### 2018 IMO SL #A1

Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f:\mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$ satisfying$$f(x^2f(y)^2)=f(x)^2f(y)$$for all $x,y\in\mathbb{Q}_{>0}$

The answer is $\boxed{f \equiv 1}$. We can easily check that this works. Let $P(x, y)$ denote the assertion. From $f\left(x,\frac{1}{f(1)}\right)=1$, we obtain $f(x)^2=f(x^2)$. Because of this, we can say that $$f(y)=\frac{f((xf(y))^2)}{f(x)^2}=\frac{f(xf(y))^2}{f(x)^2}$$
$$=\left(\frac{f(xf(y))}{f(x)}\right)^2,$$ implying that $f(y)$ is the square of some rational number. Since $f$ outputs $\mathbb{Q}_{>0}$, let $f(y)=g(y)^2$. Repeating the operation we did above once again, we find that $g(y)$ is the square of some rational number as well. Similarly, we can repeat this process indefinitely and then finally we will be able to conclude that that $f(y)=1$ is the only solution.

We are done. $\square$