Is the product of two coprime numbers coprime with a third if the three numbers are pairwise coprime?

Question

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

Answer 1

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 $a$.

Answer 2

They are co prime as well:

$(a, b) = 1, (b, c) = 1 \ \ \ (1)$

$[a, b] = a * b \ \ \ (2)$

$(1)=>([a, b], c) = 1$

$(2)=>(a * b, c) = 1$

Answer 3

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.

Answer 4

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.