Fermat's Bachet-Mordell Equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:23:55Z http://mathoverflow.net/feeds/question/36800 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36800/fermats-bachet-mordell-equation Fermat's Bachet-Mordell Equation Franz Lemmermeyer 2010-08-26T19:32:16Z 2010-08-27T10:17:21Z <p>Fermat once claimed that the only integral solutions to $y^2 = x^3 - 2$ are $(3, \pm 5)$. Fermat knew Bachet's duplication formulas (more precisely, Bachet had a formula for computing what we call $-2P$), which for $y^2 = x^3 + ax + b$ says $$x_{2P} = \frac{x^4-8bx}{4x^3+4b} = \frac x4 \cdot \frac{x^3 - 8b}{x^3+b}.$$</p> <p>Using this formula it is easy to prove the following:</p> <p><em> Consider the point $P = (3,5)$ on the elliptic curve $y^2 = x^3 - 2$. The $x$-coordinate $x_n$ of $[-2]^nP$ has a denominator divisible by $4^n$; in particular, $[-2]^nP$ has integral coordinates only if $n = 0$.</em></p> <p>In fact, writing $x_n = p_n/q_n$ for coprime integers $p_n$, $q_n$, we find $$x_{n+1} = \frac{x_n}4 \cdot \frac{x_n^3 + 16}{x_n^3 - 2} = \frac{p_n}{4q_n} \cdot \frac{p_n^3 + 16q_n^3}{p_n^3 - 2q_n^3}.$$ Since $p_n$ is odd for $n \ge 1$ and $q_n = 4^nu$ for some odd number $u$ (use induction), we deduce that the power of $2$ dividing $q_{n+1}$ is $4$ times that dividing $q_n$.</p> <p>My question is whether the general result that $kP$ has integral affine coordinates if and only if $k = \pm 1$ can be proved along similarly simple lines. The modern proofs based on the group law, if I recall it correctly, use Baker's theorem on linear forms in logarithms.</p> http://mathoverflow.net/questions/36800/fermats-bachet-mordell-equation/36868#36868 Answer by Chris Wuthrich for Fermat's Bachet-Mordell Equation Chris Wuthrich 2010-08-27T10:17:21Z 2010-08-27T10:17:21Z <p>The above result for $p=2$ can be strengthen: the denominators of $x(kP)$ and $y(kP)$ are even if and only $k$ is even. In non-elementary language, this follows from the fact that the Tamagawa number at $2$ is 1 (or that $P$ has good reduction at $2$) and that the reduction is additive; so the kernel of reduction has index $2$. It also follows easily from the duplication formula and the fact that $2$ can not divide $x$.</p> <p>Similarly, for any prime $p$, one could find a congruence for $k$ such that the denominator of $x(kP)$ is divisible by $p$. If $p$ has good reduction, i.e. $p>3$ then there is a number $M_p$ dividing $N_p=\vert\tilde E(\mathbb{F}_p)\vert$ such that $x(kP)$ is not $p$-integral if and only if $k$ is a multiple of $M_p$. </p> <p>So the answer to your question is, I guess, a "No". Just from looking at the group law, i.e. the addition and duplication formula, without using something further, one can not be certain that for any $k$ there is a $p$ such that $M_p$ divides $k$. E.g. the denominator of $x(5P)$ is the square of $29 \cdot 211\cdot 2069$.</p> <p>As Kevin comments, there are of course other elementary ways, not along these lines.</p>