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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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Have you considered the prime factorizations of $a$, $b$, and $c$? –  Jon Feb 11 '11 at 15:03
    
1=ax+by => 1=gcd(a^n,b^n) –  kira Feb 11 '11 at 15:43
    
@kira: Please ask questions, don't give orders. Also: please make your titles informative, not sentence fragments. –  Arturo Magidin Feb 11 '11 at 16:21
    
obvious assuming the fundamental theorem of arithmetic –  yoyo Feb 11 '11 at 19:00
    
@kira: The title describes your problem exactly; why did you change it back to not having mark-up and to the old phrasing? –  Arturo Magidin Feb 11 '11 at 19:46
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Of course it's trivial using unique factorization. Here's a more general proof using gcd's (or ideals) that has the benefit of giving an explicit closed form:

LEMMA $\rm\ \ c|ab,\ (a,b,c)=1\ \ \Rightarrow\ \ (a,c)^n\ (b,c)^n\ =\ (c)^n$

Proof $\rm\quad (a,c)^n\ (b,c)^n\ =\ ((a,c)\:(b,c))^n\ =\ (ab,c(a,b,c))^n\ =\ (ab,c)^n\ =\ (c)^n$

This may be considered 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. For further discussion see my post here, esp. the quote by Weil.

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Thanks! –  kira Feb 13 '11 at 20:15
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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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