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Math Help - GCD divisibility problem

  1. #1
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    GCD divisibility problem

    Back again haha. Thanks for all the great help.

    I have to prove that, for a positive integer n and any integer a, gcd(a, a+n) divides n; hence gcd(a, a+1)=1.

    Not even sure where to begin.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by glowplug19 View Post
    Back again haha. Thanks for all the great help.

    I have to prove that, for a positive integer n and any integer a, gcd(a, a+n) divides n; hence gcd(a, a+1)=1.

    Not even sure where to begin.
    Hi glowplug19.

    \gcd(a,a+n) divides both a and a+n; hence it divides (a+n)-a=n.
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