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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 $$ x_{2P} = \frac{x^4-8bx}{4x^3+4b} = \frac x4 \cdot \frac{x^3 - 8b}{x^3+b}.$$

Using this formula it is easy to prove the following:

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 $n = 0$.

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 $q_n$.

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.

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Franz---if you factorise $y^2+2$ in the integers of $\mathbf{Q}(\sqrt{-2})$ then you can find all integer points on the curve easily and deduce what you want. Is this ekementary enougn for you? – Kevin Buzzard Aug 26 '10 at 21:36
Or even elementary enough? – Kevin Buzzard Aug 26 '10 at 21:38
@Kevin, that approach must be elementary, it's in Elementary Number Theory, by Uspensky and Heaslett, pages 398-9. – Gerry Myerson Aug 27 '10 at 0:52
@Kevin: Although Weil suggests that Fermat had a solution along these lines (with the arithmetic in the ring of integers replaced by the language of binary quadratic forms), I am certain that Fermat used the approach outlined above. I just wonder if it can completed. – Franz Lemmermeyer Aug 27 '10 at 12:32
Réalis also gave a potentially useful formula: \begin{align} a^3 - b^2 &= \left(\frac{9a^4-8ab^2}{4b^2}\right)^{\!3}\! - \left(\frac{27a^6-36a^3b^2 + 8b^4}{8b^3}\right)^{\!2}. \end{align} Maybe that can be used to show what you want? – Kieren MacMillan Sep 14 '14 at 14:48

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 $x$.

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 $M_p$.

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 $29 \cdot 211\cdot 2069$.

As Kevin comments, there are of course other elementary ways, not along these lines.

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I was also thinking of showing that the absolute value of the denominator increases when you go from, say, 2kP to (2k+1)P. – Franz Lemmermeyer Aug 27 '10 at 12:34
Oh, I see. Maybe, htat works. Here the point $P$ has everywhere good reduction, so the sqrt $e(kP)$ of the denominator of $x(kP)$ satisfies $e(kP) = f_k(P)\cdot e(P)^{k^2} = f_k(P)$ where $f_k$ is the $k$-th division polynomial. – Chris Wuthrich Aug 27 '10 at 13:45

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