What is the solution of the following functional equation? (I must confess it is a headache for me.)
Find all the functions $ f : \mathbb Z \to \mathbb Q $ such that $$ f \left( \frac { x + y } 3 \right) = \frac { f ( x ) + f ( y ) } 2 $$ fora ll $x , y \in \mathbb Z $ such that $ \frac { x + y } 3 \in \mathbb Z $.
I'd appreciate your help and comments. I've tried it a lot but until today I'm not able to solve it... :/
The function $f$ is necessarily constant.
We proceed by induction. Note that $$ f(3k)=\frac{f(3k)+f(3k)}2=f\left(\frac{3k+3k}3\right)=f(2k), $$ and $$ f(k)=f\left(\frac{2k+k}3\right)=\frac{f(2k)+f(k)}2, $$ from where $f(2k)=f(k)$. In particular, $f(3)=f(2)=f(1)$.
We have $f(1)=f(\frac{0+3}3)=\frac{f(0)+f(3)}2=\frac{f(0)+f(1)}2$, so $f(0)=f(1)$. Now assume that $f(0)=f(1)=\cdots=f(k)$. Then we do the inductive step: $$ f(k+1)=f\left(\frac{3k+3}3\right)=\frac{f(3k)+f(3)}2=\frac{f(k)+f(1)}2=f(k). $$
As mentioned by Hagen, from $f(0)=\frac{f(k)+f(-k)}2$ we get $f(-k)=2f(0)-f(k)$; as $f(k)=f(0)$ for positive $k$, we get that $f(k)=f(0)$ for all $k\in \mathbb Z$.