I had a lot of trouble with this induction problem from Rosen's Discrete Mathematics and Its Applications, 8th ed.:
Use mathematical induction to show that a rectangular checkerboard with an even number of cells and two squares missing, one white and one black, can be covered by dominoes.
(We can assume the board has a black-white checkerboard coloration.)
For my partial attempt, I let $ P(n, k) $ be the claim that a $ 2n \times k $ checkerboard missing a white and a black cell can be covered by dominoes. I also noted that, in order for $ P(n, k) $ to be true, we must have $ n \geq 1 $ and $ k \geq 2 $ i.e. both sides of the checkerboard must have length at least 2.
But after that, I wasn't sure what the basis and inductive steps would be. For the basis step, I proved $ P(1, 2) $ true, but I probably should've include more base cases, just didn't know which ones.
The inductive step was the hardest part for me. I was fairly certain this would be a proof by strong induction, since the inductive step probably involves splitting up a checkerboard into smaller boards. The issue here is that at least one of these smaller boards will not have a black and a white cell missing, meaning we cannot directly apply the inductive hypothesis.
I was also feeling iffy about applying induction on a proposition containing two variables, since we only learned how to do induction on propositions of one variable. But I couldn't figure out a formulation of the claim that uses only one variable and covers all cases for the dimensions of the board.
Is there a less convoluted way of going about doing this? Did I miss something obvious?
(Of course, this question is much more easily proven by a coloring argument, but it was assigned as homework in a section on induction, so we had to use that proof method.)