IMO Shortlist 2003 problem N7


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2. travnja 2012.
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The sequence a_0, a_1, a_2, \ldots is defined as follows: a_0=2, \quad a_{k+1}=2a_k^2-1 \quad for k \geq 0. Prove that if an odd prime p divides a_n, then 2^{n+3} divides p^2-1.

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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 or 2

Thanks kiu si u


Edited by Orl.
Izvor: Međunarodna matematička olimpijada, shortlist 2003