If $(a,c)=1$ and $(b,c)=1$, prove $(ab,c)=1$.

Question

I'm posed with the problem in the title,

Let $a,b,c\in\mathbb{Z}$. Then if $(a,c)=1$ and $(b,c)=1$, prove $(ab,c)=1$.

(By the way, $(a,c)=1$ means that the greatest common divisor of $a$ and $c$ is $1$, hence they are relatively prime.)

I know that $(a,c)=1$ and $(b,c)=1$ means the following:

$$\exists x_1,x_2,x_3,x_4\in\mathbb{Z} \textrm{ such that } ax_1+cx_2 = bx_3+cx_4=1$$

So that means we have to two linear equations:

$$ \begin{align*} ax_1+cx_2&=1\\ bx_3+cx_4&=1 \end{align*} $$

And we want to manipulate them into something of the form

$$(ab)x_5+cx_6=1, \,\, \exists x_5,x_6\in\mathbb{Z}$$

because that would then prove that $(ab,c)=1$. I just can't figure out how to do it.

Is this approach alright, or is there a better one? Could you nudge me in the right direction?

Answer 1

Multiply the two equations you have and see the magic.

Answer 2

If a prime $p$ divides $c$, it does not divide $a$ or $b$. So does it divide $ab$?

Answer 3

$$1=(ax_1+cx_2)(bx_3+cx_4)=aby+cz$$ This for $$y=x_1x_3$$ and $$z=ax_1x_4+bx_2x_3+cx_2x_4$$

Answer 4

Bit different approach using factorization:

Let \begin{align} a &= p_1^{\alpha_1}\cdots p_m^{\alpha_m}\\ b &= q_1^{\beta_1}\cdots q_n^{\beta_n}\\ c &= r_1^{\gamma_1}\cdots r_o^{\gamma_o}\\ \end{align}

where $p_i, q_j,r_k \in \mathbb{N}$, for $i = 1, \ldots, m;\, j = 1, \ldots, n;\, k = 1,\ldots, o$ are primes. Then $\gcd{(a,c)}=1$ and $\gcd{(b,c)}=1$ means, that in the factorization $a$ and $c$ do not share any primes ($p$'s and $r$'s) and also $b$ and $c$ do not share any primes.

Full solution

Setting $$a b= p_1^{\alpha_1}\cdots p_m^{\alpha_m} \cdot q_1^{\alpha_1}\cdots q_n^{\alpha_n}$$does not change much and obviously $ab$ does not share any prime with $c$ too. So $\gcd{(ab,c)}=1$.