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    Please help me prove

    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
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    assume that 2^n-1 is divisible by n, i.e. 2^n-1=kn for some k\in\mathbb{Z}

    then, err... maybe consider the cases that arise according to the parity of n
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    Sorry, I give up. Just can't fathom it.
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    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?)
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    (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?
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    (Original post by methusaleh)
    Thanks for ur help pheylan, but what about k odd, n odd?
    k even, n odd also fails :sigh: better wait for someone cleverer to come along
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    (Original post by Pheylan)
    k even, n odd also fails :sigh: 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.
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    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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