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Show that for any two integers a, b , (a+b)^2 ≡ a^2 + b^2 (mod 2)

  1. Sep 25, 2010 #1
    Show that for any two integers a, b , (a+b)^2 ≡ a^2 + b^2 (mod 2)

    I have my solution below i wanted someone to help chekc if i have done anything wrong. Thank You for your help.

    The thing that is going on here is that 2x = 0 (mod 2) for any x. If x = ab, then 2ab = 0 (mod 2).
    We see that (a+b)^2 = a^2 + 2ab + b^2. Then this equals a^2 + 0 + b^2 (mod 2).

    Therfore, the Proof:

    By definition, showing (a+b)^2 = a^2 + b^2 (mod 2) is equivalent to showing that

    2 divides [(a+b)^2 - (a^2+b^2)]
    2 divides [(a^2 + 2ab + b^2 - a^2 - b^2]
    2 divides 2ab

    To show 2 divides 2ab we need to find an integer k such that 2k = 2ab. Take k = ab. Thus it is proved.
  2. jcsd
  3. Sep 25, 2010 #2


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    This looks like a good proof.

    However... you already know modular arithmetic, right?
    Then this is a complete proof.

    (For the record, you could have also proven it by exhaustion -- there are only four cases, and they're easy to check by hand. However, the argument you used generalizes to replacing 2 by any prime)
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