We are given a function $f\colon R\to R$ that satisfies $f(x+f(y))=f(x)+f(y)$ for any real $x,y$, and also $f(0)\neq0$. The equation seems to be entirely dependent on the values of $f$. It seems impossible to have anything to say about $f(x+y)$ just from manipulations with the equation; which suggests a more inventive approach should be of use. Clearly, functions of the form $x+b$ (with $b\neq 0$) satisfy mentioned properties. So, it suggests the function in the question has some properties of the latter. That is, $f(x+y)=f(x)+f(y)-b$ or $f(pf(0))=pf(0)+f(0)$ for a rational $p$.
Here is what I have found:
- $f(x)=f(x-f(0))+f(0)$,
- $f(x+nf(y))=f(x)+nf(y)$, $n\in Z$ (in particular, $f(nf(x))=f(0)+nf(x)$),
- $f(-f(0))=0$.
- Surjectivity would imply $f(pf(0))=pf(0)+f(0)$ for any $p$.
But... that's all. No continuity, no injectivity, not even anything like $f(px)=pf(x)$, like for the Cauchy's functional equation. And it seems plausible that there are pathological solutions with their graphs dense in $R^2$. Anything beyond these facts seems to be walking in circles without any meaningful conclusion. Does anyone have clues?
The original problem is from mock IMO 2023:
Find all rational numbers $q$ for which there exists a function $f\colon R\to R$ satisfying $f(x+f(y))=f(x)+f(y)$ and $f(z)\neq qz$ for all real numbers $x, y, z$.
Intuitively, I guessed that the only rational number that answers the question is $q=1$. And it is easy to infer $f(0)\neq 0$. From these conclusions I started to work things out. The matter in my question would help to prove that $q=1$ is indeed the only number to have such a function.
Edit note. The user from AoPS constructs solutions for this functional equation using additive subgroups of $R$, but with $f(0)=0$, and it can be seen that $f(x+y)=f(x)+f(y)-f(0)$ doesn't hold. It remains to verify it against $f(0)\neq 0$.
It seems that it doesn't hold. Following the construction given at AoPS, we can put \begin{align}f(x)=c+\lfloor\sin(2\pi x)\rfloor+\lfloor x\rfloor\end{align}
for an integer constant $c\neq 0$. Numerically, using GeoGebra and some values, I was able to see that this function should satisfy the functional equation. And to prove it, notice that $f(y)$ is an integer
\begin{align} f(x+f(y))&=c+\lfloor\sin(2\pi x+2\pi f(y))\rfloor+\lfloor x+f(y)\rfloor\\ &=c+\lfloor\sin(2\pi x)\rfloor + \lfloor x\rfloor +f(y)\\ &=f(x)+f(y) \end{align}
But, it doesn't satisfy $f(x+y)=f(x)+f(y)-f(0)$.
Take, for example, $c=5$, $x=-14.6$, $y=-27.8$, then we have \begin{align} f(x+y)=-39\neq -38=f(x)+f(y)-f(0) \end{align}