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Math Help - Prime number

  1. #1
    Junior Member Kanwar245's Avatar
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    Prime number

    For any prime p \ge 29. Prove that 182 | p^2 - 1
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    Re: Prime number

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    Last edited by rgrothee; March 21st 2013 at 05:22 AM. Reason: wrong
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    Re: Prime number

    Quote Originally Posted by Kanwar245 View Post
    For any prime p \ge 29. Prove that 182 | p^2 - 1
    If p= 31, p^2- 1= 960 which is NOT divisible by 182: 960/182= 5.274725...

    Further p- 1= 30 and p+ 1= 32 and neither divides 182 so I have no idea what you or rgrothee are asserting.
    Last edited by HallsofIvy; March 21st 2013 at 03:09 PM.
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    Junior Member Kanwar245's Avatar
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    Re: Prime number

    Quote Originally Posted by HallsofIvy View Post
    If p= 31, [tex]p^2- 1= 960 which is NOT divisible by 182: 960/182= 5.274725...

    Further p- 1= 30 and p+ 1= 32 and neither divides 182 so I have no idea what you or rgrothee are asserting.
    My bad, you're right. There's a typo. We need to show that 182|p^{12}-1
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    Re: Prime number

    p^{12}! Oh, Dear. Well, p^{12}- 1= (p+1)(p^2- p+ 1)(p-1)(p^2+ p+ 1)(p^6+ 1)
    182= 2(91)= 2(7)(13)

    Obviously, if p is prime, larger than or equal to 29, then both p+1, and p- 1 are even so it remains to show that one of the factors is divisible 7 and one by 13.
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    Junior Member Nehushtan's Avatar
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    Re: Prime number

    Quote Originally Posted by Kanwar245 View Post
    For any prime p \ge 29. Prove that 182 | p^2 - 1
    By Fermatís little theorem, 13 divides p^{12}-1. Also by Fermatís little theorem, 7 divides p^6-1 and hence 7 divides p^{12}-1=(p^6-1)(p^6+1). And obviously 2 divides p^2 - 1 since p is odd. The result follows.

    NB: This works for all integers p which are coprime with 2, 7, and 13.
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