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Math Help - n^3 mod 6 = n mod 6

  1. #1
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    n^3 mod 6 = n mod 6

    Hi,

    I have a problem that I haven't quite been able to solve.

    Prove for any n, n^3 mod 6 = n mod 6

    I know that a mod n = b mod n iff n|(a-b) but I can't quite get there.

    This is as far as I have gotten:

    Rewrite n^3 mod 6 = n^3 = 6q + r and n mod 6 = n = 6t + s for some integers q, t, s, r where r >= s.

    Then subtract both equations so:

    n^3 - n = (6q +r) - (6t +s) = 6(q-t) + (r-s)

    If r = s then r-s = 0 and 6|(n^3 -n)



    After that point I get stuck. Is this the right way to prove this or am I making it too complicated?

    Any help would be appreciated. Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mali916 View Post
    Hi,

    I have a problem that I haven't quite been able to solve.

    Prove for any n, n^3 mod 6 = n mod 6

    I know that a mod n = b mod n iff n|(a-b) but I can't quite get there.

    This is as far as I have gotten:

    Rewrite n^3 mod 6 = n^3 = 6q + r and n mod 6 = n = 6t + s for some integers q, t, s, r where r >= s.

    Then subtract both equations so:

    n^3 - n = (6q +r) - (6t +s) = 6(q-t) + (r-s)

    If r = s then r-s = 0 and 6|(n^3 -n)



    After that point I get stuck. Is this the right way to prove this or am I making it too complicated?

    Any help would be appreciated. Thanks.
    Since there are only six residue classes mod six, why not just check that it's true for those?

    0^3\equiv 0\text{ mod }6

    1^3\equiv 1\text{ mod }6

    2^3=8\equiv 2\text{ mod }6

    3^3=27\equiv 3\text{ mod }6

    4^3\equiv\left(-2\right)^3=-8\equiv -2\equiv 4\text{ mod }6


    5^3\equiv \left(-1\right)^3=-1\equiv 5\text{ mod }6


    ...done.
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