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Math Help - Prime

  1. #1
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    Prime

    Hi,

    p is a prime number such as p \ge 5, how to show that p^2+11 \equiv 0[12]
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  2. #2
    MHF Contributor chisigma's Avatar
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    Because is...

    p^{2} -1 = (p+1)\cdot (p-1) (1)

    ... and p is prime, p+1 and p-1 are both even numers and one of them is a multiple of 3, so that 12 devides p^{2}-1 and then 12 devides also p^{2} + 11...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by bhitroofen01 View Post
    Hi,

    p is a prime number such as p \ge 5, how to show that p^2+11 \equiv 0[12]


    Any prime number p\geq 5 equals 1,5,7\,\,\,or\,\,\,11\!\!\!\pmod {12} (why?) . Now just check that each of these ones satisfy the given congruence.

    Tonio
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