Proof
We first prove the special case
If m | n, then (Xm - 1) |
(Xn - 1).
From this we derive the general case
gcd(Xm - 1,Xn - 1) =
Xgcd(m,n) - 1.
The following identity holds in R[X]:
Xn - 1 =
(Xm - 1)(Xn-m + Xn-2m +
··· + Xm + 1).
In particular Xm - 1 divides
Xn - 1.
Suppose n = qm + r. Then, by Step 1,
gcd(Xm - 1,Xn - 1) =
gcd(Xm - 1,Xn - 1 - Xr
(Xqm - 1)) =
gcd(Xm - 1,Xr - 1).
Thus, n can be replaced by the remainder of division of n by m.
But this is the first step of Euclid's algorithm,
which can be repeated
and repeated, until one of the arguments of the gcd is
Xgcd(m,n) - 1, and the other 0.