Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$ Let $\mathbb R$ denote the set of all real numbers. Find all function $f: R\to \ R$ such that   $$f\left(x^2+f(y)\right)=y+(f(x))^2$$   It is the problem. I tried to it by putting many at the place of $x$ and $y$ but I can't proceed. Please somebody help me.
Let me summarize the proposed ideas and give a self-contained answer: Let $f : \Bbb{R} \to \Bbb{R}$ satisfy the functional equation

$$ f(x^2 + f(y)) = y + f(x)^2 \tag{*}$$

for all $x, y \in \Bbb{R}$.


  Step 1. $f(0) = 0$.


(This solution is due to @Leo163.) Let $a \in \Bbb{R}$ be such that $f(a) = 0$. For instance, plugging $y = -f(x)^2$ to $\text{(*)}$ confirms the existence of such $a$. Then plugging $(x, y) = (a, a)$ to $\text{(*)}$ gives

$$ f(a^2) = f(a^2 + f(a)) = a + f(a)^2 = a.$$

Finally, plugging $(x, y) = (0, a^2)$ to $\text{(*)}$ gives

$$ 0 = f(a) = f(f(a^2)) = a^2 + f(0)^2. $$

This shows that $f(0) = 0$.


  Step 2. $f$ satisfies various identities:
  
  
  $f(x^2) = f(x)^2$.
  $f(f(x)) = x$. In particular, $f$ is bijective.
  $f(-x) = -f(x)$.
  $f(x+y) = f(x)+f(y)$.
  


The identity 1 (resp. 2) follow by plugging $y = 0$ (resp. $x = 0$) to $\text{(*)}$. For 3, we may assume that $x \geq 0$. Then replacing $(x, y)$ in $\text{(*)}$ by $(x^{1/2}, f(-x))$, we have

$$ 0 = f(x - x) = f(-x) + f(x^{1/2})^2 = f(-x) + f(x) $$

and 3 follows. Finally, replacing $(x, y)$ in $\text{(*)}$ by $(x^{1/2}, f(y))$ gives

$$ f(x + y) = f(x) + f(y), \qquad \forall x \geq 0, \ y \in \Bbb{R}. $$

The restriction that $x \geq 0$ can be removed by combining this with 3: if $x &lt; 0$, then

$$ f(x+y) = -f(-x-y) = -(f(-x) + f(-y)) = f(x) + f(y). $$


  Step 3. $f(x) = x$.


If $x &gt; 0$ then by the first identity of the previous step,

$$ f(x) = f((x^{1/2})^2) = f(x^{1/2})^2 &gt; 0. $$

(We can exclude the equality because $f$ is bijective and $f(0) = 0$.) Combining this with the fact that $f$ is an odd function, we have:

$$ f(x) \geq 0 \quad \Longleftrightarrow \quad x \geq 0.$$

Now for any $x \in \Bbb{R}$,

$$ f(x) -x  = f(x) - f(f(x)) = f(x - f(x)). $$

Therefore $f(x) - x \geq 0$ if and only if $x - f(x) \geq 0$, which implies that $f(x) = x$.



Remark. The Cauchy functional equation has pathological solutions which are also involution, and we need to use the extra structure given by the equation $\text{(*)}$ to exclude such possibilities. The first identity of Step 2 was essential in our solution.
Observe


$f(x^2+f(0)) = [f(x)]^2$
$f(f(x)) = x+[f(0)]^2$.


Plugging $0$ into expression 1 yields
\begin{align}
f_2(0):=f(f(0)) = [f(0)]^2
\end{align}
and plugging $f(0)$ into expression 1 yields
\begin{align}
f(f(0)^2+f(0)) =[f_2(0)]^2 = [f(0)]^4. 
\end{align}
Moreover, by expression 2, we have
\begin{align}
f(f(0)^2+f(0))= f([f(0)]^2)=f(f_2(0)) = f_3(0).
\end{align}
Hence we have
\begin{align}
f_3(0) = [f(0)]^4.
\end{align}

Next, using both 1 and 2, we have
\begin{align}
f(f(x^2+f(0)))=x^2+f(0)+[f(0)]^2. \ \ (\ast)
\end{align}
Plugging $0$ into $(\ast)$ yields
\begin{align}
f_3(0) = f(0)+[f(0)]^2.
\end{align}
Combining everything yields
\begin{align}
[f(0)]^4= f_3(0) = f(0)+[f(0)]^2 \ \ \Rightarrow \ \ f(0)[f(0)^3-f(0)-1] =0.
\end{align}
Solving for $f(0)$ yields that either $f(0) = 0$ or $f(0)$ is a root of the polynomial $p(x) = x^3-x-1$, which only has one real root.

Moreover, observe
\begin{align}
f_4(0) = [f(0)]^4
\end{align}
and
\begin{align}
f_4(0) = f(f(f_2(0))) = f_2(0) + [f(0)]^2 = 2[f(0)]^2
\end{align}
which means
\begin{align}
[f(0)]^4-2[f(0)]^2 = 0.
\end{align}
Thus, $f(0)$ is also a root of $x^4-2x^2 = x^2(x^2-2)$. 

In conclusion, $f(0)$ has to be $0$. Thus, $f(f(x)) = x$ and $f(x^2) = [f(x)]^2$.