Solving $2$ Variable Diophantine Equation in Rational Numbers

Question

I am trying to solve the following diophantine equation:

$5x^4+8x^3-6x^2-8x+5=y^2$

where $x$ and $y$ are rational.

For natural $x$ and $y$ the solution is simple enough, so I am wondering about the solution in rational numbers.

ADDITION:

In order to find a solution I have used the expression: $5x^4+8x^3-6x^2-8x+1=y^2-4=(x-1)(x+1)(5x^2+8x-1)=(y-2)(y+2)$

And if we take into account that $(x-1, x+1)=1$ for even $x$ and $(x-1, x+1)=2$ for odd x. For odd $y$ the $(y-2,y+2)=1$ and for even $y=2y_1$ the $(y_1-1,y_1+1)=1$ and $2$ accordingly for odd and even $y_1$. We have some finite set of variations. I have an assumption that the only solutions are x=1 and y=2. But need to spend some time for strict prove.

Answer 1

I can give you (a partial) answer using the computer algebra system MAGMA. Your curve is of genus 1, i.e. is an elliptic curve. To see if it has finitely many rational points you have to compute its rank. If rank$\geq 1$ then it has infinitely many rational points. So we shall compute a minimal Weierstrass model first and then we shall compute its rank (which eventually is zero).

PP:=ProjectiveSpace(Rationals(),2);
C:=Curve(PP,y^2*z^2 - 5*x^4-8*x^3*z+6*x^2*z^2+8*x*z^3-5*z^4);
P0:=C![1,2,1];
E, phi:=EllipticCurve(C,P0);
Em, psi:= MinimalModel(E);
E;
Em;
Rank(Em);

Remark that I guessed the point $(2,1)$ in order the system to compute the model. So the set $E({\bf Q})$ is finite. For the code also see : Solving $2$ Variable Diophantine Equation in Rational Numbers