### Australian functional equation, 2017

**Problem.** Determine all functions $f:\mathbb{R}\to \mathbb{R}$ such that
$$f({x}^{2}+f(y))=f(xy)$$
for all real numbers $x$ and $y$.

Origin: Australian Mathematical Olympiad 2017, problem 3.

**Solution.** The first thing we need to try in problems like this (functional equations) is to set variables to some specific numbers. The most obvious number to try here is $0$. So, when we set $x=0$, we get $f(f(y))=f(0)$ for any $y\in \mathbb{R}$. Also, when we set $y=0$, we get $f({x}^{2}+f(0))=f(0)$ for any $x\in \mathbb{R}$.

Taking the latter equation, reversing its left and right sides, and replacing $f(0)$ with $f(f(y))$ from the former equation (notice that this is for an arbitrary $y$), we get $$f(0)=f({x}^{2}+f(0))=f({x}^{2}+f(f(y))).$$ Now we can apply the original equation from the problem setting (putting $f(y)$ instead of $y$ there), so we continue the chain of equalities: $$f(0)=f({x}^{2}+f(0))=f({x}^{2}+f(f(y)))=f(xf(y)).$$ So, $f(xf(y))=f(0)$ for any $x,y\in \mathbb{R}$.

Now, the idea is to represent an arbitrary $t\in \mathbb{R}$ as $xf(y)$ for some $x$ and $y$. We have two cases:

- If $f(y)=0$ for any $y\in \mathbb{R}$ (a special case), then we are done.
- If $f({y}_{0})\ne 0$ for some ${y}_{0}$, then for an arbitrary $t\in \mathbb{R}$, we have $f(t)=f(\frac{t}{f({y}_{0})}\cdot f({y}_{0}))=f(0)$. Therefore, $f(t)$ is a constant.

So, $f(x)=C$ for an arbitrary constant $C$. Notice, that this includes the special case mentioned above when $C=0$.

Finally, verifying that all such functions indeed satisfy the original condition.