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Math Help - Pythagorean triples proof

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    Pythagorean triples proof

    We have a Pythagorean triple of the form a^2+b^2=c^2 with a,b,c being positive integers. I have to show that c^2 + \frac{2}{3}ab is a composite integer.

    I get winded up in some extra long derivations that force me to make even more substitutions, but I don't really get anything that would lead me to reasonable conclusions.
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    Quote Originally Posted by atreyyu View Post
    We have a Pythagorean triple of the form a^2+b^2=c^2 with a,b,c being positive integers. I have to show that c^2 + \frac{2}{3}ab is a composite integer.

    I get winded up in some extra long derivations that force me to make even more substitutions, but I don't really get anything that would lead me to reasonable conclusions.

    English lesson: the past participle of wind is wound (it's a strong verb).
    It is only necessary to prove this for a primitive Pythagorean triple, since the general case then follows immediately. A primitive Pythagorean triple is of the form (m^2-n^2,2mn,m^2+n^2), so that c^2 + \tfrac{2}{3}ab = (m^2+n^2)^2 + \tfrac43mn(m^2-n^2).

    First check that mn(m^2-n^2) is a multiple of 3, to ensure that c^2 + \tfrac{2}{3}ab is an integer. Then check that \tfrac13\bigl(3(m^2+n^2)^2 + 4mn(m^2-n^2)\bigr) factorises as \tfrac13(3m^2-2mn+n^2)(m^2+2mn+3n^2). The only thing then remaining is to satisfy yourself that neither of those last two factors is equal to 3.
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    OK, first I managed to check whether the number is at all an integer by congruence; but I still don't know of what use is the fact that neither of these last two factors equals 3?
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    Quote Originally Posted by atreyyu View Post
    OK, first I managed to check whether the number is at all an integer by congruence; but I still don't know of what use is the fact that neither of these last two factors equals 3?
    You are trying to show that c^2+\tfrac23ab is a composite integer, so you need to find factors for it. The expression \tfrac13(3m^2-2mn+n^2)(m^2+2mn+3n^2) provides a factorisation, but you need to check that neither of the parenthesised factors is equal to 3, because it would then cancel with the 1/3. You also need to check that neither of those factors is equal to 1, but that is almost immediate.
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    Right, I just needed to think anew today. Thanks heaps
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    (editing won't work) Hmm... now I'm having a hard time seeing how this has been factorised. It's of course correct, but how do I know which factor should I take out when doing this? Or does it follow a particular method?
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