A function $f:\mathbb Z^2 \rightarrow \mathbb R$ is said to be *discrete harmonic* if it satisfies the discrete Laplacian equation
$$
\Delta f(m,n) = f(m+1,n)+f(m-1,n)+f(m,n+1)+f(m,n-1)-4f(m,n) = 0~.
$$
Let $\mathcal F$ be the set of all discrete harmonic functions on the square lattice.



> 
> **Question:** Is there another difference equation with integer (equiv. rational) coefficients that is satisfied by all discrete harmonic functions?
> 
> 
> 


If $D$ is a difference operator with integer coefficients, then any $f\in \mathcal F$ trivially satisfies
$$
(D\circ\Delta)f(m,n)=0~.
$$
My question is if there is a difference equation with integer coefficients satisfied by all $f\in\mathcal F$ that cannot be written as $D\circ \Delta$ for any $D$.


I can rephrase the above question as follows. Any function $f:\mathbb Z^2\rightarrow\mathbb R$ can be uniquely represented as
$$
\hat f(x,y) = \sum\_{m,n\in\mathbb Z} f(m,n) x^m y^n~.
$$
The discrete Laplacian operator $\Delta$ can be associated with the Laurent polynomial
$$
p(x,y) = x + x^{-1} + y + y^{-1} - 4 \in \mathbb Z[x,x^{-1},y,y^{-1}]~.
$$
Then, any $f\in\mathcal F$ satisfies
$$
p(x,y) \hat f(x,y) = 0~.
$$
My question can be rephrased as follows:



> 
> Is there a $q(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$ that is not a multiple of $p(x,y)$, and
> $$
> q(x,y)\hat f(x,y)=0~,
> $$
> for all $f\in\mathcal F$?
> 
> 
> 


Here, by a multiple of $p(x,y)$, I mean something of the form $r(x,y)p(x,y)$ where $r(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$.



Difference equation satisfied by discrete harmonic functions on square lattice
$\DeclareMathOperator\u{\mathbf u}$$\DeclareMathOperator\e{\mathbf e}$I think I know how to answer my question. First, note that $p(x,y) = x^{-1}y^{-1} \tilde p(x,y)$, where $\tilde p(x,y) = (x-1)^2y+x(y-1)^2 \in \mathbb Z[x,y]$. More generally, any $q(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$ can be written as $q(x,y) = x^a y^b \tilde q(x,y)$ for some $a,b\in\mathbb Z$ such that $\tilde q(x,y)\in\mathbb Z[x,y]$. So, my question can be rephrased as follows:



> 
> Is there a $\tilde q(x,y)\in\mathbb Z[x,y]$ that is not divisible by $\tilde p(x,y)$ such that $\tilde q(x,y) \hat f(x,y)=0$ for any $f\in\mathcal F$?
> 
> 
> 


I will show that the answer is no, i.e., any $\tilde q(x,y)$ that satisfies $\tilde q(x,y) \hat f(x,y)=0$ for any $f\in\mathcal F$ must be divisible by $\tilde p(x,y)$.




---


Choosing a [lexicographic monomial order](https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis#Monomial_ordering) with $x\succ y$ for [multivariate division](https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis#Reduction), any $\tilde q(x,y)$ can be uniquely written as
$$
\tilde q(x,y) = x^2\alpha(x) + x \beta(y) + \gamma(y) + \tilde r(x,y)\tilde p(x,y)~,\tag{1}\label{1}
$$
where $\tilde r(x,y)\in\mathbb Z[x,y]$ and $\alpha(X),\beta(X),\gamma(X)\in\mathbb Z[X]$. Let $u$, $v$, and $w$ be the degrees of $\alpha(X)$, $\beta(X)$, and $\gamma(X)$ respectively, and let
$$
\alpha(X) = \sum\_{i=0}^u a\_i X^i~,\qquad \beta(X)=\sum\_{j=0}^v b\_j X^j~,\qquad \gamma(X)=\sum\_{k=0}^w c\_k X^k~.
$$
I will show that $a\_i = b\_j = c\_k = 0$ if $\tilde q(x,y) \hat f(x,y)=0$ for all $f\in\mathcal F$.




