Vlados021 wrote: Find all real numbers $a$ for which there exists a function $f$ defined on the set of all real numbers which takes as its values all real numbers exactly once and satisfies the equality $$ f(f(x))=x^2f(x)+ax^2 $$for all real $x$. Let $P(x)$ be the assertion $f(f(x))=x^2f(x)+ax^2$ Problem statement ask for bijective solutions (as I understood) Since bijective, $\exists u$ such that $f(u)=-a$ $P(0)$ $\implies$ $f(f(0))=0$ $P(u)$ $\implies$ $f(f(u))=0$ So $u=0$ and so $f(0)=-a$ and $f(-a)=0$ $P(-a)$ $\implies$ $-a=a^3$ and so $a=0$ And since $f(x)=x|x|$ is a solution when $a=0$, we get the answer $\boxed{a=0}$

Can you find and show a solution when a=0 , satisfy f is a bijective function ?

Wangta123 wrote: Can you take a solution when a=0 , satisfy f is a bijective function ? Uhhh, did you read my post ? I gave you the example $f(x)=x|x|$ which is trivially bijective. What more are you requesting ?