# If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b

• Mar 9th 2012, 06:03 PM
math2011
If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
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.
• Mar 9th 2012, 10:28 PM
princeps
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
Quote:

Originally Posted by math2011
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.....
• Mar 10th 2012, 07:48 AM
Fernando
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
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.
• Mar 10th 2012, 04:46 PM
math2011
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
Thank you! This is the part that I could not prove.
• Mar 10th 2012, 04:57 PM
math2011
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
Quote:

Originally Posted by Fernando
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.

I don't fully understand the logic in your proof. Do you mean $\displaystyle gcd(2a,2b)$ when you say $\displaystyle (2a,2b)$? Why does $\displaystyle d | 2a$ and $\displaystyle d | 2b$ imply $\displaystyle d | gcd(2a,2b)$?
• Mar 11th 2012, 06:09 AM
Fernando
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
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.
• Mar 12th 2012, 05:56 AM
math2011
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
Quote:

Originally Posted by Fernando
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.
• Mar 13th 2012, 09:01 AM
Fernando
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
Hi

I don't know, what's the wrong? Do you think the first theorem is wrong?
• Mar 16th 2012, 06:53 AM
math2011
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
Quote:

Originally Posted by Fernando
Hi

I don't know, what's the wrong? Do you think the first theorem is wrong?

I think the first theorem is correct after thinking through it. Sorry I had not seen it before and confused it with the one that I posted.