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

Thanks,

W

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Well, looks like I can get it in the form $(2y^2-1)^2=2z^2-1$. Not sure if that helps. I'm retagging this for diophantine equations. –  Mike Aug 29 '12 at 16:18
    
$(y^2)^2+(y^2-1)^2=z^2$, One solution y=2, z=5 –  lab bhattacharjee Aug 29 '12 at 16:45
    
If you then let $x = y^2$, we have $x^2 + (x-1)^2 = z^2$: the solutions to this are leg-leg twin Pythagorean triples. –  Théophile Aug 29 '12 at 17:12
    
Using the expression of @labbhattacharjee and the general solutions of en.wikipedia.org/wiki/Pythagorean_triple hopefully will help. See also the comment of Théophile. –  vesszabo Aug 29 '12 at 17:12

1 Answer 1

up vote 5 down vote accepted

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

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