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Let $x,y \in \mathbb{Z}$ satisfying $3y^2 = 4x^3 - 1$. Does it follow that $x = 1$ and $y = \pm 1$?

Wolfram Alpha says that the answer is positive, but I am not so satisfied with an answer by a computer program since it is (most of the time) not accompanied by a proof, and even if it is, such a proof may be rather long, not illuminating, and difficult to verify.

What I would like to see most is a (reasonably short and complete) proof (or a counterexample).

I have used unique factorization in $\mathbb{Z}[\frac{-1 + \sqrt{-3}}{2}]$to show that it suffices to prove that the only $a,b \in \mathbb{Z}$ satisfying $a^3 + b^3 -6a^2b + 3ab^2 + 1 = 0$ are $(-1,0), \ (0,-1), \ (1,1)$ and verified this using Wolfram Alpha, but I am not sure whether this gets me any closer to a proof.

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    $\begingroup$ Wolfram|Alpha isn't necessarily a good tool to tell you all the integer points on a curve. For example, WA doesn't give you the integral point $(1318,47849)$ on $y^2=x^3-2x+5$. For more examples (I didn't check them all on WA, but I suppose it doesn't return them), take any curve from table 1, 2 or 3 of this paper people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2007900/… $\endgroup$
    – Wojowu
    Commented Aug 1, 2016 at 17:43
  • $\begingroup$ @Wojowu you are of course right, but I just named WA as an example of a software in order to stress that I am not looking for a computer answer. $\endgroup$
    – Pablo
    Commented Aug 1, 2016 at 17:52

1 Answer 1

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The projective form of your curve is $3y^{2} z = 4x^{3} - z^{3}$. This has three obvious points: $(1 : 1 : 1)$, $(1 : -1 : 1)$, and $(0 : 1 : 0)$.

Your curve is isomorphic over $\mathbb{Q}$ to the Fermat cubic, $x^{3} + y^{3} = z^{3}$. This latter curve has only three rational points on it: $(1 : -1 : 0)$, $(1 : 0 : 1)$ and $(0 : 1 : 1)$, and so your curve only has the three rational points named. (If you want to read a proof of the $n = 3$ case of FLT, you can read Euler's proof, which is essentially the same as the one in Hardy and Wright's ``An Introduction to the Theory of Numbers,'' see pages 193-195.)

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    $\begingroup$ What would an isomorphism over $\mathbb{Q}$ look like? And why does it necessarily give a bijection between integral points? $\endgroup$
    – Pablo
    Commented Aug 1, 2016 at 17:48
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    $\begingroup$ It doesn't necessarily give a bijection between integral points. But the NON-EXISTENCE of other rational points on one curve immediately implies the non-existence of other rational points on the other curve, and a fortiori the non-existence of other integral points. $\endgroup$
    – WhatsUp
    Commented Aug 1, 2016 at 17:53
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    $\begingroup$ @WhatsUp Right! But how do I show that the curves are isomorphic over the rationals? $\endgroup$
    – Pablo
    Commented Aug 1, 2016 at 17:55
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    $\begingroup$ The isomorphism from $3y^2z = 4x^3 - z^3$ to $X^3 + Y^3 = Z^3$ is given by $X = (y-z)/2$, $Y = x$ and $Z = (y+z)/2$. $\endgroup$
    – Jeremy Rouse
    Commented Aug 1, 2016 at 18:02
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    $\begingroup$ @Pablo and Wojowu - I used Magma, and knew off the top of my head that $y^{2} + y = x^{3} - 7$, one elliptic curve with conductor $27$, is isomorphic to the Fermat cubic. $\endgroup$
    – Jeremy Rouse
    Commented Aug 1, 2016 at 18:18

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