Proof

By hypothesis, we have $a\neq 0$, because $a\mid bc$, i.e. $a$ divides the product of the integers $b$ and $c$. Also by hypothesis, we have that $a$ and $b$ are co-prime. This means by definition that $\gcd(a,b)=1$. We have to prove that $a\mid c$.
We inspect two cases:
- Case $(1)$ $b=0$. * If $b=0$, then $a=\pm 1$ because of $\gcd(a,b)=1$. Therefore, $a$ is a trivial divisor of $c$.
- Case $(2)$ $b\neq 0$. * For the least common multiple of the natural numbers $|a|$ and $|b|$¹, it follows from the relationship between the greatest common divisor and the least common multiple that $\operatorname{lcm}(|a|,|b|)=|a|\cdot |b|.$ * On the other hand, by hypothesis, we have that $bc$ is a multiple of $|a|$ and, trivially, $bc$ is a multiple of $|b|$. * Thus, $bc$ is a common multiple of $|a|$ and $|b|$. * Because the least common multiple divides any common multiple, we have that $\operatorname{lcm}(|a|,|b|)\mid bc\Rightarrow|a||b|\mid bc.$ * By the definition of divisors, this is equivalent to $ab\mid bc,$ and from the divisibility law no. 7, it follows that $a\mid c.$

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Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927