Let $(a,b)$ denote the GCD of $a$ and $b$, and let $[a,b]$ denote the LCM of $a$ and $b$. Prove $\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$.
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Use the FTA and you can just hash it out. – Adam Hughes Jul 01 '14 at 23:10
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That was my first instinct but how? – math-sd Jul 01 '14 at 23:12
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I'll elaborate in an answer. – Adam Hughes Jul 01 '14 at 23:27
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This is the very first (1972) question on the USAMO, see for example here. – Bart Michels Jul 08 '14 at 07:20
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Hint $\ $ Let $\,\alpha,\beta,\gamma $ be the powers of the prime $\,p\,$ in the unique prime factorization of $\,a,b,c,\,$ resp. $\ $ By symmetry we may choose notation so that $\,\alpha\le \beta\le \gamma.\,$ Comparing powers of $p$ on both sides below
$$\begin{align}\frac{[a,b,\color{#90f}c]^{\large\color{#90f}2}}{[a,\color{#0a0}b][b,\color{#c00}c][\color{darkorange}c,a]} &\,=\, \frac{(a,b,c)^{\large 2}}{(a,b)(b,c)(c,a)}\\[.4em] \iff\ \ \color{#90f}{2\gamma} - (\color{#0a0}\beta + \color{#c00}\gamma + \color{darkorange}\gamma) &\,=\, 2\alpha - (\alpha + \beta + \alpha)\end{align}$$
Alternatively $ $ combine $\ (ab,bc,ca)[a,b,c]\, =\, abc\, =\, [ab,bc,ca](a,b,c)\ $ from this answer with
$$\qquad (a,b,c)(bc,ca,ab) = (a,b)(b,c)(c,a)$$
which is true since both are equal to $\ (aab,aac,abb,abc,acc,bbc,bcc)\ $ by the distributive law.
Bill Dubuque
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