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Show that if $a$, $b$, and $c$ are positive integers with $\gcd(a, b) = 1$ and $ab = c^n$, then there are positive integers $d$, and $e$ such that $a = d^n$ and $b = e^n$.

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    $\begingroup$ Have you considered the prime factorizations of $a$, $b$, and $c$? $\endgroup$ – Jon Feb 11 '11 at 15:03
  • $\begingroup$ 1=ax+by => 1=gcd(a^n,b^n) $\endgroup$ – kira Feb 11 '11 at 15:43
  • $\begingroup$ @kira: Please ask questions, don't give orders. Also: please make your titles informative, not sentence fragments. $\endgroup$ – Arturo Magidin Feb 11 '11 at 16:21
  • $\begingroup$ obvious assuming the fundamental theorem of arithmetic $\endgroup$ – yoyo Feb 11 '11 at 19:00
  • $\begingroup$ @kira: The title describes your problem exactly; why did you change it back to not having mark-up and to the old phrasing? $\endgroup$ – Arturo Magidin Feb 11 '11 at 19:46
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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 $\ \ \color{#c00}{c\mid ab},\,\ \color{#0a0}{(a,b,c)=1}\ \Rightarrow \ c = (a,c)(b,c)\ [=\, (ab,c\color{#0a0}{(a,b,c)}) = (\color{#c00}{ab,c}) = c\,]$

Remark $ $ Alternatively $\ (a,c)^n\! = (a^n,c^n) = (a^n,ab) = a(a^n,b) = a$ and $\,(b,c)^n = b\,$ by symmetry where the first equality employs the 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.

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Been a while since I did any maths, so this is probably the wrong way of going about it. I'd take logs with base $a$ of $ab=c^n$, this will give: $$1 + \log_a(b) = n\log_a(c) \rightarrow 1 = n\log_a(c) - \log_a(b).$$ For the right hand side to equal 1, we need $b = e^n$ (this wouldn't necessarily be the case if $\gcd(a,b) \neq 1$): $$1 = n(\log_a(c) - \log_a(e)).$$ I'd take the inverse log from here, giving $a = \left(\frac{c}{e}\right)^n$. Simple to explain why $\frac{c}{e}$ must be a whole number, $d$.

Bet that's the worst possible solution to this problem

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