Prove that in each coloring of a $4\times7$ board in two colors there's a square that all four of it's corners are colored by the same color.
This is a pigeon hole principle question and I have a proposed solution that I totally don't understand.
An explanation or a different approach would be appreciated.
It's not true if "square" is taken literally, e.g., $$\begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ \end{bmatrix}$$ is a counter-example (found by hand; verified computationally).
If "square" is interpreted to mean $2 \times 2$ submatrix, then we can prove the claim for $3 \times 7$ boards as follows.
In the first row, there must be $4$ cells of the same color, say blue. In the second row, directly below these $4$ cells we must have $3$ cells of the other color, say red, otherwise we have a monochromatic $2 \times 2$ submatrix. In the third row, directly below these $3$ cells we must have either $2$ red or $2$ blue cells, and in either case, we have a monochromatic $2 \times 2$ submatrix.