Let s and t be odd integers with x t = 1 and gcd(s,t) = 1. Prove that st, (s^2-t^2)/2, and (s^2+t^2)/2 are pairwise relatively prime.

Question

Let s and t be odd integers with x > t >= 1 and gcd(s,t) = 1.  Prove that st, (s^2-t^2)/2, and (s^2+t^2)/2 are pairwise relatively prime.

ganeshie8 · Answer

have you tried anything yet

anonymous · Answer

I don't even know where to start...

anonymous · Answer

FYI, those three numbers are the formulas for generating primitive pythagorean triples, with a = st, b = (s^2-t^2)/2, and c = (s^2+t^2)/2.  Also, a will be odd and b will be even.

ganeshie8 · Answer

$\large  \gcd(s, t) = 1 \iff\gcd(s^2, t^2) = 1$

agree ?

anonymous · Answer

Yes.

ganeshie8 · Answer

Next suppose $\large   d| st$ and $\large  d |   \dfrac{s^2+t^2}{2}$
that means any prime factor$\large p$ of  $d$ divides either $\large s$ or  $\large t$, but not both because $\large \gcd(s, t) = 1$

Suppose $\large p | s \implies   s = pj$,  

So  $\large  p | \dfrac{(pj)^2+t^2}{2} $
but this is not possible because  $\large p  \not |  ~t$

ganeshie8 · Answer

you can work the remaining pairs similarly

anonymous · Answer

I don't fully understand this, so I will have to study it.

ganeshie8 · Answer

sure :) you may ask me if there is a specific question

anonymous · Answer

Okay, thanks.  If I ask a question after clicking best response, will you still get a notification, or does that close the problem out?

anonymous · Answer

If I click close and ask a question, will you still get a notification?

ganeshie8 · Answer

yes for all above questions ^^

anonymous · Answer

Cool, thanks for the help.  I think my brain is tired tonight, so I will probably study this further tomorrow.  :)

ganeshie8 · Answer

I get notification everytime you reply here/tag me. It won't depend on the status of question

ganeshie8 · Answer

np :) have good sleep !

anonymous · Answer

Thanks!

anonymous · Answer

I understand, thanks!