A is the 2 x 2 matrix (a_{ij}) with a_{11} = a_{22} = 3, a_{12} = 2, a_{21} = 4 and I is the 2 x 2 unit matrix. Show that the greatest common divisor of the entries of A^{n} - I tends to infinity.

**Solution**

It is a trivial induction that A^{n} has the form (c_{ij}), where c_{11} = r, c_{12} = s, c_{21} = 2s, c_{22} = r, with r an odd integer which tends to infinity with n (for example, r > 2^{n}), and s an even integer. Hence the gcd of the entries of A^{n} - I is the gcd of r - 1 and s. Also det A = 1, hence det(A^{n}) = 1 and so r^{2} - 2s^{2} = 1, or (r - 1)(r + 1) = 2s^{2}. It follows that the gcd of r - 1 and s is either √(r - 1) or √(2r - 2). [If p is an odd prime and h is the highest power of p that divides r - 1, then p does not divide r + 1, and so h must be even and h/2 the highest power of p dividing n. Both r - 1 and r + 1 are even, but one is not divisible by 4. If h is the highest power of 2 dividing the other, then h must be even and h/2 is the highest power of 2 dividing n.] But r - 1 tends to infinity with n, so the gcd does also.

© John Scholes

jscholes@kalva.demon.co.uk

12 Dec 1998