Exercise 1.3.20 (If $(a,b,c)[a,b,c] = abc$, then $(a,b) = (b,c) = (a,c) =1$)

Question

Given $(a,b,c)[a,b,c] = abc$, prove that $(a,b) = (b,c) = (a,c) =1$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $a,b,c$ be positive integers. Write
$$a = \prod_{p \in A} p^{\alpha(p)},\qquad b = \prod_{p \in A}p^{\beta(p)}, \qquad c = \prod_{p \in A}p^{\gamma(p)},$$
where $0 \leq \alpha(p),0 \leq \beta(p), 0 \leq \gamma(p)$.

Suppose that $(a\wedge b \wedge c)(a \vee b \vee c)= abc$. By (1.7),
$$ \prod_{p \in A} p^{\min(\alpha(p),\beta(p), \gamma(p)) + \max(\alpha(p),\beta(p), \gamma(p))} =  \prod_{p \in A} p^{\alpha(p)+ \beta(p) + \gamma(p)}.$$
The fundamental theorem shows that for every $p \in A$,
$$\min(\alpha(p),\beta(p), \gamma(p)) + \max(\alpha(p),\beta(p), \gamma(p)) = \alpha(p)+ \beta(p) + \gamma(p).$$
If $\alpha(p) \leq \beta(p) \leq \gamma(p)$, then $\alpha(p) + \gamma(p) = \alpha(p)+ \beta(p) + \gamma(p)$, thus $\beta(p) = 0$. Since $0 \leq \alpha(p) \leq \beta(p)$, we obtain also $\alpha(p) = 0$.

More generally, whatever the order in which the elements $\alpha(p),\beta(p), \gamma(p)$ are, two among them are zero.

If $a \wedge b 
e 1$, there is some $p \in A$ such that $\alpha(p)>0$ and $\beta(p) >0$, so it is impossible that two exponents among $\alpha(p),\beta(p), \gamma(p)$ are zero. This is a contradiction, hence $a \wedge b = 1$. Similarly $b\wedge c = a \wedge c = 1$.

\end{proof}

Exercise 1.3.20 (If $(a,b,c)[a,b,c] = abc$, then $(a,b) = (b,c) = (a,c) =1$)

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Exercise 1.3.20 (If $(a,b,c)[a,b,c] = abc$, then $(a,b) = (b,c) = (a,c) =1$)

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