How find this all function $f(x^n+2f(y))=(f(x))^n+y+f(y)$ Question:

Given a positive integer $n\ge 2$ . Find all functions $f:R\to R$, such that
  $$f(x^n+2f(y))=(f(x))^n+y+f(y)$$

let $x=0,y=0,a=f(0)$ then
$$f(2f(0))=(f(0))^n+0+f(0)\Longrightarrow f(2a)=a^n+a=[a^{n-1}+1]a$$
if let $y\to -f(y),x=0$,then
$$f(0+2f(-f(y)))=a^n+0\Longrightarrow f(2f(-f(y)))=a^n$$
Thank you 
it is clear $f(x)=x$ is solution.so I fell 
$f(x)$ is such Cauchy's functional equation:Cauchy's functional equation
and  I found maybe can use this methods?http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=398427&p=2219933
 A: Assume $f$ is continuous, the only solution for the functional equation is $f(z) = z$.
Let $f$ be any solution of the functional equation
$$f(x^n + 2f(y)) = f(x)^n + y + f(y) \tag{*1}$$
Assume $f$ is continuous, we have


*

*$f $ is injective.
$$\begin{align}
f(y_1) = f(y_2) &\implies f(x^n + 2f(y_1)) = f(x^n+2f(y_2))\\
&\iff f(x)^n + y_1 + f(y_1) = f(x)^n + y_2 + f(y_2)\\
&\implies y_1 = y_2
\end{align}
$$

*f is surjective.  
Assume the contrary, then $f$ is either bounded from below or from above.  


*

*if $f$ is bounded from below by $m = \inf\,\{ f(z) : z \in \mathbb{R} \}$, we have
$$f(x)^n + y + f(y) = f(x^n + 2 f(y)) \ge m\tag{*2}$$
Fixing $x$ and send $y$ to $-\infty$, we will have $f(y) \to +\infty$ in order for LHS$(*2)$
stay bounded from below. Since $f$ is injective, this forces $f$ to be strictly monotonic
decreasing. Since $f$ is bounded from below, we get
$$\lim_{z\to+\infty} f(z) = \liminf_{z\to+\infty} f(z) = m$$
In $(*2)$, fixing $y$ and send $x$ to $+\infty$, we obtain
$$m^n + y + f(y) = m \quad\implies\quad f(y) = m - m^n - y$$
This contradicts with the assumption $f$ is bounded from below.

*If $f$ is bounded from above by $M = \sup\,\{ f(z) : z \in \mathbb{R} \}$, we have
$$f(x)^n + y + f(y) = f(x^n + 2 f(y)) \le M\tag{*3}$$
Fixing $x$ and send $y$ to $+\infty$, we will have $f(y) \to -\infty$ in order for LHS$(*3)$
stay bounded from above. Since $f$ is injective, this implies $f$ to be strictly monotonic
decreasing again. Since $f$ is bounded from above, we get
$$\lim_{z\to-\infty} f(z) = \limsup_{z\to-\infty} f(z) = M$$
In $(*2)$, fixing $y$ and send $x$ to $-\infty$, we obtain
$$M^n + y + f(y) = \begin{cases}-\infty, & n \text{ even }\\M, & n \text{ odd }.\end{cases}
\iff f(y) = \begin{cases} -\infty, &n \text{ even }\\M - M^n - y,&n \text{ odd }\end{cases}
$$
Independent of whether $n$ is even or odd, $f$ cannot be a function defined over $\mathbb{R}$ which is bounded from above by some $M$.
Combine above arguments, we conclude 
$$\newcommand{\mybox}[2][8pt,border: 1px solid blue]{\bbox[#1]{#2}}
\mybox{f \text{ continuous } \implies f \text{ bijective }}$$
Since $f$ is bijective, we can pick a $\alpha$ such that $f(\alpha) = 0$, we get
$$f(x^n) = f(x^n + 2f(\alpha)) = f(x)^n + \alpha + f(\alpha) = f(x)^n + \alpha\tag{*4}$$

Let us first study the case $n$ is even. 


*

*Notice,
$$f(x^n) = f((-x)^n) \implies f(x)^n + \alpha = f(-x)^n + \alpha \implies |f(x)| = |f(-x)|$$
Since $f$ is injective, this implies $f(x)$ is an odd function. 

*By continuity, we get $f(0) = 0$ and hence $\alpha = 0$. 

