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    Hi, I need to prove that if n divides any two of a, b, c for
     a^2 +  b^2 = c^2 that will also divide the third.
    (i.e. a, b, c is a pythagorean triple)

    I have shown that  n^2 divides  c^2 using a=nq and b=nq' etc, but i dont know how to progress this proof further.

    Thanks for any help
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    The second case you need to consider is n|a and n|c. Because a and b are symmetric, you can consider the other, third, case (namely, n|b, n|c) as following from this second case.

    Bundling up the other two cases by just considering the second case is what is known as a "without loss of generality" (WLOG) step.
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    Isn't it similar to replacing a and b with np and nq, but instead of getting c^2 = (n^2)(p^2 + q^2), you get a^2 = (n^2)(p^2 - q^2)?
 
 
 
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Updated: February 9, 2010
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