$s$ and $t$ are odd and relatively prime. Why is $2$ the only common factor of $s-t$ and $s+t$?

Asked Sep 08 '22 at 22:38

Active Sep 08 '22 at 22:38

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I found this in textbook Silverman (2012) A Friendly Introduction to Number Theory on page 232 without explanation. They make it sound so obvious so I must have missed something along the way.

The way I know to find common factors between two numbers is prime factorization and Euclidean algorithm (but I do not know how to apply them in this situation). I mean $s-t$ and $s+t$ must be even, this I know how to easily prove.

But I can't figure out why their only common factor must be $2$. Please help!

asked Sep 08 '22 at 22:38

twothreefive

2

Hint: if $d=\gcd(a,b)$ then $d,|,(a+b)$. – lulu Sep 08 '22 at 22:40
Does this answer your question? Prove $\gcd(a+b, a-b) = 1$ or $2,$ if $,\gcd(a,b) = 1$ - found using an Approach0 search. Since $s$ and $t$ are odd, then $s-t$ and $s+t$ are both even, as you've already noted, so this means $\gcd(s-t,s+t)=2$. Note another duplicate (among quite a few) is Suppose that$\ gcd(b, a) = 1$. Prove that $\gcd(b + a, b − a) \leq 2$. – John Omielan Sep 08 '22 at 22:42
1

Thanks lulu and John, I think I got it now :) – twothreefive Sep 08 '22 at 22:45

$s$ and $t$ are odd and relatively prime. Why is $2$ the only common factor of $s-t$ and $s+t$?

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