### Australian functional equation, 2017

September 16, 2018

Problem. Determine all functions $f:\mathbb{R}\to \mathbb{R}$$f:\mathbb{R} \to \mathbb{R}$ such that $f\left({x}^{2}+f\left(y\right)\right)=f\left(xy\right)$ for all real numbers $x$$x$ and $y$$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$$0$. So, when we set $x=0$$x=0$, we get $f\left(f\left(y\right)\right)=f\left(0\right)$$f(f(y))=f(0)$ for any $y\in \mathbb{R}$$y \in \mathbb{R}$. Also, when we set $y=0$$y=0$, we get $f\left({x}^{2}+f\left(0\right)\right)=f\left(0\right)$$f(x^2+f(0)) = f(0)$ for any $x\in \mathbb{R}$$x \in \mathbb{R}$.

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

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

• If $f\left(y\right)=0$$f(y) = 0$ for any $y\in \mathbb{R}$$y \in \mathbb{R}$ (a special case), then we are done.
• If $f\left({y}_{0}\right)\ne 0$$f(y_0) \neq 0$ for some ${y}_{0}$$y_0$, then for an arbitrary $t\in \mathbb{R}$$t \in \mathbb{R}$, we have $f\left(t\right)=f\left(\frac{t}{f\left({y}_{0}\right)}\cdot f\left({y}_{0}\right)\right)=f\left(0\right)$$f(t) = f(\frac{t}{f(y_0)} \cdot f(y_0)) = f(0)$. Therefore, $f\left(t\right)$$f(t)$ is a constant.

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

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