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Math Help - relatively prime

  1. #1
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    Exclamation relatively prime

    Show that if a and b are integers with (a,b)=1, then (a+b,a-b)=1 or 2

    Should I use the fact that (a+cb,b)=(a,b)? I've also tried to start with (a,b)=1 => 1=xa+yb, but then I was stuck, don't know how to begin. Thanks for helping me out.
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    Quote Originally Posted by suedenation View Post
    Show that if a and b are integers with (a,b)=1, then (a+b,a-b)=1 or 2

    Should I use the fact that (a+cb,b)=(a,b)? I've also tried to start with (a,b)=1 => 1=xa+yb, but then I was stuck, don't know how to begin. Thanks for helping me out.
    Well, the "2" case is easy. Consider two integers a, b such that (a,b) = 1 and both a and b are odd. Thus a+b and a-b are both even, so (a+b,a-b) = 2.

    Can't figure anything out for the other case, where a is even and b is odd. (Obviously this is the same as the case for odd a and even b. And, of course, we can't have both a and b even.)

    -Dan
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    Quote Originally Posted by suedenation View Post
    Show that if a and b are integers with (a,b)=1, then (a+b,a-b)=1 or 2

    Should I use the fact that (a+cb,b)=(a,b)? I've also tried to start with (a,b)=1 => 1=xa+yb, but then I was stuck, don't know how to begin. Thanks for helping me out.
    You need to show,
    gcd[a+b,a-b]=1
    Equivalently,
    gcd[a+b,(a+b)-2b]=1
    Say,
    gcd(a+b,a-b)=d
    Then, by definition,
    d|(a+b) and d|[(a+b)-2b]
    Using rules of divisiblity,
    d|(-2b) now, d|b would imply gcd=1
    But we can also have d|2.
    Thus, only two posibilities.
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