Diophantine equation $2y^4-2y^2 +1=z^2$

Question

How to solve the diophantine equation $2y^4-2y^2 +1=z^2$, where $(y,z) \in \mathbb{N}^2$ ?

Thanks,

W

Answer 1

Since the quartic has the rational solution $y=0,x=1$, it is birationally equivalent to an elliptic curve.

We find the curve to be $j^2=k^3+4k^2-4k$ with $y=2k/j$ and $x=k(k^2+4)/j^2$.

The curve has one finite torsion point $(0,0)$, and Denis Simon's ellrank package gives the rank to be 1 with generator $G=(2,4)$. Thus there are an infinite number of RATIONAL solutions to the equation.

For example, $5G$ gives $y=187/23$ and $x=49081/529$.

Increasing the multiples of $G$ gives rational solutions with larger and larger sizes of the numerator and denominator.

For natural numbers, $(y,x)= (0, 1), ( 1, 1), ( 2, 5)$ SEEM to be the only solutions. I would be very surprised if higher multiples of $G$ give an integer solution, but this is not a proof.

Hope I have this all correct!!

Allan MacLeod