x^{2}(f(y))^{2}\right)=(f(x))^{2} f(y), \quad \text { for all } x, y \in \mathbb{Q}^{+}
$$

$\left(\mathbb{Q}^{+}\right.$ is the notation for the set of all positive rational numbers.)
## First Solution.

Substitute $f(x)$ for $x$ in the initial equation and we get

$$
f\left(f(x)^{2} f(y)^{2}\right)=f(f(x))^{2} f(y)
$$

If we swap the roles of $x$ and $y$ in the above equality, we get

$$
f(f(y))^{2} f(x)=f\left(f(x)^{2} f(y)^{2}\right)=f(f(x))^{2} f(y)
$$

which implies

$$
\frac{f(x)}{f(y)}=\frac{(f(f(x)))^{2}}{(f(f(y)))^{2}}=\frac{\left(f^{2}(x)\right)^{2}}{\left(f^{2}(y)\right)^{2}}
$$

where $f^{k}(x)$ denotes the $k$-fold application of the function $f$.

By induction, it can be shown that

$$
\frac{f(x)}{f(y)}=\frac{\left(f^{n}(x)\right)^{2^{n-1}}}{\left(f^{n}(y)\right)^{2^{n-1}}}
$$

for all natural numbers $n$. Indeed, if we assume that the statement holds for some natural number $n$, substituting $f(x)$ and $f(y)$ for $x$ and $y$ respectively and squaring, we get

$$
\frac{f(x)}{f(y)}=\frac{f(f(x))^{2}}{f(f(y))^{2}}=\frac{\left(f^{n+1}(x)\right)^{2^{n}}}{\left(f^{n}(y)\right)^{2^{n}}}
$$

thus proving the induction statement.

This means that the number $f(x) / f(y) 2^{n}$ is a power of some rational number, for every natural number $n$. The only such positive rational number is 1. This holds for arbitrary $x$ and $y$, implying that the function is constant. Substituting into the initial equation, we see that the only constant that satisfies it is 1. Therefore, the only solution is $f(x)=1, x \in \mathbb{Q}^{+}$.