Fermat's Bachet-Mordell Equation  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.
 A: 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.
