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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2$\begingroup$ Look it up in Mordell's Diophantine Equations. $\endgroup$– Robin ChapmanCommented May 23, 2010 at 16:03
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$\begingroup$ The local university has no copy of it nor any library near me. $\endgroup$– StudentCommented May 23, 2010 at 16:14
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1$\begingroup$ Try $y^2 = x^3 + 7.$ This one really is elementary. $\endgroup$– Will JagyCommented May 23, 2010 at 18:46
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$\begingroup$ mathoverflow.net/questions/7907 $\endgroup$– Steve HuntsmanCommented May 23, 2010 at 19:05
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3 Answers
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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.
answered May 23, 2010 at 16:57
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2$\begingroup$ I should remark that if your question is a practical rather than theoretical one, the best way to proceed is: run SAGE, type E = EllipticCurve([0,1]), then type E.integral_points(both_signs=True). $\endgroup$ Commented May 23, 2010 at 17:01
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1$\begingroup$ There are other ways to show rank 0. One is to compute the analytic rank as 0 (via L-series or modular symbols) and apply Kolyvagin. A worry with SAGE, or at least PARI used by SAGE, is that it automatically assumes a GRH-like bound when computing class numbers. This is not a problem if you use Cremona's invariant based method to compute covering curves, but you need to read the details closely. I don't know which is the default in SAGE, or how to control it as an option (Simon's method). In the case where the curve has a 2-isogeny, it might use a different process altogether. $\endgroup$– JunkieCommented May 24, 2010 at 1:47
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2$\begingroup$ The idea that a curve of rank 0 has all its rational points as integral is dependent on the model. See 14a2: $y^2 + xy + y = x^3 - 36x - 70$ and note $(-9/4,5/8)$ though the rank is 0. $\endgroup$– JunkieCommented May 24, 2010 at 1:49
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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.
answered May 23, 2010 at 17:03
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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.
answered May 23, 2010 at 18:39