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 Divisible for every n (Posted on 2007-10-11)
Let x,y>1 be integers such that for all n>0, xn-1 is divisible by yn-1. Then show that x=yk, where k is a positive integer.

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 Reverse Case Comment 1 of 1

If x=y^k,

Then,

Need to show that ((y^k)^n)-1)/(y^n - 1) is an integer.

This equals (y^(kn)-1)/(y^n - 1).

Let y^n = r.

Then,

We need to show that:

(r^k -1)/(r - 1) is an integer.

I notice that this formula is the sum of a geometric progression.

Therefore,

(r^k -1)/(r - 1) = 1 + r + r^2 + ... + r^(k-1)

which is always an integer for integer values of r and k.

Since r = y^n, where y and n are integers, it has been shown that x^n -1 is divisible by y^n - 1 when x = y^k.

I have shown that setting x=y^k satisfies the requirement.  I'm not sure how to prove this is the only way to satisfy it however.

 Posted by Chris, PhD on 2012-08-11 07:17:28

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