Functional Equation From MOP 2004

[mutasimmim]

Find all functions $f:R\rightarrow R$ such that for all $x,y$, $f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$

[Phlembac Adib Hasan]

Obviously $f(x)=0$ is the only constant solution. Now we look for the other ones. Suppose $p(x,y)$ denotes the original equation. Under these assumptions, it can be easily proved that if $f(d)=0$, then $d\in \{0,1\}$.

$\bullet$ If $f(a)=f(b)\neq 0$, then $P(1,a)$ and $p(1,b)$ imply together $a^2-a=b^2-b\Longleftrightarrow a=b$ or $a+b=1$.
$\bullet$ $f(1)$ must be zero. If not, $P(x,1)$ gives a function $f(x)=f(1)x^2$ which does not satisfy $P(x,y)$.
Now $P(\frac 1 x,x)$ gives either $f(x)=x^2$ or $f(x)=x-x^2$.
Suppose $\exists c:f(c)=c^2$. $P(1,c)\Longrightarrow f(c^2)=c^4-c^3+c^2={\LARGE\{}\begin{array}{lr}c^4\\c^2-c^4\end{array}$. Solving for both values gives us $c\in \{0,1,\frac 1 2\}$. We already know $f(1)=1-1^2=0$ and $0,\frac 1 2$ are the roots of $x^2=x-x^2$. Hence $f(x)=x-x^2\;\forall x\in \mathbb R$.