Cubic Diophantine

Question

I cannot seem to find a solution to the following Diophantine Equation: $x^2-y^3=2$, where $x,y \in \mathbb{Z}.$ I thought that I could maybe reduce it to a simpler equation , maybe check for the extension $\mathbb{Q}\left[\sqrt{2}\right],$ but nothing that I have tried seems to work. Perhaps it is a known equation/curve I am not aware of? I was first introduced to the problem when my friend pointed out that he could not solve it. I would appreciate your help. Thanks, in advance.

Answer 1

Let $x$ and $y$ be integers such that $x^2-y^3=2$. Then $x$ and $y$ are both odd, and $$y^3=(x-\sqrt{2})(x+\sqrt{2}),$$ where the gcd of the two factors on the right hand side divides their sum $2\sqrt{2}=\sqrt{2}^3$, and because their product is odd we see that they are coprime. Because $\Bbb{Z}[\sqrt{2}]$ is a unique factorization domain we have $$x+\sqrt{2}=u(a+b\sqrt{2})^3,$$ for some integers $a$ and $b$, and some unit $u\in\Bbb{Z}[\sqrt{2}]^{\times}$. Then $u=\pm(1+\sqrt{2})^k$ for some integer $k$, and without loss of generality we have the $+$-sign and $k\in\{0,1,2\}$. So we distinguish three cases:

• If $k=0$ then $$x+\sqrt{2}=(a+b\sqrt{2})^3=a(a^2+6b^2)+b(3a^2+2b^2)\sqrt{2},$$ and comparing the coefficients of $\sqrt{2}$ shows that $$b(3a^2+2b^2)=1,$$ which clearly has no integral solutions.
• If $k=1$ then in a similar way we find that $$a^3+3a^2b+6ab^2+2b^3=1,$$ which is a(nother) cubic Thue equation. Its unique integral solution is $(a,b)=(1,0)$, corresponding to $x=1$.
• If $k=2$ then in a similar way we find that $$2a^3+9a^2b+12ab^2+6b^3=1,$$ which is another cubic Thue equation. Its unique integral solution is $(a,b)=(-1,1)$, corresponding to $x=-1$.

Answer 2

COMMENT.- The discriminant $\Delta = - (4a ^ 3 + 27b ^ 2)$ of the curve $x ^ 2 = y ^ 3 + 2$ considering it as a case of $y ^ 2 = x ^ 3 + ax + b$ (Weirstrass form) in which they have reversed the usual roles of the coordinates, it is different from zero so the curve is not singular or elliptic. There is an obvious solution $A = (x, y) = (1, -1)$.

It is doubtful that the tangent to the curve in this integer point (that is, $2A$ in the arithmetic of the elliptic curve) produces another integer point (in fact graphically it does not give it) but if this point $A$ is not torsion then it will produce an infinity of rational points on the curve.

We do not apply the operations that define an additive group law on the curve because this is not the case here.

Regarding solving the equation $x^2-y ^ 3 = 2$ by other means, this has the savor of the Catalan's conjecture that lasted almost two centuries to be proven by Mihăilescu's so we better stop here this comment.

Answer 3

This is an elliptic curve, and can be solved by a computer algebra system.

I use the following code on sage cell server, and you can also try it yourself.

E = EllipticCurve([0, 2])
print(E.integral_points())

The output:

[(-1 : 1 : 1)]

Therefore this is (up to sign) the only non-trivial solution.

An interesting fact is that this curve actually has Mordell-Weil rank $1$, meaning that it has infinitely many rational points.

Examples:

(17/4 : -71/8 : 1)
(127/441 : 13175/9261 : 1)
(66113/80656 : -36583777/22906304 : 1)

etc.

All these calculations are done with Sage.

How does Sage do that, and how can Sage be sure that the results are correct? This is a much deeper subject. A good starting point is the GTM book "The Arithmetic of Elliptic Curves" by Joe Silverman.

Answer 4

$x^2-y^3=2\quad\implies\quad y=\sqrt[\large3]{x^2-2}\quad\lor\quad x=\sqrt{y^3+2}\quad$

Both equations are true only when $\quad (x,y)=(\pm1,-1)$

For larger values, no squares and cubes differ by only $2$.