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Math Help - Prime number and congruence modulo24

  1. #1
    Super Member dhiab's Avatar
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    Prime number and congruence modulo24

    n is a prime number.
    Prove :
    \begin{array}{l}<br />
\forall n \in {\rm N}:n \ge 5 \\ <br />
n^2 - 1 \equiv 0\left[ {\bmod 24} \right] \\ <br />
\end{array}
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  2. #2
    Super Member PaulRS's Avatar
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    Since n is odd we have that n-1 and n+1 are consecutive-evem numbers -they differ in 2-, so one of them must be divisible by at 2 -and not 4- and the other divisible by at least 4, thus, since n^2-1=(n-1)(n+1) , we must have that n^2-1 is a multiple of 8.

    On the other hand, since 3 doesn't divide n - because n is a prime greater than 4 - either n=3k+1 or 3k-1 for some k in \mathbb{Z} thus 3|(n-1) or 3|(n+1) hence 3|(n^2-1)

    Now, since 3 and 8 divide n^2-1 , and 3 and 8 are coprime, then (3\cdot 8)|(n^2-1) i.e. 24|(n^2-1)
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