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Math Help - Factor Theorem.

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    Factor Theorem.

    Use the factor theorem to show that if 2^p - 1, where p does not equal 3, is a prime number, then p is neither divisible by 4 or divisible by 3. {Alternatively, prove that if p is divisible by 4 or 3, the 2^p - 1 is divisible by some number other than +/- itself or +/-1.}
    Last edited by mr fantastic; June 17th 2010 at 09:31 PM. Reason: Deleted irrelevant 'HELP' in post title.
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    Quote Originally Posted by julia78 View Post
    Use factor theorem to show that if 2^p-1, where p does not equal 3, is a prime number, then p is neither divisible by 4 or 3
    See the first theorem here (proof given).
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    OMG, the answer to the question seems complicated! Thanks for the link anyway!

    I am hoping that maybe someone will know an easier way of solving this problem.... THANK YOU!!!
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by julia78 View Post
    OMG, the answer to the question seems complicated! Thanks for the link anyway!

    I am hoping that maybe someone will know an easier way of solving this problem.... THANK YOU!!!
    Well I can write it out in another way, not sure whether you'll find it easier. The formatting won't be very nice because LaTeX isn't quite working yet.

    Multiples of 3

    2^(3k) - 1 = (2^3)^k - 1 = 8^k - 1

    Consider modulo 7.

    8^k - 1 ≡ 1^k - 1 ≡ 0 (mod 7)

    Thus the number is divisible by 7. If k = 1, then the number is 7 and thus prime, otherwise it is greater than 7 and thus composite.

    Multiples of 4

    2^(4k) - 1 = (2^4)^k - 1 = 16^k - 1

    Consider modulo 15.

    16^k - 1 ≡ 1^k - 1 ≡ 0 (mod 15)

    Thus the number is divisible by 15 and always composite.
    Last edited by undefined; June 17th 2010 at 05:42 PM. Reason: typo
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  5. #5
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    Thank you so so much! Very much appreciated!
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