huyhoang wrote: Find all functions $f: \mathbb{R}\to [0;+\infty)$ such that: \[f(x^2+y^2)=f(x^2-y^2)+f(2xy)\] for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion $f(x^2+y^2)=f(x^2-y^2)+f(2xy)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(0,x)$ $\implies$ $f(x^2)=f(-x^2)$ and so $f(x)$ is even. Let $x\ge y\ge z\ge 0$ (a) : $P(\sqrt{\frac {x+y}2},\sqrt{\frac{x-y}2})$ $\implies$ $f(x)=f(y)+f(\sqrt{x^2-y^2})$ (b) : $P(\sqrt{\frac {y+z}2},\sqrt{\frac{y-z}2})$ $\implies$ $f(y)=f(z)+f(\sqrt{y^2-z^2})$ (c) : $P(\sqrt{\frac {x+z}2},\sqrt{\frac{x-z}2})$ $\implies$ $f(x)=f(z)+f(\sqrt{x^2-z^2})$ (a)+(b)-(c) : $f(\sqrt{x^2-z^2})=f(\sqrt{x^2-y^2})+f(\sqrt{y^2-z^2})$ Writing $f(x)=g(x^2)$, this becomes $g(x+y)=g(x)+g(y)$ $\forall x,y\ge 0$ And since $g(x)\ge 0$, we get $g(x)=ax$ and so $f(x)=ax^2$ $\forall x\ge 0$ and for some $a\ge 0$ And since $f(x)$ is even, we get $\boxed{f(x)=ax^2}$ $\forall x$ and for any real $a\ge 0$ which indeed is a solution.

nice solution, dear pco

Also posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=425331 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=51585

Note that $\forall a,b \exists x,y :x^2-y^2=a, 2xy=b.$ Then the original equation becomes $f(a)+f(b)=f(\sqrt{a^2+b^2}).$ Trivial to get that $f$ is even. For non-negative $a$ setting $f(\sqrt{a})=g(a) \implies g(a^2)+g(b^2)=g(a^2+b^2).$ It's clear that $g$ is a non-negative additive function, thus by Cauchy, $g(x)=kx.$ We have that $\boxed{f(x)=kx^2} \forall x \in \mathbb{R}, k\geq 0,$ which works.