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Math Help - 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 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.
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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 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.....
    Last edited by princeps; March 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 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.
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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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  5. #5
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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 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)?
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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 ! 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.
    Last edited by Fernando; March 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 ! 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.
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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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