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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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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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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obvious assuming the fundamental theorem of arithmetic |
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