How prove this $f=C$ if $4f(x,y)=f(x-1,y)+f(x,y-1)+f(x+1,y)+f(x,y+1)$ Question:
if $f:\mathbb{Z}^2\to \mathbb{R}$ is bounded ,and for any $x,y\in \mathbb{Z}$,we have
$$4f(x,y)=f(x-1,y)+f(x,y-1)+f(x+1,y)+f(x,y+1)$$
show that
$$f\equiv C$$
where $C$ is constant.
My try:let $x=y=0$,then we have
$$4f(0,0)=f(-1,0)+f(0,-1)+f(1,0)+f(0,1)$$
and let
$x=0,y=1$,then
$$4f(0,1)=f(-1,1)+f(0,0)+f(1,1)+f(0,2)$$
let 
$x=1,y=0$,
then
$$4f(1,0)=f(0,0)+f(1,-1)+f(2,0)+f(1,1)$$
$$\cdots\cdots$$
Then I fell very ugly, so I can't works,maybe have other methods,and this problem is from  a middle school student mathematics exercises
Thank you  very much!
 A: Apparently the solution below was the official solution to this problem, which appeared on the 2003 Bulgarian TST. I edited this from dgrosev's solution found here.

Assume on the contrary that $f(x,y)$ is a non constant bounded harmonic function (assume $0\leq f \leq 1$). Then there exist two consecutive points (along the $x$ or $y$ axes) in which $f$ has different values. Rotating the plane, if necessary, we may assume without loss of generality that $f(x_0+1,y_0) > f(x_0,y_0)$.
Let consider $g(x,y)=f(x+1,y)-f(x,y)$. Then $g$ is not identical to $0$ and $g$ is bounded. So if we take
  $M= \sup g(x,y) \implies 0< M < \infty$, 
  it is easy to see that $g$ is also a harmonic function.
Let $\varepsilon >0$ and take $(x,y)$ so that: $g(x,y) > M-\varepsilon$. Because $g$ is harmonic, it follows:
  $g(x+1,y)  > M-4\varepsilon$. Applying this argument several times implies that
  $g(x+n,y) > M-4^n \varepsilon$.
  The definition of $g$ shows that
  $$f(x+n,y)-f(x,y) = \sum_{k=0}^{n-1} g(x+k,y)\tag{1}$$
Note that LHS of $(1)$ is between $-1$ and $1$. However, if we take $n>\frac{2}{M}$, then take sufficiently small $\varepsilon$ and $(x,y)$ so that $f(x,y)>M-\varepsilon$, the RHS of $(1)$ is greater than $1$. This is a contradiction, and so $g\equiv 0$ and so $f$ is constant.

