if gcd(n,m) = d then prove that gcd((a^n)-1,(a^m)-1)=(a^d)-1.

Help would be very appreciated.

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- Nov 14th 2013, 06:57 AM #1

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- Nov 14th 2013, 08:29 AM #2

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## Re: Problem with GCD

Let be an integer that divides both and . Then and for some pair of integers . So,

Similarly, .

So, clearly divides both and for any common divisor of . So, is a common divisor of both and . All that is left is to show that any other common divisor of and must divide

- Nov 16th 2013, 09:35 AM #3

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## Re: Problem with GCD

..... to show that any other common divisor of a^n-1 and a^m-1 must divide a^{\gcd(m,n)}-1

If u is such a divisor then and are each congruent to 1 (mod u)

We need to show that is congruent to 1 (mod u) where d=gcd(m,n)

This follows from the fact that d is a linear combination of m and n

For example, say there are positive integers x and y such that then

- Nov 17th 2013, 06:58 AM #4

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