---


Consider the function
$$
f(m,n;t) = (-1)^m t^{-m+n} \left(\frac{1+t}{1-t}\right)^{-m-n}~.
$$
It is easy to check that it is a discrete harmonic function on a square lattice for any $t\ne 0,\pm 1$ (see page 13 of [this note](https://web.cs.elte.hu/%7Elovasz/analytic.pdf) for a related function). Now, using \eqref{1}, the equation $\tilde q(x,y) \hat f(x,y;t) = 0$ becomes
$$
x\_t^2 \alpha(x\_t) + x\_t \beta(y\_t) + \gamma(y\_t) = 0~,
$$
where $x\_t := t\left(\frac{1+t}{1-t}\right)$ and $y\_t := t\left(\frac{1-t}{1+t}\right)$. Let $\nu := \max(v-1,w)$. Multiplying by $(1-t)^{u+2} (1+t)^\nu$ gives a polynomial in $t$ given by
$$
\sum\_{i=0}^u a\_i t^{i+2} (1+t)^{\nu+i+2} (1-t)^{u-i} + \sum\_{j=0}^v b\_j t^{j+1} (1+t)^{\nu-j+1} (1-t)^{u+j+1} + \sum\_{k=0}^w c\_k t^k (1+t)^{\nu-k} (1-t)^{u+k+2} = 0~.\tag{2}\label{2}
$$
Since this equation holds for any $t\ne0,\pm 1$, the polynomial in $t$ must vanish identically (even at $t=0,\pm1$).




---


We can write \eqref{2} as
$$
\sum\_{i=0}^u a\_i \alpha\_i(t;\u) + \sum\_{j=0}^v b\_j \beta\_j(t;\u) + \sum\_{k=0}^w c\_k \gamma\_k(t;\u) = 0~.\tag{3}\label{3}
$$
where $\u:=(u,v,w)$. I will proceed by induction to show that the set $S(\u)$ of polynomials $\alpha\_i(t;\u)$'s, $\beta\_j(t;\u)$'s, and $\gamma\_k(t;\u)$'s is linearly independent for any $\u$:


* **Base cases:** For $\u=\mathbf 0$, $S(\mathbf 0)$ contains
$$
\alpha\_0(t;\mathbf 0) = t^2(1+t)^2~, \qquad \beta\_0(t;\mathbf 0) = t(1-t^2)~,\qquad \gamma\_0(t;\mathbf 0) = (1-t)^2~,
$$
which are clearly linearly independent. Similarly, for $\u=(0,1,0)$, $S(0,1,0)$ contains
$$
\alpha\_0(t;\u) = t^2(1+t)^2~, \qquad \beta\_0(t;\u) = t(1-t^2)~, \qquad \gamma\_0(t;\u) = (1-t)^2~,
\\
\beta\_1(t;\u) = t^2(1-t)^2~,
$$
which are also linearly independent.
* **Induction step:** Assume that $S(\u)$ is linearly independent.


	1. For $\u+\e\_1$, where $\e\_1:=(1,0,0)$, $S(\u+\e\_1)$ contains
	$$
	\alpha\_i(t;\u+\e\_1) = \begin{cases}
	(1-t)\alpha\_i(t;\u)~,&\text{for } i = 0,\ldots,u~,
	\\
	t^{u+3} (1+t)^{\nu+u+3}~,&\text{for } i=u+1~,
	\end{cases}
	\\
	\beta\_j(t;\u+\e\_1) = (1-t) \beta\_j(t;\u)~,\quad\text{for } j = 0,\ldots,v~,
	\\
	\gamma\_k(t;\u+\e\_1) = (1-t) \gamma\_k(t;\u)~,\quad\text{for } k = 0,\ldots,w~.
	$$
	Except for $\alpha\_{u+1}(t;\u+\e\_1)$, all the other polynomials are linearly independent because they are all $(1-t)$ times polynomials that are linearly independent by hypothesis. On the other hand, $\alpha\_{u+1}(1;\u+\e\_1)\ne0$, so it is linearly independent of the other polynomials, which vanish at $t=1$. Therefore, $S(\u+\e\_1)$ is linearly independent.
	2. For $\u+\e\_2$ and $\u+\e\_3$, the above argument is a bit subtle. Instead, we can proceed as follows. Say $S(u,v\_0,v\_0-1)$ is a linearly independent set. Then $S(u,v\_0,w)$ with $w<v\_0-1$, and $S(u,v,v\_0-1)$ with $v<v\_0$ are also linearly independent sets because $S(u,v,v\_0-1)\subset S(u,v\_0,v\_0-1)$ and $S(u,v\_0,w)\subset S(u,v\_0,v\_0-1)$. So, all I need to show is that if $S(u,v\_0,v\_0-1)$ is linearly independent, then $S(u,v\_0+1,v\_0)$ is also linearly independent. This can be shown using an argument similar to the one above (it's only slightly more complicated).


Since the polynomials $\alpha\_i(t;\u)$'s, $\beta\_j(t;\u)$'s, and $\gamma\_k(t;\u)$'s are all linearly independent for any $\u$, it follows from \eqref{3} that $a\_i = b\_j = c\_k = 0$. Therefore, $\tilde q(x,y)$ is divisible by $\tilde p(x,y)$.


