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I'm just starting with integer division and congruence in an algebra course and I have this problem:

Let $a$ be an odd integer. Prove that $\forall n \in \Bbb N$:

$$2^{n+2}\ |\ a^{2^n} - 1$$

I've tried a few things but got nowhere. For instance, if I try by induction, I end up with this:

First I check $P(1):$

$$2^3\ |\ a^2-1$$ Being an odd number: $$8\ |\ (2k+1)^2-1$$ $$8\ |\ 4k^2+4k$$ But how can I continue? Or can you think of another way of proving it? Thanks a lot.

jrs
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    Note that $k(k+1)$ is the multiplication of two consecutive integers, which must always be even. – abiessu May 05 '16 at 02:57
  • So could I say that $P(1)$ works because $4k^2+4k=8h$ equals $k(k+1)=2h$? Thanks @abiessu – jrs May 05 '16 at 03:03
  • I proved this in the following answer, and also gave some other results: http://math.stackexchange.com/questions/1559844/order-of-5-in-bbbz-2k/1560650#1560650 – user236182 May 05 '16 at 03:14

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