BxMO Team Selection Test, March 2020

$$2f(y^2)=f(y{)}^2.(4)$$ Using this, we can cancel the $2f(y^2)$ on the left hand side of the original functional equation against the $f(y{)}^2$ on the right hand side: $$f(x^{2}y)=x^{2}f(y).$$ Substituting $y=1$ in this equation yields $f(x^2)=x^{2}f(1)$, and substituting $y=-1$ yields $f(-x^2)=x^{2}f(-1)$. Because $x^2$ takes on all non-negative numbers as value when $x\in \mathbb{R}$, we get $$f(x)=\{\begin{array}{ll}cx& \text{als }x\ge 0,\\ dx& \text{als }x<0,\end{array}$$ with $c=f(1)$ and $d=-f(-1)$. Now substitute $y=1$ in equation (4), which yields $2f(1)=f(1{)}^2$, hence $2c=c^2$. It follows that $c=0$ or $c=2$. If we actually substitute $y=-1$ in equation (4), then we find that $2f(1)=f(-1{)}^2$, hence $2c=(-d{)}^2$. For $c=0$, we get $d=0$, and for $c=2$, we get $d=2$ or $d=-2$. So, there are three cases: $c=0,d=0$: then $f(x)=0$ for all $x$;

$c=2,d=2$: then $f(x)=2x$ for all $x$; * $c=2,d=-2$: then $f(x)=2x$ for $x\ge 0$, and $f(x)=-2x$ for $x<0$, or, in other words, $f(x)=2|x|$ for all $x$. Using the first function, both sides of the functional equation become 0, so this function is a solution. Using the second function, we get $2x^{2}y+4y^2$ on the left hand side, and $(x^2+2y)\cdot 2y=2x^{2}y+4y^2$ on the right hand side, so this function is a solution as well. Using the third function, we get $2|x^{2}y|+4|y^2|=2x^2|y|+4y^2$ on the left hand side, and $(x^2+2|y|)\cdot 2|y|=2x^2|y|+4|y{|}^2=2x^2|y|+4y^2$ on the right hand side, so also this function is a solution. Altogether, we found the three solutions: $f(x)=0$, $f(x)=2x$, and $f(x)=2|x|$. $\square$