设 d(x) = (f(x),g(x))
f(x) = d(x)f1(x)
g(x) = d(x)g1(x)
则:(f1(x),g1(x)) = 1
∴ (f1(x)^n ,g1(x)^n) = 1
(f(x)^n,g(x)^n)
=(d(x)^nf1(x)^n ,d(x)^ng1(x)^n)
=d(x)^n(f1(x)^n ,g1(x)^n)
=d(x)^n
=(f(x),g(x))^n
设 d(x) = (f(x),g(x))
f(x) = d(x)f1(x)
g(x) = d(x)g1(x)
则:(f1(x),g1(x)) = 1
∴ (f1(x)^n ,g1(x)^n) = 1
(f(x)^n,g(x)^n)
=(d(x)^nf1(x)^n ,d(x)^ng1(x)^n)
=d(x)^n(f1(x)^n ,g1(x)^n)
=d(x)^n
=(f(x),g(x))^n