Good catch. Presumably, they were using the Mutilated Chessboard Problem where diagonally opposite corners are removed. If you removed a red and a black corner (same side), it would be easy to cover the resulting board in 31 dominoes.
No doubt, the puzzle is stated a little incorrectly. It’s a well-known puzzle, and when I’ve come across it before, they have always removed two squares at each end of the same diagonal, so both were the same color. In those versions don’t think the dominos are required to cover a red/black combo, any two squares you want. But since covering a red/black combo with a domino is the only possibility, that is the proof why it can’t be done.
Here’s an interesting puzzle. Math theory usually says you have to have as many equations as you have unknowns to find a unique answer. But it is possible sometimes to find a unique answer even with just one equation and three unknowns. For example
a x b x c =30?
“x” means multiplication.
One equation, three unknowns, a, b, & c are all whole numbers. clue: very easy
Oops… you caught me on that one … lol … rigorously speaking I guess the answer I was thinking of (second in your list) isn’t “unique” b/c a=2, b=3, c=5 is just one solution. a=3, b=5, c=2 would work too, etc. .
Good recollection. Square roots always need to be removed from the denominator of a fraction because the square root of a perfect square (1, 4, 9, 16, etc.) is a whole number which is easily removed. And, the square root of any non-perfect square is an irrational number. You can’t have an irrational number in the denominator because it has no definitive value.