Mike wrote:

> With 3 divisors, each must be the square of a prime, of course.

> Are there any such chains of length > 1? (I haven't searched.)

Yes, there are a few easy examples. There are no examples of

such chains with length > 2; this can probably be proven in

several ways, but a simple analysis of the solutions to the

Pell equations (x^2 = 2*y^2 +/-1) suffices.

Of length 2, the following very small examples exist:

k=5, n=(0,1), k*2^n-1 == (4,9) (Only example of form k*2^n-1)

k=3, n=(3,4), k*2^n+1 == (25,49)

k=105, n=(3,4), k*2^n+1 == (841,1681)

And some bigger ones:

k=248204571168918457275, n=(3,4), k*2^n+1

k=22980046115755920139039424449209532289506425, n=(3,4), k*2^n+1

I'm not sure if there are any more. An heuristic argument would

indicate that the total number of such chains is finite.