1.3: Problem 20: Given \((a,b,c)[a,b,c] = abc\), prove that \((a,b) = (b,c) = (a,c) = 1\).
Proof
Suppose \((a,b,c)=d\). Then, we can write \(a = da'\), \(b = d'b\) and \(c = c'a\). This also implies that \((a',b',c') = 1\) (otherwise, \(d\) is not the gcd). From this, we can see that \([a,b,c] = [d'a, db', dc'] = d[a',b',c']\). Thus, the left hand side of the given statement is
$$
\begin{align*}
(a,b,c)[a,b,c] = d \cdot d[a',b',c'] = d^2 \cdot [a',b',c']
\end{align*}
$$
while the right hand side is
$$
\begin{align*}
abc = da' \cdot db' \cdot dc' = d^3 \cdot a'b'c'
\end{align*}
$$
So if the right hand side is equal to the left hand side in
$$
\begin{align*}
d^2 \cdot [a',b',c'] &= d^3 \cdot a'b'c' \\
[a',b',c'] &= d \cdot a'b'c' \\
\end{align*}
$$
then, the equality can only hold if and only if \(d = 1\) which means that we must have \((a,b)=(b,c)=(a,c)=1\). \(\ \blacksquare\)