Find all functions $ f:\mathbb{R} \mapsto \mathbb{R} $ such that $f[x^2+f(y)] = y +[f(x)]^2$ I read this question in the book 'Problem Solving Strategies ' by Arthur Engel. This question was asked in IMO of 1992. Here's how it goes
Find all the functions  $ f:\mathbb{R} \mapsto \mathbb{R} $ that satisfy
$f[x^2+f(y)] = y +[f(x)]^2$ $ \ \ \ \ x,y \in \mathbb{R}$
The book simply states that it has no solution but it is not a sufficient answer. How can I prove that no such function exists ?


This partially answers the amended question with a different requirement: $$f(x^2+f(y)) = y+f(x)^2$$
Let $x=0$ to obtain $f(f(y)) = y + f(0)^2$. Let $y=0$ to obtain $f(x^2+f(0)) = f(x)^2$.

Suppose $f$ has a fixed point: $y=f(y)$. Then $f(x^2+y) = y+f(x)^2$, so $f(y) = y+f(0)^2$ by letting $x=0$, and hence $f(0) = 0$. Therefore $f(x^2) = f(x)^2$ for all $x$, and $f(f(y)) = y$ for all $y$; in particular, $f$ is bijective.
But then $f(-x)^2 = f((-x)^2) = f(x^2) = f(x)^2$, so $f(-x) = \pm f(x)$; since $f$ is injective, that means $f(-x) = -f(x)$, so $f$ is odd.
Now letting $y=f(u)$, obtain $f(x^2+u) = f(u)+f(x)^2$, which is to say $f(x^2+u) = f(u) + f(x^2)$; therefore $f(z+u) = f(z) + f(u)$ whenever $z$ is nonnegative. By oddness of $f$, this also holds when both $z$ and $u$ are negative; so we have shown that $f$ is self-inverse, has $f(0) = 0$, has $f(x)^2 = f(x^2)$, and distributes over $+$. (These conditions imply the original requirement; but we are still operating under the assumption that $f$ has a fixed point.)
Since $f$ is odd, that means $f(x) = \sqrt{f(x^2)}$ when $x \geq 0$, and $-\sqrt{f(x^2)}$ when $x < 0$.
Noting then that $f(1)^2 = f(1)$, we have $f(1) = \pm 1$; but $f$ is nonnegative when given nonnegative input since $f(x)^2 = f(x^2)$, so $f$ is the identity on the rationals (by following the reasoning of my answer to the other version of this question). Therefore it is also the identity on the inductively defined set $X = \cup X_n$ where $X_1 = \mathbb{Q}, X_{n+1} = X_n \cup \{\sqrt{x}: x \in X_n\} \cup \{x+q : x \in X_n\} \cup \{x^2 : x \in X_n\}$. But (for example) $\pi$ is not in this set, and I think $f$'s value is not determined on $\pi$.

If $f$ does not have a fixed point, I'm afraid I've got nowhere.
