LCM equals Product iff Coprime

Theorem

Let $a, b \in \Z_{>0}$ be strictly positive integers.

Then:

$\lcm \set {a, b} = a b$

if and only if:

$a$ and $b$ are coprime

where $\lcm$ denotes the lowest common multiple.

Proof

Necessary Condition

Let $a$ and $b$ be coprime.

Then:

\(\ds \lcm \set {a, b}\)

\(=\)

\(\ds \frac {a b} {\gcd \set {a, b} }\)

Product of GCD and LCM

\(\ds \)

\(=\)

\(\ds \frac {a b} 1\)

Definition of Coprime Integers

\(\ds \)

\(=\)

\(\ds a b\)

$\Box$

Sufficient Condition

Let $\lcm \set {a, b} = a b$.

Then:

\(\ds \lcm \set {a, b} \gcd \set {a, b}\)

\(=\)

\(\ds a b\)

Product of GCD and LCM

\(\ds \leadsto \ \ \)

\(\ds \gcd \set {a, b}\)

\(=\)

\(\ds \frac {a b} {\lcm \set {a, b} }\)

\(\ds \leadsto \ \ \)

\(\ds \gcd \set {a, b}\)

\(=\)

\(\ds \frac {a b} {a b}\)

by hypothesis

\(\ds \)

\(=\)

\(\ds 1\)

$\blacksquare$

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm