1. ## 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.

2. Originally Posted by atreyyu
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.

3. 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?

4. Originally Posted by atreyyu
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.

5. Right, I just needed to think anew today. Thanks heaps

6. (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?