Show that if $\displaystyle gcd(a,b) = 1$ then $\displaystyle gcd(a-b,a+b) = 1$ or $\displaystyle 2$ for all integers $\displaystyle a,b$.
I don't know how to start.
suppose $\displaystyle ~\gcd(a-b,a+b)=k~$ where $\displaystyle ~k>2~$ then :
$\displaystyle a-b=k\cdot m ~\text {and}~ a+b=k \cdot n $ , hence :
$\displaystyle 2a=k\cdot m+k\cdot n \Rightarrow a = \frac{k}{2} \cdot(m+n) $
$\displaystyle 2b=k\cdot n-k\cdot m \Rightarrow b = \frac{k}{2} \cdot(n-m) $
If $\displaystyle k $ is an even number then
$\displaystyle \gcd(a,b) \geq 2 ~$, contradiction.....
If $\displaystyle k $ is an odd number then
$\displaystyle (m+n) ~\text {and}~ (n-m) ~$ are even numbers , hence :
$\displaystyle \gcd(a,b) \geq 3 ~$, contradiction.....
Hello.
You could do following:
Let $\displaystyle d$ be such that $\displaystyle d=(a+b,a-b)$, then we're goning to show that $\displaystyle d=2$.
If $\displaystyle d=(a+b,a-b)$, $\displaystyle d|(a+b)$ and $\displaystyle d|(a-b)$, so $\displaystyle d|(a+b)(1)+(a-b)(1)=2a$. On the other hand, $\displaystyle d|(a+b)(1)+(a-b)(-1)=2b$.
Hence $\displaystyle d|(2a,2b)=2(a,b)$, but $\displaystyle (a,b)=1$, then $\displaystyle d|2$ therefore $\displaystyle d=1$ or $\displaystyle d=2$
Best regards.
Hi, yeah ! $\displaystyle gcd(a,b)=(a,b)$ it's my notation.
OK. There is a theorem states that: Let $\displaystyle a$ and $\displaystyle b$ be integers differents of zero. Then $\displaystyle d=(a,b)$ iff $\displaystyle d$ holds following:
$\displaystyle 1.$ $\displaystyle d>0$
$\displaystyle 2.$ $\displaystyle d|a$ and$\displaystyle d|b$
$\displaystyle 3.$ If $\displaystyle f|a$ and $\displaystyle f|b$ then $\displaystyle f|d$.
So, by the theorem We can assert if $\displaystyle d|2b$ and $\displaystyle d|2b$ then $\displaystyle d=(2a,2b)$ and I guess you could continue...
Best regards.
Do you mean the following theorem?
Let $\displaystyle a,b \in \mathbb{Z}$ and not both zero. Then a positive integer $\displaystyle g$ is the greatest common divisor of $\displaystyle a$ and $\displaystyle b$ if and only if the following equivalence holds:
$\displaystyle \forall d \in \mathbb{Z}, d | g \text{ if and only if } d | a \text{ and } d | b$.
This is different to what you stated.