21 - Problem F: Discussion and Generalisations

Summary

We assume that a p × p board can be covered, and that the monomino is at the square labelled x^ay^b. Then there are polynomials g(x, y) and h(x, y) such that (2) in Part III holds. Again, we let ω = e^2πi/3, and we consider the two cases p ≡ 1 and p ≡ 2 (mod 3).

Case 1: p ≡ 1 (mod 3). In this caseω^p = ω. Thus if we put x = y = ω in (2) we obtain ω^a+b = 1. This implies that a + b ≡ 0 (mod 3). Next, we put x = ω and y = ω² in (2) and obtain a + 2b ≡ 0 (mod 3). The only solution to these simultaneous congruences is a ≡ b ≡ 0 (mod 3), so that a covering of a p × p square with p ≡ 1 (mod 3) is only possible if the monomino is placed at a square with a label x^3sy^3t. In particular, if p = 4 then the monomino must be placed in a corner of the board. Although it is easy to give an ad hoc proof of this fact, it is always preferable to give a proof which is likely to be generalised.

Case 2: p ≡ 2 (mod 3). Again we take x = y = ω and we obtain a + b ≡ 1 (mod 3). If we take x = ω and y = ω2 we obtain a + 2b ≡ 0 (mod 3), so we find that a ≡ b ≡ 2 (mod 3). Thus in this case the monomino must be placed at a square labelled x^3s+2y^3t+2. In particular, if p = 5 then the monomino must be placed at the centre of the board.

• • Does a covering exist in Case 1 for any admissible choice of s and t? Does a covering exist in Case 2 for any admissible choice of s and t?

• • Is it possible to cover a 3 × 7 board using one monomino and ten 2 × 1 tiles? Is it possible to cover a 5 × 5 board using one 4 × 1 tile and seven triominoes?