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20 December, 00:39

Prove that for any integer n and any integer a, gcd (a, a+n) divides n; hence, gcd (a, a+1) = 1

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  1. 20 December, 04:25
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    Let g=gcd (a, a+n)

    a=gc, a+n=gd, where gcd (c, d) = 1

    Put a=gc into a+n=gd,

    we have gc+n=gd

    Then g (d-c) = n, that is, g divides n.
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