Solving the functional equation $f(x^2+f(y))=(f(x))^2+y$ Find all $ f : R\rightarrow R $ such that $f(x^2+f(y))=(f(x))^2+y, \forall\text{ x,y}\in R$
Thanks in advance!
 A: This type of equation is called functional equations and are very well known to be posted in Olympiads of mathematics, solving them does not require heavy prerequisites it requires only technical methods.

Solution The given equation has only one solutions $f=Id_\Bbb R$.

In order to solve your equation, assume that $f$ is a function such that:
$$\begin{align} \forall x,y\in \Bbb R^2&& f(x^2+f(y))=y +f(x)^2 \tag 1\end{align}$$
From $(1)$ we have $f$ is onto hence we can find an element $a$ such that $f(a)=0$, it follows from $(1)$ that $y=f(a^2+f(y))=f((-a)^2+f(y))=y+f(-a)^2$ hence $f(-a)=0$ now by taking $x=0, y=\mp a$ in $(1)$ we have $f(0)=\mp a+f(0)^2$ this possible only if $a=0$ and hence $f(0)=0$.
Now if we take $x=0$ in $(1)$ we have:
$$\begin{align} \forall x,y\in \Bbb R^2&& f\left(f\left(y\right)\right)=y+f(0)^2=y\tag2\end{align}$$
and we use $(2)$ to expand $(1)$  when $y$ is replaced by $f(y)$ so that:
$$\begin{align} \forall x,y\in \Bbb R^2&&  f\left(x^2+y\right)=f(y)+f(x)^2\tag{3}\end{align}$$
and from $(1)$ again with $y=0$ we have $$f(x^2)=(f(x))^2 \tag 4$$  now given $t$ positive $t=x^2$ in $(3)$ gives :
$$f(t+y)=f(t)+f(y)\tag 5$$
but this equation can be extended to negative reals $t$ using $(4)$ ( To made things clear from $4$ we have $(f(t)^2)=f(-t)^2=f(t^2)$ ad the injectivity from $(2)$ gives us that $f(-t)=-f(t)$)
Return to $(5)$ the functions solutions verify for $f(x)=f(1)x$ for $x\in \Bbb Q$, the relation $(3)$ justifies that $f(x)$ increases monotonously so that $f(x)=f(1)x$ for every real $x\in \mathbb R$. using $(4)$ we have $f(1)\geq 0$ and from $2$ we have $f(1)^2=1$ hence $f(1)=1$.

Exercise : For a given natural number $k > 2$, find all functions $f:\mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$, $f[x^k + f(y)] = y +[f(x)]^k$.

