# Pythagorean Triples using odd integers

I am trying to understand how the Proof by Contradiction can be applied to deriving the generalized rule for Pythagorean triples. I understand the general idea behind setting:

\begin{align} a &= 2pq\\ b &= p^2 - q^2\\ c &= p^2 + q^2 \end{align}

but not sure how I could use this to imply that a and b are odd? Any suggestions on how to go about this? or should I try using a variation of the definition of an odd number?

• $a$ isn't odd. It is quite obviously divisible by $2$. Why do you say it is? – Thomas Andrews Oct 12 '15 at 23:10
• I'm trying to imply a variation that would allow me to use A and B as odd to show either A or B must be even, but I am slightly puzzled on how to do that... – John Doe Oct 12 '15 at 23:13
• Try to be clear what the hypothesis and conclusion is in the implication you are trying to prove. Perhaps you are trying to prove that if $a^2+b^2=c^2$ and if $c^2$ is even, then $a$ and $b$ must both be even? As it currently reads, you seem to be saying the hypothesis is "if $a$ and $b$ are odd then $a$ or $b$ is even" which is of course a contradiction. – JMoravitz Oct 12 '15 at 23:15
• In the triple $(a,b,c)$ we cannot have $a$ and $b$ both odd. For the square of an odd number is congruent to $1$ modulo $4$. So if $a$ and $b$ were both odd we would have $a^2+b^2\equiv 2\pmod{4}$. But there is no integer $c$ such that $c^2\equiv 2\pmod{4}$. – André Nicolas Oct 12 '15 at 23:17
• I was trying to write a simplistic proof through means of contradiction to imply " suppose a,b,c>0. If a^2+b^2=c^2 ,then neither A and B is even." as a hypothesis considering that either a or b must be even in order for it to be a pythagorean triple. and was wondering if there was a way algebraic to show this other then using mods ( no number theory) ? If anyone could help on that understanding . – John Doe Oct 12 '15 at 23:25