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A Theorem on Squares and Congruences

  1. Oct 21, 2012 #1
    Hey guys --

    I was wondering whether the following theorem was already established, or if it was something novel. I couldn't find it in my abstract algebra textbook and a Google search was not productive ("a list of congruence theorems" just turns up elementary geometry theorems on triangle congruences.)

    Thm For any n in ℕ, if a+b=n, then a2 ≡ b2 (mod n).

    Proof: Let a+b=n. Then a=b-n.

    a2 = (b-n)2 = b2 - 2bn + n2 = b2 - n*(2b)+n) ≡ b2 (mod n)
     
  2. jcsd
  3. Oct 21, 2012 #2


    Well, this is a trivial statement. Observe (and this also serves as a proof) that

    $$a^2=b^2\pmod n\Longleftrightarrow (a-b)(a+b)=0\pmod n$$

    so if [itex]\,a+b=0\pmod n\,[/itex] then we can go leftwards in the above and get the result.

    DonAntonio
     
  4. Oct 21, 2012 #3
    Thanks DonAntonio for your explanation.

    I agree that it is a rather trivial result, but I thought it was "nice" in a way. Just curious if it's already been proven/noticed. (Perhaps no one thought to write it down because it is so trivial and it's relegated instead to the Exercises section of some introductory textbook.) Please forgive me if I'm writing the equivalent of "Hey guys, I discovered that the sky is blue!" but I'm only an undergraduate and my breadth of knowledge is still pretty thin.
     
  5. Oct 21, 2012 #4
    Yes, this has certainly been proven before. However, it's great that you're discovering little results like this own your own--keep it up!
     
  6. Oct 21, 2012 #5
    Thanks for the encouragement!
     
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