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

  1. #1
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    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.
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  2. #2
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    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 View Post
    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.....
    Last edited by princeps; Mar 9th 2012 at 10:35 PM.
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  3. #3
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    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.
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  4. #4
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    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.
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    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 View Post
    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 $\displaystyle d | gcd(2a,2b)$?
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  6. #6
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    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.
    Last edited by Fernando; Mar 11th 2012 at 06:11 AM. Reason: Grammar
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  7. #7
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    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 View Post
    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.
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  8. #8
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    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?
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  9. #9
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    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 View Post
    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.
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