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Prime factorisation: Show that if $a|b\cdot c$ and $\gcd(a,c) = 1 \implies a|b$.

$a|b\cdot c$ means "$a$ divides $b\cdot c$"

And I also read that if the greatest common divisor (aka $\gcd$) is $1$, then this means that $a$ and $c$ are mutually prime.

Now somehow apply this knowledge here, if it is relevant at all.

Let $a,b,c,x \in \mathbb{Z}$ then we have that:

$$\frac{b \cdot c}{a}=x$$

$$c=\frac{a \cdot x}{b}$$

Hmm but now I don't see how to get to $a|b$ using this way. Any ideas how to do it better?

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Since $(a,b)=1$, you have $ua+vc=1. Since $a divides $bc$ there exists $d$ such that $bc=da$ this implies $bua+bvc=b=a(ub+dv)$.

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  • $\begingroup$ Thank you for answer. Can you please add a little more details? All variables here are in $\mathbb{Z}$ is that right? $\endgroup$ – berndgr Oct 31 '16 at 16:31
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$a|b.c$ implies all the prime factors of $a $ get cancelled out by prime factors of $b.c$, and there are no common prime factors between $ a$ and $ c$. So all the prime factors of $ a$ get cancelled by prime factors of $b,$ meaning $ a|b$ .You can write it more mathematically.

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