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
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