Problem (Algebra):
Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}_{>0} \rightarrow \mathbb{Q}_{>0}$ satisfying

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

for all $x, y \in \mathbb{Q}_{>0}$.

Solution:
Take any $a, b \in \mathbb{Q}_{>0}$. By substituting $x=f(a), y=b$ and $x=f(b), y=a$ into $(*)$ we get

$$
f(f(a))^{2} f(b)=f\left(f(a)^{2} f(b)^{2}\right)=f(f(b))^{2} f(a)
$$

which yields

$$
\frac{f(f(a))^{2}}{f(a)}=\frac{f(f(b))^{2}}{f(b)} \quad \text { for all } a, b \in \mathbb{Q}>0
$$

In other words, this shows that there exists a constant $C \in \mathbb{Q}_{>0}$ such that $f(f(a))^{2}=C f(a)$, or

$$
\left(\frac{f(f(a))}{C}\right)^{2}=\frac{f(a)}{C} \quad \text { for all } a \in \mathbb{Q}_{>0}
\tag{1}
$$

Denote by $f^{n}(x)=\underbrace{f(f(\ldots(f}_{n}(x)) \ldots))$ the $n^{\text {th }}$ iteration of $f$. Equality (1) yields

$$
\frac{f(a)}{C}=\left(\frac{f^{2}(a)}{C}\right)^{2}=\left(\frac{f^{3}(a)}{C}\right)^{4}=\cdots=\left(\frac{f^{n+1}(a)}{C}\right)^{2^{n}}
$$

for all positive integer $n$. So, $f(a) / C$ is the $2^{n}$-th power of a rational number for all positive integer $n$. This is impossible unless $f(a) / C=1$, since otherwise the exponent of some prime in the prime decomposition of $f(a) / C$ is not divisible by sufficiently large powers of 2 . Therefore, $f(a)=C$ for all $a \in \mathbb{Q}_{>0}$.

Finally, after substituting $f \equiv C$ into $(*)$ we get $C=C^{3}$, whence $C=1$. So $f(x) \equiv 1$ is the unique function satisfying $(*)$.

Final answer: $f(x)=1$