ISL 2003 N7

The sequence $a_0$ , $a_1$ , $a_2,$ $\ldots$ is defined as follows: $$a_0=2, a_{k+1}=2a_k^2-1 \text{for }k \geq 0.$$ Prove that if an odd prime $p$ divides $a_n$ , then $2^{n+3}$ divides $p^2-1$ .

Here is a nice problem:

Let be given a sequence $a_n$ such that $a_0=2$ and $a_{n+1}=2a_n^2-1$ . Show that if $p$ is an odd prime such that $p|a_n$ then we have $p^2\equiv 1\pmod{2^{n+3}}$ Here are some futher question proposed by me :Prove or disprove that :

1. $gcd(n,a_n)=1$ 2) for every odd prime number $p$ we have $a_m\equiv \pm 1\pmod{p}$ where $m=\frac{p^2-1}{2^k}$ where $k=1$ or $2$ Thanks kiu si u

Edited by Orl.

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