Ovchinnikov Denis wrote: Found all functions $f: \mathbb{R} \to \mathbb{R}$, such that for any $x,y \in \mathbb{R}$ holds next equality: $f(x^2+xy+f(y))=f^2(x)+xf(y)+y$. Let $P(x,y)$ be the assertion $f(x^2+xy+f(y))=f^2(x)+xf(y)+y$ 1) $f(0)=0$ =========== Let $f(0)=a$ (a) : $P(4a,0)$ $\implies$ $f(16a^2+a)=f^2(4a)+4a^2$ (b) : $P(-4a,0)$ $\implies$ $f(16a^2+a)=f^2(-4a)-4a^2$ (c) : $P(-4a,4a)$ $\implies$ $f(f(4a))=f^2(-4a)-4af(4a)+4a$ (d) : $P(0,4a)$ $\implies$ $f(f(4a))=4a+a^2$ (a)-(b)+(c)-(d) : $0=(f(4a)-2a)^2+3a^2$ and so $a=0$ Q.E.D. 2) $f(x)=x$ =========== (a) : $P(0,x)$ $\implies$ $f(f(x))=x$ and so $f(x)$ is a bijection (b) : $P(-x,x)$ $\implies$ $f(f(x))=f^2(-x)-xf(x)+x$ (c) : $P(x,0)$ $\implies$ $f(x^2)=f^2(x)$ (d) : $P(-x,0)$ $\implies$ $f(x^2)=f^2(-x)$ -(a)+(b)+(c)-(d) : $0=f(x)(f(x)-x)$ Since $f(0)=0$ and $f(x)$ is a bijection, we have $f(x)\ne 0$ $\forall x\ne 0$ and the above equality becomes $f(x)=x$ $\forall x\ne 0$ And so $f(x)=x$ $\forall x$ which indeed is a solution.