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Prove that the square of any integer, when divided by 3. only by odd and even.

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data

    I know you could prove this by stating every integer is either 3m, 3m+1 or 3m+2. However I am trying to prove this just using either even numbers or odd numbers.

    so for example, when I try:
    (2x+1)^2
    = 4x^2 + 4x + 1 - expand
    = 3x^2 + x^2 + 3x + x + 1 - group like terms
    = 3x^2 + 3x + x^2 + x + 1
    = 3*(x^2 + x) + x(x+1) + 1

    x(x+1) + 1 is an odd number

    so
    = 3*(x^2 + x) + 2p + 1

    thats as far as I can go.

    thanks
     
  2. jcsd
  3. Nov 6, 2012 #2

    Mark44

    Staff: Mentor

    What are you trying to prove?
    Your thread title seems to be missing a few words
    Prove that the square of any integer, when divided by 3, does what?

    The problem template has a problem statement section titled, and this is where the statement of the problem should go. Use it.
     
  4. Nov 6, 2012 #3
    sorry mate.

    Prove that the square of any integer, when divided by 3, leaves remainder 0 or 1 but never 2.
    thanks
     
  5. Nov 6, 2012 #4

    haruspex

    User Avatar
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    What possibilities do you need to consider for the integer?
     
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