Is the product of two coprime numbers coprime with a third if the three numbers are pairwise coprime? I have some trouble finding a proof for the following:
Let $a, b$ and $ c $   be three pairwise coprime integers.
Is $ab$  coprime with $c$? If yes, how to prove it?
Thank you in advance
 A: More generally, if $\gcd(a,c)=1$ and $\gcd(b,c)=1$ then $\gcd(ab,c)=1$. This can be proved by solving:
$$ax+cy=1\\bu+cv=1\tag{1}$$
then multiply these two equations to get:
$$ab(xu)+c(axv+byu+cyv)=1$$
(1) can be solved by Bézout's identity.
A: Another way is to invoke the fact that the GCD function is multiplicative (ProofWiki has one proof): given $b$ and $c$ coprime, $$\gcd(a, bc) = \gcd(a, b) \gcd(a, c).$$ Therefore, if $\gcd(a, b) = \gcd(a, c) = \gcd(b, c) = 1$, then $\gcd(ab, c) = 1$ as well.
A: They are co prime as well:
$(a, b) = 1, (b, c) = 1 \ \ \ (1)$
$[a, b] = a * b \ \ \ (2)$
$(1)=>([a, b], c) = 1$
$(2)=>(a * b, c) = 1$
A: Yes. Let $a_1,a_2,\ldots,a_i$ be the prime factors of $a$,  $b_1,b_2,\ldots,b_j$ be the prime factors of $b$, $c_1,c_2,\ldots,c_k$ be the prime factors of $c$. All of these primes are distinct (because $a,b,c$ are pairwise coprime). When you multiply $b \times c$ the new product has the prime factors $b_1,b_2,\ldots,b_j, c_1, c_2, \ldots, c_k$. None of these prime factors are in $a_1, a_2, \ldots, a_i$. Therefore, $b\times c$ is coprime with $a$.
