https://artofproblemsolving.com/community/q1_%22f(f(x%5E2)%2By%2Bf(y))%22

CatalinBordea wrote: Find all nondecreasing functions $ f:\mathbb{R}\longrightarrow\mathbb{R} $ that verify the relation $$ f\left( f\left( x^2 \right) +y+f(y) \right) =x^2+2f(y) , $$for any real numbers $ x,y. $ Another unpublished (seems to me) rather simple solution : Let $P(x,y)$ be the assertion $f(f(x^2)+y+f(y))=x^2+2f(y)$ If $\exists m$ such that $f(x)\ge m\quad\forall x$, then $\exists c=\lim_{x\to-\infty}f(x)$ (since nondecreasing) Setting then $y\to-\infty$ in $P(x,y)$, we get $LHS\to c$ while $RHS\to x^2+2c$, impossible So $f(x)$ is not lowerbounded So $P(x,y)$ implies surjectivity Let then $x\in\mathbb R$ and $y$ such that $f(y)=-f(x^2)$ (since surjective) $P(x,y)$ $\implies$ $f(y)=-x^2$ and so $f(x^2)=x^2$ So $f(x)=x\quad\forall x\ge 0$ Setting then in $P(y,x)$ $y^2\ge -x-f(x)$, we get $\boxed{f(x)=x\quad\forall x}$ Which indeed is a solution

I claim that $f(x) = x$ is our only answer. First off, I will show that $f$ is surjective by showing there is no upper bound $\mathcal{U}$ and no lower bound $\mathcal{L}$. Suppose there is an upper bound $\mathcal{U}$. Then if we fix $x = 0$ and choose $y \to \infty$, we see that $f(y)$ approaches $\mathcal{U}$, so we get that the LHS becomes\[f(f(0) + y + f(y)) \to f(\infty) \to \mathcal{U}\]and the RHS approaches $2\mathcal{U}$ so $\mathcal{U}$ must be equal to $2\mathcal{U}$ thus $\mathcal{U} = 0$. However, we get a contradiction when we pick a finite $y$ and infinitely large $x$ to get that\[f(f(x^2) + y + f(y)) = x^2 + 2f(y) > 0\]so the upper bound is in fact not $0$, a contradiction. Next, similarly, suppose there is a lower bound $\mathcal{L}$. Again fix $x = 0$ and choose $y \to -\infty$, so that $f(y)$ approaces $\mathcal{L}$ so the LHS becomes\[f(f(0) + y + f(y)) \to f(-\infty) \to \mathcal{L}\]and the RHS approaches $2\mathcal{L}$ so $\mathcal{L} = 2\mathcal{L} \implies \mathcal{L} = 0$. Rearrange to get\[f(y) = \tfrac12(f(f(x^2) + y + f(y)) - x^2).\]Take $x$ to be a large constant and $y \to -\infty$. Then $f(y)$ actually approaches $\tfrac12(f(-\infty) - x^2) < 0$ hence $0$ is not the lower bound, a contradiction. Thus $f$ is unbounded and is therefore surjective. So we can plug in $(x, y)$ for which $f(x^2) + f(y) = 0$. This yields\[f(y) = x^2 + 2f(y) \implies f(y) = -x^2 \implies f(x^2) = x^2.\]This is possible for all real $x$ so thus $f(c) = c$ for all $c \geq 0$. Now for each $y = -c$ for some positive $c$, choose $x$ to be large. We have\[x^2 - c + f(-c) = x^2 + 2f(-c) \implies f(-c) = -c\]as desired. Hence $f(c) = c$ for all real $c$, as desired. $\blacksquare$

Turkey National MO 2012 p3