Fermat challenged Frenicle with finding a pythagorean triple (a,b,c) where $(a-b)^2-2b^2$ is itself a square. By making the substitution $a=m^2-n^2$, $b=2mn$, and $c=m^2+n^2$ into $(a-b)^2-2b^2=d^2$ we obtain the following quartic:
The goal of the exercise is to somehow obtain a curve that corresponds to this equation, and to then use another curve, namely a quadratic intersecting it with this curve to generate another solution from the solution (1,0,1).
This method was used to show that there also exists a pythagorean triple (a,b,c) where c and a-b are both squares. Using the curve $y^2=2x^4-1$ and using an intersecting quadratic to generate a nontrivial solution.
Any hints or tips would be appreciated.