*From this, we find
$$n \text{ even } \implies f(x) \text{ odd } \wedge f(x^n) = f(x)^n$$

*Replace $x$ by $1$, we get $f(1) = f(1)^{n}$. 

*Since $f$ injective implies $f(1) \ne f(0) = 0$, we find $f(1) = 1$.

*For all $z \ge 0$, substitute $x = z^{1/n}$ in $(*1)$, we get
$$f(z + 2) = f(x^n + 2f(1)) = f(x)^n + 2 = f(z)+2\tag{*5a}$$
Together with $f(0) = 0$, this implies $f(2p) = 2p$ for any $p \in \mathbb{N}$.

*For any $p, q \in \mathbb{Z}_{+}$. Pick a $y$ such that $f(y) = \frac{p}{q}$, we have
$$f\left(z + \frac{2p}{q}\right) = f(x^n + 2f(y)) = f(z) + y + \frac{p}{q}$$
Applying this $q$ times, we get
$$f(z+2p) = f(z) + q\left(y + \frac{p}{q}\right) = f(z) + qy + p\tag{*5b}$$
Compare $(*5a)$ with $(*5b)$, we get $y = \frac{p}{q}$. 

*Together with the fact $f$ is odd, we find $f(z) = z$ for all $z \in \mathbb{Q}$ 

*By continuity, we can extend the equality
to whole $\mathbb{R}$. 


Conclusion:
$$\mybox{ n \text{ even } \implies f(z) = z \text{ for all } z \in \mathbb{R}}$$

Let us switch to the case $n$ is odd.
Pick $\beta$ such that $f(\beta) = 1$. Since $n$ is odd, for any $z \in \mathbb{R}$, we can find a $x$ such that $z = x^n$.  


*

*For any $p \in \mathbb{Z}_{+}$, we have
$$f(z+2) = f(x^n+2f(\beta)) = f(x)^n + \beta + f(\beta) = f(z) + (\beta - \alpha + 1)$$
Apply this $p$ times, we get
$$f(z+2p) = f(z) + p(\beta - \alpha + 1 )\tag{*6a}$$
It is clear this relation remains valid when $p$ is allowed to vary over $\mathbb{Z}$.

*For any $p \in \mathbb{Z}, q \in \mathbb{Z}_{+}$, pick a $y$ such that $\displaystyle\;f(y) = \frac{p}{q}$, we have
$$f\left(z + \frac{2p}{q}\right) = f(x^n + 2f(y)) = f(x)^n + y + \frac{p}{q} = f(z) + \left(y - \alpha + \frac{p}{q}\right)$$
Apply this $q$ times, we get
$$f(z + 2p) = f(z) + q (y - \alpha) + p\tag{*6b}$$

*Combine $(*6a)$ with $(*6b)$, we get
$$\frac{p}{q} = \frac{y - \alpha}{\beta - \alpha}$$
This means the set of $y \in \mathbb{R}$ which satisfies 
$$f(y) = \frac{y - \alpha}{\beta - \alpha}\tag{*7}$$
is dense in $\mathbb{R}$. By continuity, $(*7)$ is true over all $\mathbb{R}$.

