# Quartic Diophantine equation $2 x^4 - 2 x^2 = 3 (y^2 - 1)$

$$2 x^4 - 2 x^2 = 3 (y^2 - 1)$$

On oeis.org/A180445 it says that all positive solutions $(x,y)$ are: $$(1,1)\ \ (2,3)\ \ (3,7) \ \ (6,29)\ \ (91,6761)$$

I am looking for a simple proof for that (if any simple proof exists, because I know that these kind of equations can be very difficult). Introducing $X=2x^2-1$, it becomes

$$6 y^2-X^2 = 5$$ which may be more tractable.

EDIT: Indeed, here the solutions are infinitely many: there are two families, for all $k\ge 1$. Let,

$$u_1 =(5 - 2 \sqrt6)^k\\ u_2 = (5 + 2 \sqrt6)^k$$

then,

$$X_k=\frac{ ( 1 + \sqrt6) u_1 +(1 - \sqrt6) u_2}{2}$$ $$y_k=\frac{ (6 + \sqrt6) u_1 + (6 - \sqrt6) u_2}{12}$$ and $$X_k=\frac{ (1 - \sqrt6) u_1 + (1 + \sqrt6) u_2}{2}$$ $$y_k=\frac{ (6 - \sqrt6) u_1+ (6+ \sqrt6) u_2}{12}$$

but still I don't know how to sort out those corresponding to solutions to the original equation.

• I think you should check this: en.wikipedia.org/wiki/Pell's_equation – peter.petrov Mar 28 '16 at 19:18
• ... and this one: en.wikipedia.org/wiki/Chakravala_method – peter.petrov Mar 28 '16 at 19:20
• Thanks, It explains how to solve the $6y^2-X^2=5$. I will make an edit, accordingly. But there are infinitely many solutions, and I don't quite see how to sort out those corresponding to the original equation. – René Gy Mar 28 '16 at 22:05
• A relation $F(x)=y^2$, where $F(x)$ is a cubic or quartic, with an initial rational point is birationally equivalent to an elliptic curve. Thus, it has only finitely many integer points. To prove your solution set is complete in the positive integers is a bit more difficult. – Tito Piezas III Mar 29 '16 at 4:32
• @Tito Piezas III. would you kindly explain what " birationally equivalent to an elliptic curve" means? (If not too complicated). Also, I thought maybe a specific proof might exist for the solutions to this equation because it can be factored like $$x \binom{x+1}{3}=2 \binom{ \frac{y+1}{2} }{2}$$ involving binomial coefficients, and using known properties of the binomial coefficients. But I could not achieve that. – René Gy Mar 29 '16 at 16:32

$$2 x^4 - 2 x^2 = 3 (y^2 - 1)\tag{1}$$
$$v^2 = 6x^4-6x^2+9\tag{2}$$
IntegralQuarticPoints($$[6,0,-6,0,9]$$);
It says that all integral points of equation $$(2)$$ are $$[(0,3),(\pm1,\pm3),(\pm2,\pm9),(\pm3,\pm21),(\pm6,\pm87),(\pm91,\pm20283)]$$.
Hence all positive integral points of equation $$(1)$$ are $$(x,y)=(1,1),(2,3),(3,7),(6,29),(91,6761).$$