# 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!

• 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. – Redundant Aunt Jun 18 '18 at 11:10
• Although power of y is big, we only use $y=1,0,-1$ so on. So this problem is same to equation of $y^{2018}→y^2$. – 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 $y \in \mathbb{R}$.

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.

• Do you agree that $f$ might not be differentiable anywhere? – user2661923 Nov 5 '20 at 2:14

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 is showing 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 this, 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.

• Are you aware of any online pdf's that one could study to facilitate attacking this kind of problem? – user2661923 Nov 5 '20 at 2:19
• @user2661923 I don't know anything particularly about "this kind of problem", but these two PDFs by Evan Chen are very good sources for learning general techniques for solving functional equations: FuncEq-Inro and Monsters. I also suggest taking a look at numerous post on this site and AOPS where you can learn a lot. You can search for specific formulas to find specific problems on these two sites using this. – Mohsen Shahriari Nov 5 '20 at 4:04