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.