# Find all functions of defined on the set of all real numbers with real values

Find all functions of defined on the set of all real numbers with real values, such that
$$f(x^2+f(y)) = y + (f(x))^2$$

My attempt:
putting $x = 0$, $f(f(y)) = y + f^2(0)$ and $f(x^2 + f(0)) = f^2(x)$
but i can't go any further.

Suppose $a$ is such that $f(a)=0$ (we know there is such $a$, it is enough to consider $y=-f(x)^2$ for some $x$), then substituting $x=a$ and $y=a$, we have $$f(a^2)=a.$$ Then, set $x=0$ and $y=a^2$, to obtain $$0=f(f(a^2))=a^2+f(0)^2.$$ Since we are only dealing with real numbers, it means that $f(0)=0$, and no other number has $0$ as image.
As consequences, we have $f(f(y))=y$ and $f(y^2)=f(y)^2$ for every $y\in\mathbb{R}$.
This shows $f$ is a bijection. Now plug $y=f^{-1}(\zeta)$, we see $$f(x^2+\zeta)=f^{-1}(\zeta)+(f(x))^2\geq f^{-1}(\zeta)=f(\zeta)$$ Hence $f$ is monotone increasing. Now suppose for some $x_0$ we have $f(x_0)> x_0$, then $x_0=f(f(x_0))\geq f(x_0)> x_0$, a contradiction. Thus we have $f(x)\leq x$. Similarly we have $f(x)\geq x$, so $f(x)=x$ is the only solution.
• $f(f(y)) = y$, so $f \circ f$ is a bijection, hence $f$ is a bijection. – B. Mehta Mar 17 '18 at 19:25