Evolving Polynomial

On the Cartesian field, each lattice point $(x,y)$ is initially assigned the number $x^2+y^2-1$.

At each step, if $A(x,y)$ represents the number on $(x,y)$, that number is increased by: $$\frac{(x-y) \left( A(x+1,y) + A(x,y-1) - A(x-1,y) - A(x,y+1) \right)}{4}$$

All the additions are done simultaneously.

1. Prove that the numbers at each point are always integers

2. Prove that, aside from the initial configuration, there is never a number that is divisible by 3

Solution

The recursion formula guarantees that $A(x,y)$ is always a polynomial in $x,y$, with real coefficients. In the initial configuration, $B(x,y) = A(x,y) + 1 = x^2+y^2$ is homogeneous in $x,y$ with degree 2. That is, $B(kx,ky) = k^2 A(x,y)$. We will show that this homogeneity is maintained by the recursion formula, and therefore will hold true at all times.

Suppose $A(x,y) = P.x^2 + Q.xy + R.y^2 - 1$ for some real numbers $P,Q,R$ then the recursion formula maintains it. This is because $A(x+1,y) - A(x-1,y)$ cancels out the $x^2,xy,y^2,1$ terms, so only $x,y$ terms are left. Similarly $A(x,y+1) - A(x,y-1)$ will also only preserve $x,y$ terms. So the second factor is homogenous with degree 1, but the factor $(x-y)$ is also homogenous with degree 1, so the added number is homogeneous with degree 2. Therefore, the final number is still of the form $A(x,y) = P'.x^2 + Q'.xy + R'.y^2 - 1$ (for some $P',Q',R'$ constants that may be different from the previous iteration).

Now we also notice that $A(x,y) = A(y,x)$ for all time. This property is preserved in the beginning, but also preserved by the recursion, so the numbers are symmetric with respect to $x,y$. Therefore $P = R$.

Next, we notice that for $x=y$, the additions are always zeros, so $A(x,x) = 2x^2-1$ for all time. Therefore $A(x,x) = (2P+Q)x^2-1 = 2x^2-1$ so $2P+Q = 2$.

This means that, because $Q = 2- 2P$, $A(x,y) = Px^2 + (2-2P)xy + Py^2 - 1= (x^2+y^2 - 1) + (P-1)(x-y)^2$ for some $P$ real (not necessarily integer) that depends on $t$. So we've shown that $A(x,y)$ has the form $(x^2+y^2-1) + K_t(x-y)^2$ for some $K_t$, where $K_0 = 0$.

Then, plugging in to the original recursion formula, we can see: $$(K_{t+1} - K_t)(x-y)^2 = \frac{x-y}{4} ( 4x - 4y +K_t(2(x-y+1)^2 -2(x-y-1)^2))$$ $$(K_{t+1} - K_t)(x-y)^2 = \frac{x-y}{4} ( 4x - 4y +8K_t(x-y))$$ $$K_{t+1} = 3K_t + 1$$ From there, it's a simple matter to see that $K_t = \frac{3^t-1}{2}$ because $K_0 = 0$, but we didn't need to solve the recursion. The fact that $K_0 = 0$ is enough to prove that $K_t$ is always an integer, and that, except for $t=0$, it's always $\equiv 1 \mod 3$.

Because $K_t$ is always an integer then $A(x,y)$ is also integral everywhere. And because $K_t \equiv 1 \mod 3$ then we can check that for any combinations of $x,y \mod 3$, $A(x,y)$ will never be $\equiv 0 \mod 3$.