Thue equation $ x^4 - 6 x^3 y - x^2 y^2 + 6 x y^3 - y^4=-1 $

Question

I need help solving the Thue equation $ x^4 - 6 x^3 y - x^2 y^2 + 6 x y^3 - y^4=-1 $. It can be written as $ x(x-y)(x+y)(x-6y) = (y-1)(y+1)( y^2 +1) $. From this I found 8 solutions (0,1),(0,-1),(1,1),(-1,-1),(-1,1),(1,-1),(6,1) and (-6,-1). But there are two more solutions (15,17) and (-15,-17), and I don't know how to get them. I would be grateful for any kind of suggestions.

Answer 1

OP seems to be looking for a simple method that would ensure an easy discovery of all solutions listed. One such simple method is to let $y$ run through $\pm 1, \pm 2, \pm 3, ...$ and calculate $(y−1)(y+1)(y^2+1) = y^4-1$. From the left hand side, we can see that $x$ must be a divisor of $y^4-1$ and we simply check all divisors of $y^4-1$. For $ y = \pm 17$ we get that $y^4-1 = 83520$ which is divisible by $15$.