http://www.kalva.demon.co.uk/irish/soln/sol952.html [ edit: new link since the death of Kalva: http://web.archive.org/web/20070611004423/http://www.kalva.demon.co.uk/irish/soln/sol952.html ]

http://www.mathlinks.ro/viewtopic.php?t=103217

Peter's link doesn't work anymore and chess64's link doesn't seem to provide a correct solution so I'll provide one here. Clearly a = 0 doesn't work, and WLOG let a be negative. Note that (x, y) = (1, a) is a solution and we can Vieta jump to infinitely many solutions, namely if (x, y) is a valid solution, then so is (ay - x, y). For example, the solution (1, a) yields the solution (a, a^2 - 1) which in turn yields (a^2 - 1, a^3 - 2a) and so on. It is clear that there are then infinitely many solutions for each negative integer a except a = -1, which clearly has finitely many solutions, so every integer a whose absolute value is greater than 1 works. I don't know Latex, so if anyone could rewrite my solution (and add the actual method of vieta-jumping) that would be highly appreciated.

my solution : if we got \[x^{2}+axy+y^{2}=1 \] for $ x , y \in Z , (x,y \neq 0)$ we make $ k , t $ respectively \[x(1-a) + 2ay - a^{3}y , -y + a(ay + x)\] and we take $ 2y - a^{2}y - ax = n \Rightarrow $ we have : $ (na + x)^{2} + a(na + x)(y-n) + (y-n)^{2} = 1 $ and this sentence is right (try it !). for example : $(1,-a) , a > 1 $ is right in the equation $(1+ a(-a) + a^{2} = 1)$ and also $(1- 3a^{2} + a^{4} , 2a - a^{3}) .$ for $a = 2$ either $(1,-2)$ and $(5,-4)$ are right .