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Proof by contrapositive; if (m^2+n^2) div by 4, then m,n are even numbers

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Let m and n be two integers. Prove that if m^2 + n^2 is divisible by 4, then both m and n are even numbers

    2. Relevant equations



    3. The attempt at a solution

    Proof by Contrapositive. Assume m, n are odd numbers, showing that m^2 + n^2 is not divisible by 4.

    let:
    m= 2a + 1 (a,b are integers)
    n=2b+1

    m^2+n^2 = (2a+1)^2 + (2b+1)^2 = 4a^2 +4a + 1 + 4b^2 + 4b +1

    let: 4(a^2 + a + b^2 + b) = 4q (q an integer)

    m^2 + n^2 = 4q + 2

    with 4q + 2 not divisible by 4 since 4 divides 4q + 2 with a remainder of 2.
    ==> if m and n are odd numbers, m^2 + n^2 is not divisible 4, which by contrapositive reasoning proves that if m^2 + n^2 is divisible by 4, then m and n are odd numbers.

    ----------------------------------

    I finished my proof there, but had a note underneath that said that it was incomplete, since I was missing "other cases". Does this mean I have to also show that m and n cant be of opposite parity either? I've been under the impression that the negation of "m and n are even numbers" was "m and n are odd numbers". Is this logic wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 12, 2012 #2
    Thenegation of "both m and n are even" is "at least 1 of m or n is odd".
     
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