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# Hard maths proof help! Watch

2^(n) - 1 is indivisible by n

There are infinitely many n that divide 2^(n)+1

Please spoiler solutions. I need you to help me help myself. Thanks
2. bump
3. assume that is divisible by n, i.e. for some

then, err... maybe consider the cases that arise according to the parity of n
4. Sorry, I give up. Just can't fathom it.
5. k even

if n is even: 2^n - 1 is odd, kn is even*even=even, this is a contradiction. (you've learnt proof by contradiction, right?)
6. (Original post by Pheylan)
k even

if n is even: 2^n - 1 is odd, kn is even*even=even, this is a contradiction. (you've learnt proof by contradiction, right?)
Thanks for ur help pheylan, but what about k odd, n odd?
7. (Original post by methusaleh)
Thanks for ur help pheylan, but what about k odd, n odd?
k even, n odd also fails better wait for someone cleverer to come along
8. (Original post by Pheylan)
k even, n odd also fails better wait for someone cleverer to come along
i have tried almost every trick in the book ( an admittedly small book comprising my knowledge of the subject). Last night I tried to solve this problem before going to bed...and believed I had done it. When I woke up in the morning i couldnt remember my solution lol. I think it was some kind of contradiction.
9. Oh I have just realised a^p = a mod p. Fermat's little theorem
therefore 2^n= 2 mod n
therefore 2^(n)-1=1 mod n.

QED

now to prove fermats little theorem.

However this result seems to contradict the result of the second question, so I think I hve made a mistake.
p must be a prime in fermats little theorem. Back to the drawing board.

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Updated: December 9, 2010
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