Find all functions: $f: \mathbb{R}\rightarrow \mathbb{R}$ such that:

$$f\left ( x+ f\left ( y \right ) \right )= f\left ( x+ y^{2018} \right )+ f\left ( y^{2018}- f\left ( y \right ) \right ),\,\forall x,\,y \in \mathbb{R}$$

I tried the standard way: $x=0,\,x=y,\,x=1,\,...$ 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 corse of it, I found interesting works of Aczel, Erdos and even Putnum, but they are not directly related, I guess.

Any idea? I am interested in this problem but I couldn't solve!

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    $\begingroup$ Is it from a contest, i.e. do you know whether it has only 'nice' solutions? I'm asking this because a functional equation from $\mathbb{R}$ to $\mathbb{R}$ where the variables only appear inside of the function, and where no furder constraints are given, has often also pathological solutions. This is just my general experience, but it would come quite surprising to me if this functional equation is solvable in the 'olympiad'-sense. $\endgroup$ – Redundant Aunt Jun 18 '18 at 11:10
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    $\begingroup$ Although power of y is big, we only use $y=1,0,-1$ so on. So this problem is same to equation of $y^{2018}→y^2$. $\endgroup$ – Takahiro Waki Jul 31 '18 at 8:13

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 $y \in \mathbb{R}$.


(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.


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