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The sequence

a_0

a_1

a_2,

\ldots

is defined as follows: a_0=2, \qquad 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

.
[hide="comment"] Hi guys ,
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

2

Thanks kiu si u
Edited by Orl.

Problem Statement