Find all functions $f:\mathbb R\to\mathbb R$ such that $f\big(x+f(y)\big)=f(x+y^n)+f\big(y^n-f(y)\big),\ \forall x,y\in\mathbb R$ 
Let $ n = 2018 $. Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \big( x + f ( y ) \big) = f ( x + y ^ n ) + f \big( y ^ n - f ( y ) \big) ,\ \forall x , y \in \mathbb R \text . $$

I tried the standard way: $ x = 0 $, $ x = y $, $ x = 1 $, $ \dots $, but without any success. I spent quite some time trying to solve it but without success!
I tried to reduce it to Cauchy's $ 1 - 4 $ equations but didn't succeed. In the course of it, I found interesting works of Aczél, Erdős and even Putnam, but they are not directly related, I guess.
Any ideas? I am interested in this problem but I couldn't solve it!
 A: Without any additional restrictions, this seems to be hopeless for me (Of course, I might be wrong). 
However, some things can be said and maybe this might help you:
For $x = -f(y)$ you obtain $f(y^{2018} - f(y)) = \frac{1}{2}f(0)$.
So you have 
$$ f(x + f(y)) =  f(x + y^{2018}) + \frac{1}{2}f(0) = f(x + y^{2018} + f(0))$$
where the second equality follows from your equation with $x' = x + y^{2018}$ and $y' = 0$. 
In particular, if $f$ were injective, you would get $f(y) = y^{2018} + f(0)$ and it is easy to see that your identity implies $f(0) = 0$ in this case.
However, it gets much more difficult if $f$ is not injective. The only thing I also noticed was that for $x = y^{2018} - 2 f(y)$ you get $f(2(y^{2018} - f(y))) = 0$ for all $y \in \mathbb{R}$.
A: (Not a complete answer)  
Partial Differentiating both sides with respect to $x$ we get $$f'(x+f(y))=f'(x+y^{2018})$$
Putting $x=0$ we get $$f'(f(y))=f'(y^{2018})$$
Now the most obvious function popping out which I noticed is $$f(y)=y^{2018}$$ but couldn't find any other leading function. Hope it helped a bit.
A: It can be shown that for any integer $ n \ge 2 $, the only functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \big( x + f ( y ) \big) = f ( x + y ^ n ) + f \big( y ^ n - f ( y ) \big) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ are $ f ( x ) = 0 $ and $ f ( x ) = x ^ n $. It's easy to see that these functions are solutions. We prove that they are the only ones.
Letting $ x = - f ( y ) $ in \eqref{0} we have
$$ f \big( y ^ n - f ( y ) \big) = \frac 1 2 f ( 0 ) \text . \tag 1 \label 1 $$
\eqref{1} can be used to rewrite \eqref{0} as
$$ f \big( x + f ( y ) \big) = f ( x + y ^ n ) + \frac 1 2 f ( 0 ) \text . \tag 2 \label 2 $$
\eqref{2} can be used to derive
$$ f \left( x - \frac 1 2 f ( 0 ) + y ^ n \right) + \frac 1 2 f ( 0 ) = f \left( x - \frac 1 2 f ( 0 ) + f ( y ) \right) \\
= f \left( x - \frac 1 2 f ( 0 ) + f \Big( \big( y - f ( z ) \big) + f ( z ) \Big) \right) = f \Big( x + f \big( y - f ( z ) + z ^ n \big) \Big) \\
= f \Big( x + \big( y - f ( z ) + z ^ n \big) ^ n \Big) + \frac 1 2 f ( 0 ) \text , $$
which by substituting $ x - y ^ n + \frac 1 2 f ( 0 ) $ for $ x $ yields
$$ f ( x ) = f \left( x + \big( y - f ( z ) + z ^ n \big) ^ n - y ^ n + \frac 1 2 f ( 0 ) \right) \text . \tag 3 \label 3 $$
The last step implies that if $ f ( z ) \ne z ^ n $ for any $ z \in \mathbb R $, then $ f $ is constant. This will prove what is desired, as the only constant function $ f $ satisfying \eqref{0} is clearly the zero function. To see why the claimed implication holds, choose $ z $ as such, and define $ g : \mathbb R \to \mathbb R $ with $ g ( y ) = \big( y - f ( z ) + z ^ n \big) ^ n - y ^ n + \frac 1 2 f ( 0 ) $. As $ g $ is a nonconstant polynomial, $ \operatorname {ran} g $ is either $ \mathbb R $, $ [ a , + \infty ) $ for some $ a \in \mathbb R $ or $ ( - \infty , a ] $ for some $ a \in \mathbb R $. This will depend on parity of $ n $, and also on whether $ f ( z ) < z ^ n $ or $ f ( z ) > z ^ n $. In any case, for any $ x , y \in \mathbb R $, we can choose $ z \in \mathbb R $ so that both $ z - x $ and $ z - y $ are in $ \operatorname {ran} g $, which by \eqref{3} shows that
$$ f ( x ) = f \big( x + ( z - x ) \big) = f ( z ) = f \big( y + ( z - y ) \big) = f ( y ) \text , $$
and hence $ f $ is constant.
