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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.

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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.

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  • $\begingroup$ $f(f(y)) = y$, so $f \circ f$ is a bijection, hence $f$ is a bijection. $\endgroup$ – B. Mehta Mar 17 '18 at 19:25

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