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Math proof

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data

    prove:6 divides (n^3-n) for all integers n.

    2. Relevant equations

    3. The attempt at a solution
    tried to use direct proof. Then used cases that involed n=2k for some integer k and n=2k+1 for some integer k. However, i could not get it so that 6 was factored from either odd/even of n^3-n i.e. n^3-n=6m for some integer m.
    Just a hint please.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 10, 2008 #2
    What's the remainder on division by 3 of the 3 factors you have exhibited?
  4. Mar 10, 2008 #3


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    Science Advisor

    One of any two consecutive number is even. One of any three consecutive numbers is a multiple of 3.
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