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?
D. T. Walker, ``On the Diophantine Equation $mX^2 - nY^2 = \pm 1$, American Mathematical Monthly, Volume 74, Issue 5, (May 1967), pp. 504-513, has a theorem that answers Professor Wilson's query. Consider (1) $aX^2 - bY^2 = \pm 1$. Theorem 9 in the above is (a and b are positive integers, not squares (and whose product is not a square)): If (1) is solvable for both of a, b > 1 and has r \sqrt{a} + s \sqrt{b} as its smallest solution; and if i is a nonnegative integer, then the formula r_i \sqrt{a} + s_i \sqrt{b} = (r \sqrt{a} + s \sqrt{b} )^{2i + 1} gives all possible solutions of (1). So r will always divide r_i and s will always divide s_i. Any a, b missing from the list in Professor Clark's reply apparently are not there because either the equation has no solution for those values, or because at least one of r, s is 1. John John Robertson 973-331-9978 jpr2718@aol.com