A chessboard is infinite in all directions. We write an integer in each black cell. Thereafter, for each white cell $W$, let $a,b$ be the two numbers in the two cells horizontally adjacent to $W$, and $c,d$ be the two numbers in the two black cells vertically adjacent to $W$.
Is it possible that $|ab-cd|=1$ for all white cells?
If it were possible, I think it would be through repeating patterns in the sense that we repeatedly use the same block of cells throughout the chessboard. So then it would suffice to show that the $|ab-cd|=1$ condition holds for the cells in this pattern, along with those adjacent to the next block.
Lemma 1. If in a square matrix we replace the first row with the sum between the first and second row, its determinant stays the same.
Lemma 2. If $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}}$ are two consecutive convergents of the continued fraction of $\alpha$, $$\det\begin{pmatrix}p_n & p_{n+1} \\ q_n & q_{n+1}\end{pmatrix}\in\{-1,+1\}$$
From now on, let rows be black diagonals in the original chessboard.
Lemma 3. By Lemma 1,
$$\begin{array}{l} \ldots\verb \76543234567\\ldots\\ \ldots\verb \65432123456\\ldots\\\ldots\verb \11111111111\\ldots \\\ldots \verb \65432123456\\ldots\\ \ldots\verb \76543234567\\ldots\\ \end{array}$$
works. Just to be clear, the previous and next rows are given by: $$ \ldots \verb \13 11 9 7 5 3 5 7 9 11 13 \ \ldots$$
but the repeating pattern $$\begin{array}{l} \ldots\verb \11111111111\\ldots\\ \ldots\verb \21212121212\\ldots\\\ldots\verb \11111111111\\ldots \\\ldots \verb \21212121212\\ldots\\ \ldots\verb \11111111111\\ldots\\ \end{array}$$ works just as fine.
Another interesting chance is given by placing the Fibonacci number $F_{i+j}$ at the position $(i,j)$ - the extension of the Fibonacci sequence to the negative indices is given, as usual, by the relation $F_n=F_{n+2}-F_{n+1}$, so that $F_{-n} = (-1)^{n+1} F_n$. This works by Lemma 2 or Lemma 1.