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.