We call a map $f:{\mathbb Z}\times {\mathbb Z} \to {\mathbb Z}$ an *[additive grid](https://mathoverflow.net/questions/292383/fibonacci-grids)* 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?



Injective additive grids
Yes, such a grid can be constructed easily by induction.
As can be seen from the [answer given to your previous question](https://mathoverflow.net/a/292400/955), 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 $|i|\le k$.


