Given the elliptic curve $E:y^2=x^3-4x+4$.


(a) How to prove that the group of rational points $E(\mathbb{Q})$ is generated by $P=(2,2)$.


(b) If we consider the piece of curve on the region $0<x<2, 0<y<2$ with the aid of Magma we find the points 
\begin{eqnarray\*}
7P &=& \left(\dfrac{10}{9},\dfrac{26}{27}\right),\\
-10P &=& \left(\dfrac{88}{49},\dfrac{554}{343}\right),
\end{eqnarray\*}
and $13P, -15P, 18P$, etcetera. What can be said about the sequence $7,-10,13,-15,18,...$?



A particular interesting elliptic curve
(a) This isn't really a MathOverflow level question. See for example Cremona's book for a description of how to do a 2-descent, which is probably all that's needed for this curve.


(b) There is a real analytic isomorphism $E(\mathbb R)\to\mathbb R/\mathbb Z$, i.e., the real locus is isomorphic to the circle group. The point $P=(2,2)$ corresponds to an irrational number $\alpha\in\mathbb R/\mathbb Z$. The part of $E(\mathbb R)$ inside the box $0<x<2$ and $0<y<2$ corresponds to an interval $I\subset\mathbb R/\mathbb Z$. So your set of integers is exactly the set
$$
 \{ n\in\mathbb Z : n\alpha\bmod\mathbb Z\in I \}.
$$
This reduces you to a Diophantine approximation-type problem that really has little to do with elliptic curves. The numbers $n\alpha\bmod\mathbb Z$ are equidistributed in $\mathbb R/\mathbb Z$, but the specific set of $n$'s depends on $\alpha$. Presumably $\alpha$ has no special qualities (I think it's known to be transcendental).


