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Prove True/False that n^3-n is Always Divisible By 6

  1. Apr 30, 2017 #1
    1. The problem statement, all variables and given/known data
    For all natural numbers n, prove whether the following is true or false:
    n3-n is always divisible by 6.

    From SQA Advanced Higher Mathematics 2006 Exam Paper
    2. Relevant equations
    I can choose from the following types of proof:
    Direct proof
    Proof by contradiction
    Proof by contrapositive
    Proof by induction

    3. The attempt at a solution
    I know the statement is true, but proving it has been more difficult than I thought it would be!
    I tried proof by induction, but got stuck with trying to prove true for n=k+1. I then tried proving the statement true for n=2k (even number) and n=2m+1 (odd number), but again, I didn't seem to be getting anywhere.
    Am I along the right lines, or should I be trying something different?
  2. jcsd
  3. Apr 30, 2017 #2


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    Staff: Mentor

    Try to factor ##n^3-n## and see what you can say about the factors.
  4. Apr 30, 2017 #3


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    A simple direct proof is possible. Start by factorising ##n^3-n## into three integer factors.

    EDIT: Now I am jinxed and will have to wait for somebody to say my name before I can speak aloud again.
  5. Apr 30, 2017 #4
    Okay, so I've factorised it and found that ##n^3-n## is always even and therefore always divisible by 2. In order for it to be divisible by 6, it must be divisible by 2 and 3, but I'm not sure how to go about proving that it's divisible by 3.

    EDIT: I looked up the divisibility rule for 3, and found it expressed as ##n(n+1)(n-1)## which is exactly what I have from factorising ##n^3-n##!
    Last edited: Apr 30, 2017
  6. Apr 30, 2017 #5


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    Staff: Mentor

    What kind of numbers are ##n-1\, , \,n\, , \,n+1##?
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