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Math Help - Prove

  1. #1
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    Exclamation Prove

    Prove that is p (greater than or equal to) 5 is a prime then (p^2) is congruent to 1(mod24).

    I have a hint that a positive integer k divides on of k sequential integers n,(n+1),(n+2),...,(n+k-1), where n is an integer. the product of any k sequential integers n(n+1)(n+2)...(n+k-1) is congruent to 0(mod k)
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  2. #2
    Super Member PaulRS's Avatar
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    p^2-1=(p-1)\cdot{(p+1)}\equiv{0(\bmod.24)}

    We will show that (p-1)\cdot{(p+1)} is multiple of 3 and 8, which is equivalent to show that p^2\equiv{1}(\bmod.24)

    Indeed, since p is prime and p>3 we have p\equiv{1;2}(\bmod.3) thus p^2\equiv{1}(\bmod.3) and (p-1)\cdot{(p+1)} is multiple of 3

    p is prime and p>2 so p\equiv{1;3;5;7}(\bmod.8) thus p^2\equiv{1}(\bmod.8) (try each case)

    Therefore (p-1)\cdot{(p+1)} is multiple of 3 and 8
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