If a product of relatively prime integers is an $n$th power, then each is an $n$th power

Solution 1:

Of course it is easy using existence and uniqueness of prime factorizations. Below is a more general proof using gcd's (or ideals) that has the benefit of giving an explicit closed form.

$ab=c^n \overset{\rm Lemma}\Rightarrow c=(a,c)(b,c) \,\Rightarrow\, ab = (a,c)^n(b,c)^n\Rightarrow \dfrac{a}{(b,c)^n}\! = \dfrac{(a,c)^n}b$ $\,\Rightarrow\begin{align} a &= (a,c)^n\\ b &= (b,c)^n\end{align}$

where the last inference uses Unique Fractionization [both fractions are irreducible by $(a,b)\!=\!1$]

Lemma $\ \ \color{#c00}{c\mid ab},\,\ \color{#0a0}{(a,b,c)=1}\ \Rightarrow \ c = (a,c)(b,c)\ [=\, (ab,c\color{#0a0}{(a,b,c)}) = (\color{#c00}{ab,c}) = c\,],\,$ where the braced proof uses gcd "polynomial" arithmetic, i.e. associative, commutative, distributive laws.

Alternatively $\ (a,c)^n\! \overset{\rm\color{#C00}F}= (a^n,c^n) = (a^n,ab) = a(a^n,b) = a$ and $\,(b,c)^n = b\,$ by symmetry, where we have invoked $\rm\color{#c00}F$ = GCD Binomial Theorem (Freshman's Dream).

As $ $ Weil remarks, $ $ this result can be viewed as the essence of Fermat's method of infinite descent. $ $ It generalizes to rings of algebraic integers but depends upon much deeper results in this more general context, viz. the finiteness of the class number and Dirichlet's unit theorem.