Given functional equation $f(x,y)=f(x,y+1)+f(x+1,y)+f(x-1,y)+f(x,y-1)$, show that $f(0,0)=0$ Let $f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{R}$, such that
for all $x,y$
$$f(x,y)=f(x,y+1)+f(x+1,y)+f(x-1,y)+f(x,y-1),$$
and if $m,n\in \mathbb{Z},mn\neq 0$,we have $f(2m,2n)=0$.
Show that
$$f(0,0)=0.$$

Sketching a solution since there is a lot of tedium required. This method finds all solutions $f$; perhaps there is a shortcut to show only $f(0, 0) = 0$?


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*Show that if $f$, $g$ are solutions, then $f + g$ is a solution as well. Easy from the condition.

*Show that if$$f(3, 4) = f(4, 3) = f(3, 6) = f(6, 3) = 0,$$then $f(x, y) = 0$ for all $x$, $y$. To show this, notice that$$f(3, 3) + f(5, 3) = 0$$from $(x, y) = (4, 3)$ in the condition, and$$f(3, 3) + f(3, 5) = 0$$from $(x, y) = (3, 4)$ in the condition. So $f(3, 5) = f(5, 3)$. Then since$$f(5, 3) + f(5, 5) = f(5, 4)$$and $$f(3, 5) + f(5, 5) = f(4, 5),$$we have $f(5, 4) = f(4, 5)$. Finally,$$f(5, 4) + f(4, 5) = 0$$from $(x, y) = (4, 4)$ in the condition, so$$f(5, 4) = f(4, 5)=0.$$This logic is the hardest step needed; showing all the other values have to be $0$ is long, but each step is either repeating what we just did or applying the original condition where four of the terms are $0$.

*Show that if we treat $f$ as a function on the plane, then the following $4 \times 6$ array is a solution when tessellated infinitely, aligned so the top left corner is $(0, 0)$. Here, $a$ is an arbitrary real.$$\begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & a & a & 0 & -a & -a \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -a & -a & 0 & a & a\end{matrix}$$

*Show that if $f(3, 4)$, $f(4, 3)$, $f(3, 6)$, $f(6, 3)$ are arbitrarily set to real numbers, then there exists a solution. This is easy using step 1 and various transposes/translations of step 3.

*Using steps 1 and 2, we have that if $f(3, 4)$, $f(4, 3)$, $f(3, 6)$, $f(6, 3)$ are specified, then there is at most one solution $f$. This means that step 4 describes all possible solutions. These solutions all satisfy $f(2m, 2n) = 0$ for all $m$, $n$, so we are done.

