If $|a_{n+1} - 2a_n| = 2$ and $|a_n| \leq 2$ and $a_1$ is rational, then it is as periodic sequence. 
If $|a_{n+1} - 2a_n| = 2$ and $|a_n| \leq 2$ and $a_1$ is rational, then it is as periodic sequence.

Periodic sequence means that there exists $p\in \mathbb{N}$ such that $a_n = a_{n+p}$ for all values of $n.$
I have no idea how to start at all.
Any hint is appreciated.
 A: Since $a_1$ is rational, we know that $qa_1 = p$ for some integers $p,q$.
Now, note that either $a_{n+1} = 2a_n - 2$ if $a_n > 0$, and $a_{n+1} = 2 + 2a_n$ otherwise.
Using this, one sees by induction that $qa_{n}$ is an integer for all $n$. However, because $|a_n| \leq 2$ , we have $|qa_n| \leq 2q$ for all $n$.
Therefore, the sequence $a_n$ takes values only in the set $\{\frac kq : -2q \leq k \leq 2q\}$. By the pigeonhole principle, $a_i = a_j$ for some $i<j$ finite (Take $a_1,...,a_{2q+2}$). 
Now, once this happens, noting that $a_n \to a_{n+1}$ is an injective function (the inverse is clear from my restatement of the mapping above) tells you that the sequence is periodic, because $a_i = a_j$ implies $a_{k} = a_l $ for all $k,l$ differing by $j-i$. (For $k < i$, use injectivity. For $k>i$, repeatedly apply the recurrence).
EDIT :Suppose $a_1 \neq 0$. Then, if $|a_1| < 2$ we can use the recurrence to conclude that $|a_n | < 2$ for all $n$. In this case, injectivity is obtained. If that $a_1 = \pm 2$. Then, the sequence looks like $2,-2,2,-2,...$ or the negative of each term. The only problem occurs when some $a_n = 0$, in which case the sequence looks like $0,2,-2,2,-2,...$ or its negative after that. Either way, the sequence is eventually periodic.
