State and Prove Euclid's lemma. Prove that if $a ~|~ b \cdot c$, with $ gcd(a, b) = 1$, then $a ~|~ c$.

 Lemma:  If $a ~|~ b \cdot c$, with $ gcd(a, b) = 1$, then $a ~|~ c$.

Proof:  Since $gcd(a, b) = 1$, by Theorem, there exist integers $x$ and $y$ such that $a \cdot x + b \cdot y = 1$. Multiplying this equation by $c$, we get $c = c \cdot 1 = c~(a \cdot x + b \cdot y) = a \cdot c \cdot x + b \cdot c \cdot y$. As $ a ~|~ a \cdot c$ and  $ a ~|~ b \cdot c$, by part $7.$ of Theorem, $d ~|~ (a \cdot c \cdot x + b \cdot c \cdot y)$. Therefore $d ~|~ c$.




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