EhsanNamdari wrote: Find all functions $f:R \to R$ such that: $f(f(x)-y^{2})=f(x)^{2}-2f(x)y^{2}+f(f(y))$ $f(x)=0$ $\forall x$ is a solution and let us from now look for non all zero solutions. Let $P(x,y)$ be the assertion $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(f(y))$ Let $A=f(\mathbb R)$ $P(0,0)$ $\implies$ $f(f(0))=f(0)^2+f(f(0))$ and so $f(0)=0$ $P(y,0)$ $\implies$ $f(f(y))=f(y)^2$ and so $f(x)=x^2$ $\forall x\in A$ So $P(x,y)$ may be written $f(f(x)-y^2)=f(x)^2-2f(x)y^2+f(y)^2$ Let $y=f(z)\in A$ : $f(y)=y^2$ and so the above equality may be written : $f(f(x)-f(y))=f(x)^2-2f(x)f(y)+f(y)^2$ And so $f(f(x)-f(y))=(f(x)-f(y))^2$ $\forall x\in \mathbb R,y\in A$ Let then $u$ such that $f(u)\ne 0$ : $f(f(u))=f(u)^2$ and so $f(v)>0$ where $v=f(u)$ Let then $t\le 0$ and $s$ such that $s^2=\frac{f(v)^2-t}{2f(v)}$ $P(v,s)$ $\implies$ $f(f(v)-s^2)-f(f(s))=t$ and so any non positive real $t$ may be written as $t=f(x)-f(y)$ with $x\in\mathbb R$ and $y\in A$ And since we previously got that $f(f(x)-f(y))=(f(x)-f(y))^2$ $\forall x\in \mathbb R,y\in A$, we now have : $f(x)=x^2$ $\forall x\le 0$ But then $x\in A$ $\forall x\ge 0$ since $x=f(-\sqrt x)$ and since we know that $f(x)=x^2$ $\forall x\in A$, we get $f(x)=x^2$ $\forall x\ge 0$ And so $f(x)=x^2$ $\forall x$ which indeed is a solution. Hence the solutions : $\boxed{f(x)=0}$ and $\boxed{f(x)=x^2}$

EhsanNamdari wrote: Find all functions $f:R \to R$ such that: $f(f(x)-y^{2})=f(x)^{2}-2f(x)y^{2}+f(f(y))$ I forgot to say that this problem was proposed by Mohammad Jafari!

My solution from WOOT: Let $P(x,y)$ denote the given assertion. $\boxed{f(x)=0}$ is a solution, otherwise $\exists j:f(j)\ne0$. $P(0,0)\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(f(x))=f(x)^2$ $P(f(j),x)\Rightarrow f(f(j)^2-x^2)-f(x)^2=f(j)^4-2f(j)^2x^2$, and so every real $x\in(-\infty,f(j)^4]$ can be written as: $$x=f(u)-f(v)^2$$for some $u,v\in\mathbb R$. $P(u,f(v))\Rightarrow f(x)=x^2\forall x\in(-\infty,f(j)^4]$ For any $x$, we repeat this with $f(j)$ sufficiently negative. Since $x$ is contained in $(-\infty,f(j)^4)$, we have $\boxed{f(x)=x^2}$ for all $x$, which fits.