Injective additive grids

Question

We call a map $f:{\mathbb Z}\times {\mathbb Z} \to {\mathbb Z}$ an additive grid if for all $x,y \in {\mathbb Z}$ we have that $f(x,y)$ is the sum of the neighboring values, that is, $$f(x,y) = f(x-1,y)+f(x+1,y) + f(x,y-1) + f(x,y+1).$$ To me it feels like there cannot be injective additive grids - but I haven't been able to come up with a proof. So: are there injective additive grids?

Answer 1

Yes, such a grid can be constructed easily by induction. As can be seen from the answer given to your previous question, the adjacent rows can be chosen arbitrarily. Denote the numbers in these rows by $\ldots,x_{-1},x_0,x_1,\ldots$ and $\ldots,y_{-1},y_0,y_1,\ldots$. How does the rest of the grid change if we change $x_0$? The elements that are in the infinite triangle whose sides are the halflines that go North-West and North-East from $x_0$. The coefficients can be also computed, along the sides they are always $\pm x_0$, while inside the triangle some other numbers (possibly $0$). Let us pick the numbers in the order $x_0,x_1,x_{-1},x_2,\ldots$ such that they grow very fast. I claim that this guarantees that all elements of the grid are different. (Let's ignore the $y_i$'s - it's easy to make the argument work for them too.)

Suppose that there are two numbers that are equal. For both, take the most significant $x_i$ that influenced them. This will be one that is diagonally South-West or South-East. If these $x_i$ differ, the numbers also differ. If they are the same, then the coefficient of how $x_i$ influences are numbers might be still different. If it's also the same, take the second most significant $x_j$ etc. Sooner or later we get a different $x_j$ or coefficient, and we are done.

Note that in the above argument we have used that for any $k$ there are finitely many numbers whose most significant $x_i$ has $|i|\le k$.