Coprime cofactors of n'th powers are n'th powers, up to associates, for Gaussian integers & UFD

Solution 1:

The proof for naturals (or integers) via prime factorization immediately generalizes to any UFD, but we need to account for unit (invertible) factors, so we work up to associates (unit multiples). [ENT readers: $ $ in $R = \Bbb Z,\,$ $\,u\,$ is unit $\!\iff\! u=\pm1,\,$ so $\,m,n\,$ are associate $\!\iff\! m = \pm n$]

Theorem $\ $If $R\,$ is a UFD and coprime $\,a,b\in R\,$ satisfy $\,ab=c^n$ for some $\,0\ne c\in R,\ n \ge 1,\,$ then $\,a=u\,r^n$ and $\,b=u^{-1}s^n$ for some $\,r,s\in R\,$ and for some unit (i.e. invertible) $u\in R.\,$ Therefore both factors $\,a\,$ and $\,b\,$ are ― like $\,c^n$ ― associates of $\,n$'th powers in $R$.

Proof $\ $ We induct on $\,k =\,$ number of prime factors of $\,c.\,$ If $\,k=0\,$ then $\,c\,$ is a unit, so $\,a,b\,$ are units, so $\,a = a\cdot 1^n,\ b = a^{-1}c^n$ works. Else $\,k\ge 1,\,$ so a prime $\,p\mid c,\,$ so $\,p^n\mid c^n\! = ab\,$ hence $\,p^n\mid a\ {\rm or}\ b\,$ by $\,a,b$ coprime, $R\,$ UFD. Wlog $\,p^n\!\mid b\,$ so canceling $\,p^n$ we obtain $\,a(b/p^n) = (c/p)^n.\,$ $\,c/p\,$ has fewer prime factors than $c\,$ so induction $\Rightarrow a = ur^n,\ b/p^n\! = u^{-1} s^n,\,$ so $\,b = u^{-1}(ps)^n$.

Remark $\ $ For generalizations, see here for a proof using gcds (or ideals), and see here for Weil's remarks on the relationship with Fermat's method of infinite descent.