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

• Mar 9th 2012, 07:03 PM
math2011
If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
Show that if $gcd(a,b) = 1$ then $gcd(a-b,a+b) = 1$ or $2$ for all integers $a,b$.

I don't know how to start.
• Mar 9th 2012, 11:28 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 $gcd(a,b) = 1$ then $gcd(a-b,a+b) = 1$ or $2$ for all integers $a,b$.

I don't know how to start.

suppose $~\gcd(a-b,a+b)=k~$ where $~k>2~$ then :

$a-b=k\cdot m ~\text {and}~ a+b=k \cdot n$ , hence :

$2a=k\cdot m+k\cdot n \Rightarrow a = \frac{k}{2} \cdot(m+n)$

$2b=k\cdot n-k\cdot m \Rightarrow b = \frac{k}{2} \cdot(n-m)$

If $k$ is an even number then

$\gcd(a,b) \geq 2 ~$, contradiction.....

If $k$ is an odd number then

$(m+n) ~\text {and}~ (n-m) ~$ are even numbers , hence :

$\gcd(a,b) \geq 3 ~$, contradiction.....
• Mar 10th 2012, 08:48 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 $d$ be such that $d=(a+b,a-b)$, then we're goning to show that $d=2$.

If $d=(a+b,a-b)$, $d|(a+b)$ and $d|(a-b)$, so $d|(a+b)(1)+(a-b)(1)=2a$. On the other hand, $d|(a+b)(1)+(a-b)(-1)=2b$.

Hence $d|(2a,2b)=2(a,b)$, but $(a,b)=1$, then $d|2$ therefore $d=1$ or $d=2$

Best regards.
• Mar 10th 2012, 05:46 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, 05:57 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 $d$ be such that $d=(a+b,a-b)$, then we're goning to show that $d=2$.

If $d=(a+b,a-b)$, $d|(a+b)$ and $d|(a-b)$, so $d|(a+b)(1)+(a-b)(1)=2a$. On the other hand, $d|(a+b)(1)+(a-b)(-1)=2b$.

Hence $d|(2a,2b)=2(a,b)$, but $(a,b)=1$, then $d|2$ therefore $d=1$ or $d=2$

Best regards.

I don't fully understand the logic in your proof. Do you mean $gcd(2a,2b)$ when you say $(2a,2b)$? Why does $d | 2a$ and $d | 2b$ imply $d | gcd(2a,2b)$?
• Mar 11th 2012, 07:09 AM
Fernando
Re: If gcd(aa,b)=1, then gcd(a-b,a+b)=1 or 2 for all integers a,b
Hi, yeah ! $gcd(a,b)=(a,b)$ it's my notation.

OK. There is a theorem states that: Let $a$ and $b$ be integers differents of zero. Then $d=(a,b)$ iff $d$ holds following:

$1.$ $d>0$
$2.$ $d|a$ and $d|b$
$3.$ If $f|a$ and $f|b$ then $f|d$.

So, by the theorem We can assert if $d|2b$ and $d|2b$ then $d=(2a,2b)$ and I guess you could continue...

Best regards.
• Mar 12th 2012, 06:56 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 ! $gcd(a,b)=(a,b)$ it's my notation.

OK. There is a theorem states that: Let $a$ and $b$ be integers differents of zero. Then $d=(a,b)$ iff $d$ holds following:

$1.$ $d>0$
$2.$ $d|a$ and $d|b$
$3.$ If $f|a$ and $f|b$ then $f|d$.

So, by the theorem We can assert if $d|2b$ and $d|2b$ then $d=(2a,2b)$ and I guess you could continue...

Best regards.

Do you mean the following theorem?

Let $a,b \in \mathbb{Z}$ and not both zero. Then a positive integer $g$ is the greatest common divisor of $a$ and $b$ if and only if the following equivalence holds:
$\forall d \in \mathbb{Z}, d | g \text{ if and only if } d | a \text{ and } d | b$.

This is different to what you stated.
• Mar 13th 2012, 10:01 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, 07:53 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.