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(Proof) Square of integer is 3k or 3k+1

  1. Jan 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that the square of any integer a is either of the form 3k or of the form 3k+1 for some integer k.

    2. Relevant equations
    The Division Algorithm: Let a,b be integers with b>0. Then there exists unique integers q and r such that a = bq + r and 0<=r<b

    3. The attempt at a solution
    I know from the division algorithm that any integer a can be written as 3q, 3q+1, or 3q+2, so
    a^2=(3q)^2=9q^2=3(3q^2), or
    a^2=(3q+1)^2=9q^2+6q+1, or
    but i don't understand how the 3q+1 or 3q+2 case helps me.

    Can anyone give me some hints or point me further in the right direction? thanks.
    Last edited: Jan 14, 2008
  2. jcsd
  3. Jan 14, 2008 #2

    Tom Mattson

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    Staff Emeritus
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    Gold Member

    Well it's obviously not going to be of the form 3k, right? So try to fit it to the other form, namely 3k+1. That means strip a +1 off from the +4 at the end. It should fall right out.
  4. Jan 14, 2008 #3
    Good Work, you are basically done. Copying and pasting what you wrote,

    a^2=(3q)^2=9q^2=3(3q^2), or
    a^2=(3q+1)^2=9q^2+6q+1 = 3(3q^2 + 2q) + 1, or
    a^2=(3q+2)^2=9q^2+12q+4 = 3(3q^2 + 4q + 1) + 1,

    so you are done.
  5. Jan 14, 2008 #4
    ah, simple arithmetic :) thank you!
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