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Integer expression

  1. Dec 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that the expression [tex] \frac{a(a^2 + a)}{3} [/tex] is an integer for all integers [tex] \geq 1 [/tex]

    2. Relevant equations

    3. The attempt at a solution

    [tex] a(a^2 + a ) = 3q + r [/tex]

    r can be:

    [tex] r = 0,1,2 [/tex]

    for r = 0

    [tex] \frac{a(a^2 + a)}{3} = q [/tex]

    q is an integer by the division algorithm.

    When I try this attempt with 3q+1 and 3q+2 I can't get it to work.

    Thanks in advance.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 14, 2008 #2
    This is not true for all positive integers. For example, setting a = 1 gives 2/3.
  4. Dec 14, 2008 #3
    Sorry there is a correction [tex] \frac{a^3 -a}{3} [/tex]
  5. Dec 14, 2008 #4


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    Gold Member

    You might want to factor [itex]a^3 -a[/itex], and show that for all integers [itex]a\geq1[/itex], one of the factors will be divisible by 3.
  6. Dec 14, 2008 #5


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    Where did q and r come from? :confused: Whenever you introduce new variables, you need to say what they are, and why they should have the properties you say they have. You need to say something like "By the division algorithm, we can select integers q and r such that a^3 - a = 3q + r, where 0 <= r < 3". If you don't explain the introduction of new variables, then your work is just nonsense.

    Yes, I was able to figure out what you meant, but that's irrelevant; it would be completely analogous to saying "me food want" in your English classes.

    Of course not. If r=1 or r=2, then (a^3-a)/3 is not an integer. That's what 'remainder' means, right? And r is supposed to be a remainder?
  7. Dec 15, 2008 #6
    Fermat's little theorem: a^2 Mod 3 = 1 if a is not a multiple of 3. So we always have that a^3 = a Mod 3.
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