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Prove by Contradiction: For all Prime Numbers a, b, and c

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove by Contradiction: For all Prime Numbers a, b, and c, a^2 + b^2 =/= c^2

    2. Relevant equations
    Prime number is a number whose only factors are one and itself.
    Proof by contradiction means that you take a statement's negation as a starting point, and find a contradiction.


    3. The attempt at a solution
    The statement's negation is:
    There exists prime numbers a, b, and c, such that a^2 + b^2 = c^2
    Rearrange it:
    a^2 = c^2 - b^2
    a^2 = (c - b) (c + b)
    a = √(c-b)(c+b)

    I'm stuck here. To show that it's a contradiction I would have to show that it's factors are not equal to 1 or a, and I've been staring at this a little too long, my head just keeps going in circles.. some help would be appreciated!
     
  2. jcsd
  3. Aug 1, 2012 #2

    HallsofIvy

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    Staff Emeritus
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    Up to here, great. Since a is prime, the only factor of a^2 is a.
    So you must have c- b= 1 and c+ b= a^2 or c- b= a and c+ b= a. Now if all of a, b, and c are odd that is impossible.

     
    Last edited: Aug 1, 2012
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