I had two questions I can't figure out...
1) prove that if gcd(a,b) = 1 then gcd(a+b,a*b) = 1
2) let gcd(a,b) = 1. prove that d= gcd(a+b, a-b) = 1 or 2
(hint: prove d<2)
Thanks!
Let d=gcd(a+b,a*b)
Then d divides (a+b)x+(a*b)y, where x and y can be any integers.
If you let x=a and y=-1, you get : a²+ab-ab=a². Thus d divides a²
If you let x=b and y=-1, you get : ab+b²-ab=b². Thus d divides b²
---> d=1.
d divides (a+b) and (a-b) and thus divides (a+b)+(a-b)=2a and (a+b)-(a-b)=2b.2) let gcd(a,b) = 1. prove that d= gcd(a+b, a-b) = 1 or 2
(hint: prove d<2)
Thanks!
So d divides 2 or d divides a. If it doesn't divide 2, it divides a and b. Then...
If it divides 2, then...
That would only be true if d = 1 or a prime. It is not true for general integers d.
$\displaystyle \gcd(a,b)=1$ $\displaystyle \Rightarrow$ $\displaystyle \exists\,r,s\in\mathbb{Z}$ with $\displaystyle ra+sb=1$.
Hence $\displaystyle r\left[(a+b)+(a-b)\right]+s\left[(a+b)-(a-b)\right]$ $\displaystyle =$ $\displaystyle 2(ra+sb)$ $\displaystyle =$ 2.
$\displaystyle \therefore\ d$ divides 2 since $\displaystyle d$ divides the LHS.
Suppose $\displaystyle \gcd(a+b,ab) \ne 1$, and $\displaystyle p$ be a prime divisor of $\displaystyle a+b$ and $\displaystyle ab$. Then as $\displaystyle \gcd(a,b)=1$ $\displaystyle p|a$ and $\displaystyle p \not| b$ or $\displaystyle p \not |a$ and $\displaystyle p| b$.
Without loss of generality suppose the first of these is the case. Then $\displaystyle p \not| (a+b) $, a contradiction so no such prime can exist and $\displaystyle \gcd(a+b, ab)=1$.
RonL