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From: Christopher Kolago on 3 Nov 2009 08:22
Why does:

2^n != 1 (mod n)

for every n > 1? Is there any "simple" proof of this fact?

Chris
From: Pubkeybreaker on 3 Nov 2009 18:39
On Nov 3, 6:22�pm, Christopher Kolago <krzysztof.kol...(a)uj.edu.pl>
wrote:
> Why does:
>
> 2^n != 1 (mod n)
>
> for every n > 1? Is there any "simple" proof of this fact?

Hint:

Lagrange's Theorem. The order of any element must
divide the order of the group. The group is Z/nZ*.
I assume you know how to compute its order?
Now ask if n divides its order......
From: master1729 on 3 Nov 2009 08:30
> Why does:
>
> 2^n != 1 (mod n)
>
> for every n > 1? Is there any "simple" proof of this
> fact?
>
> Chris

lol

how is 2^2n ! = 1 mod 2n ?
From: Christopher Kolago on 3 Nov 2009 11:08
> Hint:
>
> Lagrange's Theorem. The order of any element must
> divide the order of the group. The group is Z/nZ*.
> I assume you know how to compute its order?
> Now ask if n divides its order......

I know that order(Z/pZ*)=p-1, but it is a group only if p is a prime number. I can't see how this leads to solutions of my problem. Thanks anyway! :)

By the meantime I proved that 2^n != 1 (mod n) using Euler's theorem. :D

Chris
From: Christopher Kolago on 3 Nov 2009 11:14
> > Why does:
> >
> > 2^n != 1 (mod n)
> >
> > for every n > 1? Is there any "simple" proof of
> this
> > fact?
> >
> > Chris
>
> lol
>
> how is 2^2n ! = 1 mod 2n ?

Is wish it was so easy, but it's not. ;P

In the meantime I managed to prove that 2^n != 1 (mod n) using the Euler's theorem.

Chris
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