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Thread: Congruence

  1. #1
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    Congruence

    Hi,

    How tow show that ($\displaystyle \forall n \in$ N) $\displaystyle 5n^3+n \equiv 0[6]$????
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  2. #2
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    Quote Originally Posted by bhitroofen01 View Post
    Hi,

    How tow show that ($\displaystyle \forall n \in$ N) $\displaystyle 5n^3+n \equiv 0[6]$????
    This congruence can be rewritten as:
    $\displaystyle -n^3+n \equiv 0 \pmod 6$
    $\displaystyle n^3 \equiv n \pmod 6$

    One way is to try all possibilities: $\displaystyle n\equiv 0,\ldots,n\equiv 5$.
    Another possibility is to notice that this is equivalent to two congruences
    $\displaystyle n^3 \equiv n \pmod 2$
    $\displaystyle n^3 \equiv n \pmod 3$
    and verify the validity of them.

    You can notice that one of them is exactly Fermat's little theorem for exponent 3.

    I do not think that the original congruence can be obtained directly from Euler's theorem, but it might be useful for you to notice, that Euler's theorem implies this congruence for all n coprime to 6.
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  3. #3
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    $\displaystyle 5n^3 + n \equiv -n^3 + n = -n(n^2-1) = -(n-1)(n)(n+1) (mod6)$

    we know that the product of $\displaystyle k $ consecutive integers is the multiple of $\displaystyle k! $ so in this case , it is the multiple of $\displaystyle 3! = 6 $ , $\displaystyle -(n-1)(n)(n+1) \equiv 0 (mod6) $
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