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
$$f(0,0)=0.$$
 A: If this is true then there is a "finitary" proof, i.e. you can use the relations to express $f(0,0)$ as a finite linear combination of the $f(2n,2m)$.
First, here is a way to rephrase the problem in terms only of even entries of $f$, i.e. how to find a relation between $f(x,y),f(x+2,y),f(x,y+2),\ldots$
Consider $R = \Bbb R[X,X^{-1},Y,Y^{-1}]$ and let $P = (X+Y+X^{-1}+Y^{-1}-1)$.
$R$ is the set whose elements are finite linear combinations of monomials of the form $X^{\pm n} Y^{\pm m}$, and you can rephrase the goal by asking if there is an element $Q \in R$ such that $Q \times P$ has a nonzero constant coefficient and whose other nonzero coefficients occur only for exponents $(2n,2m)$ with nonzero $nm$.
Since $Q \times P$ is in $\Bbb R(X^2,Y^2)$ it is invariant by the Galois group $G$ of $\Bbb R(X,Y)$ over $\Bbb R(X^2,Y^2)$. $G$ has order $4$ and is given by the automorphisms $(X,Y) \mapsto (\pm X, \pm Y)$.
This tells us that $Q \times P$ has to be of the form $Q'(X^2,Y^2) \times P'(X^2,Y^2)$ where $P'(X^2,Y^2) = Norm(P) = P(X,Y)P(-X,Y)P(X,-Y)P(-X,-Y) $
You can easily compute $P'(X^2,Y^2) = (X^4 + Y^4 + X^{-4} + Y^{-4}) - 2(X^2Y^2 + X^2Y^{-2} + X^{-2}Y^{-2} + X^{-2}Y^2) - 2(X^2+Y^2+X^{-2}+Y^{-2}) -3$.

This gives you the relations between the even entries of $f$, and now with this new polynomial, the problem reduces to finding $Q'$ such that $Q' \times P'$ has a nonzero constant coefficient and a zero coefficient in front of $X^n$ and $Y^m$.
Start with $1 \times P'$. It has a nonzero constant coefficient (-3) so we just have to "push out" the $8$ extra unwanted nonzero coefficients.
To do this, first add some $-\frac 29(X+Y+X^{-1}+Y^{-1}) \times P'$. This should make the $4$ entries near the constant coefficient zero while still keeping the constant coefficient nonzero (it is $-11/9$).
Next, by symmetry reasons you can remove the $X^{\pm2},Y^{\pm2}$ coefficients by adding a suitable multiple of $(X^2Y^2+X^{-2}Y^{-2}) \times P'$ without touching the constant and neighbouring coefficients.
Finally you can remove the remaining nonzero coefficients of $X^n$ one at a time by adding multiples of $X^nY^2  \times P'$ and similarly for $Y^n$.

Computing the resulting polynomial gives you an explicit relation between $f(0,0)$ and a finite linear combination of some $f(2n,2m)$ with $nm \neq 0$, as needed.
A: 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 $f(0, 0) = 0$?


*

*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.