*Substitute $(*7)$ into $(*4)$, we get
$$\frac{x^n - \alpha}{\beta - \alpha} = \left(\frac{x - \alpha}{\beta-\alpha}\right)^n + \alpha$$
By comparing the coefficient of $x^{n-1}$, we find $\alpha = 0$ and hence 
$$\beta^{n-1} = 1\quad\implies\quad \beta = \pm 1 \quad\implies\quad f(z) = \pm z$$
Since $f(z) = -z$ is not a solution of $(*1)$, we find:
$$\mybox{ n \text{ odd } \implies f(z) = z \text{ for all } z \in \mathbb{R}}$$
A: This is more of an extended comment than a full answer, but here goes.
I will add assumptions along the way.
Assume $n$ is odd.
Denote $a=f(0)$ and $b=(-2a)^{1/n}$.
If we set $x=b$ and $y=0$, the original equation gives $f(b)=0$.
If we choose $y=b$, the original equation gives $f(x^n)=f(x)^n+b$ ($\star$).
Assume $f$ is continuously differentiable.
Differentiate ($\star$) and set $x=0$ to get $f(0)=0$ or $f'(0)=0$.
Differentiating the original equation with respect to $y$ and putting $x=0$ gives 
$1=f'(y)(2f'(2f(y))-1)$.
This implies $f'(0)\neq0$, so we must have $f(0)=0$.
Thus in the earlier notation $a=b=0$.
Now ($\star$) gives $f(x^n)=f(x)^n$.
Differentiating this $n$ times and letting $x=0$ (and remembering $f(0)=0$) gives $f'(0)=f'(0)^n$, whence $f'(0)=\pm1$.
But we got $1=f'(y)(2f'(2f(y))-1)$, and at $y=0$ this yields $1=f'(0)(2f'(0)-1)$, so that $f'(0)=1$.
Using the equation $1=f'(y)(2f'(2f(y))-1)$ again we see that $f'(y)>0$ and $f'(2f(y))>\frac12$ for all $y$.
Thus $f$ must be strictly increasing (and diffeomorphic to its image).
Let $S=\sup f$ and suppose $S<\infty$.
Now for all $t\in(0,2S)$ we have $f'(t)>\frac12$, so by the fundamental theorem of calculus $f(2S)>S$, a contradiction.
A similar argument works in the other direction.
Thus $f(\mathbb R)=\mathbb R$ and $f'>\frac12$ everywhere.
(And so $\pm2f(y)>\pm y$ for $y>0$ if that helps.)
Assume $f$ is twice continuously differentiable.
Differentiating the equation $1=f'(y)(2f'(2f(y))-1)$ gives
$$
(2f'(2f(y))-1)f''(y)+4f'(y)^2f''(2f(y))=0.
$$
At $y=0$ we obtain $f''(0)=0$.
Note that the coefficients $(2f'(2f(y))-1)$ and $4f'(y)^2$ are always strictly positive.
Therefore $f''(y)f''(2f(y))\geq0$ implies $f''(y)=f''(2f(y))=0$.
I was hoping that something like this could show that $f''$ vanishes identically and thus $f(x)=x$, but I don't know how to proceed.
Summary:
If $f\in C^1$ and $n$ is odd, then $f:\mathbb R\to\mathbb R$ is a bijection satisfying $f'(x)>\frac12$, $f(0)=0$, $f'(0)=1$ and $f(x^n)=f(x)^n$.
Also, if $f\in C^2$ and $f''$ has constant sign, then $f(x)=x$ for all $x$.
A: Any continuously differentiable solution to that equation must be the trivial $f(t)=t$.
Assume $f$ is a continuously differentiable solution. Akin to Joonas Ilmavirta's answer we may differentiate with respect to $y$ (keeping $x$ constant) to obtain $$ f'(y)(2f'(x^n+2f(y))-1)=1$$ from which it follows in particular that $f'(y)\ne 0$ for all $y$. On the other hand, differentiating with respect to $x$ we also have $$x^{n-1}f'(x^n+2f(y))=f(x)^{n-1}f'(x). $$ Using the above observation and substituting $x=0$ we obtain $f(0)=0$. From the first equation we also have (setting $y=x=0$) that $f'(0)[2f'(0)-1]=1$, that is, $f'(0)=1$ or $-1/2$. 
Letting $y=0$ in the first equation we immediately have $$f'(0)[2f'(x^n)-1]=1, $$ or $$f'(x^n)=\frac{1}{2}\left( \frac{1}{f'(0)}+1\right) =f'(0).$$
Thus at least $f(t)=f'(0)t$ for all $t\ge 0$ (regardless of whether $n$ is odd or even). Checking against the original equation we may rule out the possibility $f'(0)=-1/2$. 
Dividing by $x^{n-1}$ ($x>0$) in the second equation and letting $x\to 0$ we get $$f'(2f(y))=f'(0).$$  
Let $m=\inf_{\mathbb R}f\le 0$. We want to show that in fact $m<0$. Suppose otherwise, namely that $f\ge 0$ everywhere. Then $x^n+2f(y)\ge 0$ and so the original equation becomes $$x^n+y+f(y)=f(x^n+2f(y))=x^n+2f(y) $$ for all $x,y$, giving $f(y)=y$, a contradition. 
Now since $m<0$ it follows that $f'(2t)=f'(0)=1$ for all $t>m$, in particular thus $(2m,\infty)\subset range(f)$. Since $2m<m$ this contradicts the fact that $m$ is the infimum. Therefore $m=-\infty$, meaning that $f$ is surjective, and we have $f'(t)=1$ everywhere. 
