I would like to find those integers $x,y$ that satisfies $y^2=x^3+1$. Is there some elementary way to find those?
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I'm not sure the following counts as "elementary" but it's certainly not too difficult. First show your curve has rank 0. To do this, I asked SAGE, but it's not so hard to do by hand. If you rewrite your curve $C$ as $y^2 = x^3-3x^2+3x$ by renaming $x+1$ as $x$, then you can write down the 2-isogeneous curve $\overline{C}:y^2=x^3+6x^2-3x$. You are now in the setup to apply the algorithm discussed in Ch. III.6 of the book "Rational Points on Elliptic Curves" by Silverman-Tate. (Which book I highly recommend for learning how to solve this sort of problem.) The upshot is just that $C$ has rank 0, which implies by general theory (Nagell-Lutz) that all its rational points are integral. Moreover, the Nagell-Lutz theorem even says what possible y-coordinates can occur: either $y$ is zero (for points of order 2) or else $y$ divides the discriminant. In your case this says that for integral points, $y=0$ or $y$ divides $-27$. The rest is easy. |
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Euler's arguments (see Robin's answer) are reproduced and discussed in "A note on Pépin's counter examples to the Hasse principle for curves of genus 1", which can be found here. |
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This curve has rank zero, which one can prove by descent, so all rational solutions are integer solutions. This was originally proved by Euler, and his argument (in Latin) is reproduced in Rene Schoof's book on Catalan's Conjecture. |
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