Proof
(related to Proposition: Divisors of a Product Of Two Factors, CoPrime to One Factor Divide the Other Factor)
 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 coprime. 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 $\operatorname{lcm}(a,b)\mid bc\Rightarrowab\mid bc.$
* 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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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927
Footnotes