Square Chessboard Problem Show that there is a $6$ x $4$ board whose squares are all black or white, where no rectangle has the four vertex squares of the same color. Also show that on each $7$ 
 x $4$ board whose squares are all black or white, there is always a rectangle whose four vertex squares are the same color. Point: On the board, adjacent houses do not necessarily have different colors.
I suspect it involves Pigeonhole principle or something discrete-mathematics, but I don't know how to solve it

Let's say the rows are $4$ wide, and let's try and get as many columns while avoiding the unicolor-vertex-square. Now, if there is a row with $4$ black squares, then the other rows can have at most $1$ other black square. But that means that two of those other rows will have at least $2$ white squares in the same columns, and so you cannot have any row with $4$ black squares, and by symmetry no row with $4$ white squares either if you want at least $4$ columns.
If the first row has $3$ black squares, then no other rows can have $3$ black squares. Likewise, you can have at most $1$ row with $3$ white squares. All other rows must therefore have $2$ black and $2$ white squares. 
