How to show that ab and c are coprime when $(a,c)=(b,c)=1$?

Question

Let $a$, $b$ and $c$ be non-zero integers. Suppose $a$ and $c$ are coprime. And suppose $b$ and $c$ are coprime. How can I then show that $ab$ and $c$ are coprime?

From what I know so far this means $ra + sc = 1$ (since a and c are coprime) Likewise: $tb + uc = 1$

If $ab$ and $c$ are coprime it follows: $v(ab) + pc = 1$

Beyond that I am rather stuck- I have tried rearranging in terms of $a$ and $b$ but with no success

Answer 1

Suppose that $ab$ and $c$ were not coprime. Then there must be some prime number $p$ that divides both $ab$ and $c$. (Be sure to understand why).

But since $p$ is prime and $p\mid ab$ that implies that $p\mid a$ or $p\mid b$.

In the first case, we have then $p\mid a$ and $p\mid c$ contradicting that $\gcd(a,c)=1$, and in the second case we have $p\mid b$ and $p\mid c$ contradicting that $\gcd(b,c)=1$.

Answer 2

HINT

If $ab$ and $c$ are not coprime, there is a $d$ which divides both $ab$ and $c$. Thus, there exists a prime p that divides both $ab$ and $c$ (either $d$ is prime and then $p=d$ or $d$ is not prime and take any one of its prime factors).

Since $p|(ab)$, either $p|a$ or $p|b$. Can you finish this reaching the contradiction?

Answer 3

Suppose there is a prime $p$ that divide $c$ and $ab$. But then $p$ divide also $a$ or $b$. So if $p$ divide $a$ then $p$ divide $a$ and $c$. A contradiction. The same contradiction we get if $p$ divide $b$.