Question:
Suppose $a,b \in \Bbb N$, $\gcd (a,n) = \gcd(b,n) = 1$. The question is to prove or give a counterexample: $\gcd(ab,n) = 1$.
My Work:
This is what I have so far (for $\alpha, \beta, \gamma, \delta \in \Bbb Z$): \begin{align*} \gcd(a,n) = 1 \ &\Rightarrow 1 = \alpha a + \beta n\\ \gcd(b,n) = 1 \ &\Rightarrow 1 = \gamma b + \delta n \end{align*} Multiplying the top equation by $b$, and the bottom by $a$, I have $$ b + a = (\alpha + \gamma)ab + (\beta b + \delta a)n $$
Here is where I am stuck. I now know that you can write a linear combination of $ab, n$ in this form, where all coefficients are integers, but I think I may have gone down the wrong road in this proof in multiplying by $a,b$. Hints would be appreciated.