David Wilson asked:

For which (a, b) do all solutions of a x^2 - b y^2 = [1] have the form (r k, s j) for some fixed r, s > 1?

Consider the binary quadratic form equation Ax^2 + Bxy + Cy^2 = N, with discriminant D = B^2 - 4AC. Any two solutions x, y and x', y' are equivalent if one can be derived from the other by x = ((u-Bv)/2)x' - Cvy', y = Avx' + ((u+Bv)/2)y', where u, v is any solution to X^2 - DY^2 = 4. When N = 1, any two solutions are equivalent. (See the section ``2 The structure of solutions'' in Keith Matthews, The Diophantine equation ax^2 + bxy + cy^2 = N, D = b^2 - 4ac > 0, Journal de Theorie des Nombres de Bordeaux, 14 (2002) 257-270.) Let r, s be the minimal positive solution to ax^2 - by^2 = 1. Then (4ar^2 - 2)^2 - 4ab(2rs)^2 = 4. Assume, for the moment, that 4ar^2-2, 2rs is the minimal positive solution to X^2 - 4abY^2 = 4. If x', y' is a solution to ax^2 - by^2 = 1, and if x' is a multiple of r and y' is a multiple of s (as they are for the minimal solution), then x and y from the above formulas are also multiples of r and s respectively (because v = 2rs). So, by induction, all solutions are multiples of r and s. This says that solutions to ax^2 - by^2 = 1 can fail to have x and y be multiples of r, s only when 4ar^2-2, 2rs is not the minimal positive solution to X^2 - 4abY^2 = 4. I do not know whether this can occur. But, instead of Professor Wilson's question, we have the question as to whether there are any equations ax^2 - by^2 = 1 where x and y for any solution are _not_ multiples of those for the minimal solution. Richard Mollin probably discusses this, and it seems to me there is a paper in the Monthly from the 1950s that may address this issue. John John Robertson 973-331-9978 jpr2718@aol.com