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. 
 A: 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 $x < 0$, then
$$ f(x+y) = -f(-x-y) = -(f(-x) + f(-y)) = f(x) + f(y). $$

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

If $x > 0$ then by the first identity of the previous step,
$$ f(x) = f((x^{1/2})^2) = f(x^{1/2})^2 > 0. $$
(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.
A: Here is a short approach without Cauchy equation. Starting from $f(0)=0$ by @Leo163, we obtain $f(f(x))=x$ by plugging $y=0$. 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 $f(x)=x$.
A: 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 $y\in\mathbb{R}$. 
A: 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 $f(x^2) = [f(x)]^2$. 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{equation}
\mrm{f}\pars{x^{2} + \mrm{f}\pars{y}} =y + \bracks{\mrm{f}\pars{x}}^{2}
\label{1}\tag{1}
\end{equation}

Deriving both members of \eqref{1} respect of $\ds{x}$ and $\ds{y}$ yield, respectively:
\begin{equation}
\left\{\begin{array}{rcl}
\ds{\mrm{f}'\pars{x^{2} + \mrm{f}\pars{y}}\pars{2x}} & \ds{=} &
\ds{2\mrm{f}\pars{x}\mrm{f}'\pars{x}}
\\[2mm]
\ds{\mrm{f}'\pars{x^{2} + \mrm{f}\pars{y}}\mrm{f}'\pars{y}} & \ds{=} & \ds{1}
\end{array}\right.
\label{2}\tag{2}
\end{equation}

By comparing both expressions in \eqref{2}:
\begin{equation}
{\mrm{f}\pars{x}\mrm{f}'\pars{x} \over x} = {1 \over \mrm{f}'\pars{y}} = \alpha\,,
\qquad
\pars{~\alpha\ \mbox{is independent of}\ x\ \mbox{and}\ y~}
\label{2.a}\tag{2.a}
\end{equation}

The second one is quite trivial:
\begin{equation}
{1 \over \mrm{f}'\pars{y}} = \alpha \implies
\mrm{f}\pars{y} = {1 \over \alpha}\, y + \beta\,,\qquad
\pars{~\beta\ \mbox{is a constant}~}\label{3}\tag{3}
\end{equation}

\eqref{2.a} and \eqref{3} yield:
\begin{align}
&{\mrm{f}\pars{x}\mrm{f}'\pars{x} \over x} = \alpha
\,\,\,\stackrel{\mrm{see}\ \eqref{3}}{\implies}\,\,\,
{\bracks{x/\alpha + \beta}/\alpha \over x} = \alpha \implies
{1 \over \alpha^{2}}\, x + {\beta \over \alpha} = \alpha x \implies
\left\{\begin{array}{rcl}
\ds{1 \over \alpha^{2}} & \ds{=} & \ds{\alpha}
\\[2mm]
\ds{\beta \over \alpha} & \ds{=} & \ds{0}
\end{array}\right.
\\[5mm] &\ \stackrel{\mrm{see}\ \eqref{3}}{\implies}\,\,\,
\bbx{\ds{\mrm{f}\pars{x} = {x \over \alpha}\,,\qquad \alpha^{3} = 1}}
\label{4}\tag{4}
\end{align}

By inserting \eqref{4} into \eqref{1} we conclude that $\ds{\alpha = 1}$:
$$
\bbx{\ds{\mrm{f}\pars{x} = x}}
$$
