We assume that a p × p board can be covered, and that the monomino is at the square labelled xayb. Then there are polynomials g(x, y) and h(x, y) such that (2) in Part III holds. Again, we let ω = e2πi/3, and we consider the two cases p ≡ 1 and p ≡ 2 (mod 3).

Case 1: p ≡ 1 (mod 3). In this caseωp = ω. Thus if we put x = y = ω in (2) we obtain ωa+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 x3sy3t. 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 x3s+2y3t+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?