For an elliptic curve $E(\mathbb{Q})$ if $nP$ has integer coordinates does $P$ as well? Consider a point $P=(x,y)$ on an elliptic curve $E$ over $\mathbb{Q}$.
If a multiple of $P$ has integer coordinates, does this necessarily mean $x$ and $y$ are integers as well?
Is there a simple way to prove this?
EDIT: If the answer depends on the torsion group, what about the case of a curve with trivial torsion group?
 A: Let $E: y^2+xy=x^3+4x+1$ and let $P=(-1/4,1/8)$. Then, $P$ is a point of two-torsion. Thus, $2P=\mathcal{O}=[0,1,0]$, so $2P$ has integral coordinates while $P$ does not. 
If the elliptic curve has trivial torsion subgroup, there are probably other examples that one could come up with, particularly if you allow the coefficients of your model to be rational numbers. So you probably want to restrict your question to elliptic curves given by a minimal integral model. 
If $E$ is given by a minimal model, here is a result towards your question. Suppose $P=(x_0,y_0)$ has a multiple with integral coordinates, say $nP$, and let $q$ be a prime dividing a denominator of $x_0$ or $y_0$ (and assume $q$ is a prime of good reduction for $E$). Then, $P \equiv \mathcal{O} \bmod q$ and $nP$ does not reduce to the identity (because its coordiantes are integral), but $nP \equiv n\mathcal{O} \equiv \mathcal{O} \bmod q$, and that's a contradiction. So this shows that if such a point $P$ exists, the primes dividing the denominators of $x_0$ and $y_0$ are all primes of bad reduction. 
A: No. Take $E$ having a rational point $P$ with of finite order, which has not integer coordinates. If $n$ is the order of $P$ then $nP=(0,0)$.
