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Pythagorean triples

  1. Aug 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Let a,b,c be integers such that a2+b2=c2.

    Is it always true that at least one of {a,b} is even?

    2. Relevant equations



    3. The attempt at a solution
    I say yes, and I am going to try and prove it with a proof by contrdiction.

    Suppose a,b,c be integers such that a2+b2=c2 and that a and b are odd. By definition of odd a=2m+1, and b=2n+1, for some m,n in Z. By substitution we get c2= (2m+1)2+(2n+1)2. By simple arithmatic we get c2=2(2m2+2n2+2m+2n+1).
    c=[tex]\sqrt{2}[/tex][tex]\sqrt{2m^2+2n^2+2m+2n+1}[/tex].
    Because the second square root has an odd number in it means that we can not pull out a 1/[tex]\sqrt{2}[/tex] to cancel out the [tex]\sqrt{2}[/tex]. This means that we will have an irrational answer, rather than an integer for all m,n in Z. Thus we have reached a contradiction.//

    I was wondering if my logic is correct on this.

    Thank you for your time.
     
  2. jcsd
  3. Aug 11, 2010 #2

    lanedance

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    look reasonable to me, I think it is enough to note
    [tex] c^2=2(2m^2+2n^2+2m+2n+1) = c^2=2(2k+1)[/tex]

    where k is the integer given by
    [tex] k = m^2+n^2+m+n[/tex]
     
  4. Aug 11, 2010 #3

    Dick

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    I think it would be clearer if you skip the square root arguments and just talk about remainders mod 4 which is what I think lanedance is getting at.
     
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