Show that there exist two strictly increasing sequences a1, a2, a3, ... and b1, b2, b3, ... such that an(an + 1) divides bn2 + 1 for each n.
Solution
A little experimentation suggests that such pairs are relatively rare. The smallest b seems to be 57, with 25.26 dividing 572 + 1 (and a more extensive search than one would want to carry out by hand suggest there are no other b under 200).
Suppose d2 divides c2 + 1. Put b = c(d2 + 1) + d3. Then b2 + 1 = c2(d2 + 1)2 + 2cd3(d2 + 1) + d6 + 1. Since d6 + 1 = (d2 + 1)(d4 - d2 + 1), it is clear that b2 + 1 is divisible by d2 + 1. Since c2 + 1 is divisible by d2, b2 + 1 is also divisible by d2. So if we can find c and d, then we can take a = d2 and b = c(d2 + 1) + d3.
That is more promising. We can specialise down to c2 + 1 = 2d2. We may remember that solutions to the Pell equation (and its variants) generate further solutions as simple linear combinations. So starting with (m, n) satisfying m2 + 1 = 2n2 we get M2 + 1 = 2N2 by taking M = 3m + 4n, N = 2m + 3n. So starting with the solution (7, 5) we get (41, 29), (239, 169), ... . That gives an increasing sequence (c, d) and hence an increasing sequence (a, b).
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 2002