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Using the follow theorem:

Let $a$ and $b$ two positive integers numbers so $gcd(a,b) = 1 \Leftrightarrow \exists s,t \in \mathbb{Z} : sa+tb = 1$

I want to prove that if $a\mid bc \mbox{ and } gcd(a,c) = 1 \rightarrow a\mid b$

Using the definition of divisibility I arrived at: $\exists k \in \mathbb{Z} : ak = bc$. Using this and the theorem I isolate the c, I arrived in $ak = b\frac{1-sa}{t}$. But I can not get anywhere from there,I can not see a way to do it the way I'm thinking

Matheus
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1 Answers1

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You use the theorem to represent 1 in terms of a and c. Then multiply both sides of the equation by b. On the left, you see that a | ab (obviously), and you are given that a | bc. See where this goes?

PMar
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