Given an integer sequence $a_{n+1}=a_{n}^{2}-a_{n}-1$, prove that $\forall n\in \mathbb{Z}$, $a_{n+1}$ and $2n+1$ are coprime This is a very interesting infinite integer sequence problem I came across. The crux of the matter lies in comparing the values and the indices - we are supposed to check whether $a_{n+1}$ and $2n+1$ are coprime as opposed to any two members of the sequence itself. I tried plugging in a few numbers. In the cases when $a_{1}$ is $-1$,$0$,$1$ or $2$ the sequence starts oscillating between $-1$ and $1$ and the proof becomes trivial. In general, however, the numbers obviously become large really fast. What kind of approach should be taken?
Thank you for the help
 A: The very  first observation is that if for some prime number $p$ and positive integer $n$ we have $p|a_n$ then $a_m$ is not divisible by $p$ for all $m>n$ (*) (the modulo will be $0,-1,1,-1,1,\dots$). We will return to this observation later in this post.
Let $P(x)=x^2-x-1$ 
For any odd prime number $p$, consider the sequence $b_n = (a_n \mod p)$ for all $n \ge 0$.
Note that for any number $a$, $P(a) \equiv P(1-a) (\mod p)$ hence the modulo of $P(a)$ mod $p$ has at most $\frac{p+1}{2}$ different values .
So by pigeonhole principle, there are two distinct numbers $1 \le m<n \le \frac{p+1}{2}+1$ such that $b_m=b_n$, i.e.
From the position $m$, the sequence $(b_k)$ repeats with the period $n-m$ (1).
We will show next there is not any $k$ that lies between $m,n$ such that $b_k=0$(2).
Indeed, if its true, that leads to $b_{k+m-n}=0$, which means $a_k$ and $a_{k+m-n}$ are divisible by $p$. This is a contradiction to our observation (*).

From (1) and (2), we conclude that $b_k \ne 0$ for all $k \ge m$. Besides $m \le \frac{p+1}{2}$, thus in general, $b_k \ne 0 $ with all $k \ge \frac{p+1}{2}$. 
In other words, $a_k$ is not divisible by $p$ for all $k\ge \frac{p+1}{2}$. Hence the conclusion.
