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.
Best Answer
The following is an answer to the question in the title, using the technique that you tried to use. Of course the answer has nothing to do with the question in the body of the post.
Because $a$ and $n$ are relatively prime, there exist integers $q$ and $r$ such that $qa+rn=1$. Similarly, there exist integers $s$ and $t$ such that $sb+tn=1$. Rewrite these equations as $qa=1-rn$ and $sb=1-tn$. Multiply. We obtain $$(qa)(sb)=(1-rn)(1-tn).$$ Expand and rearrange a bit. We get $$(qs)ab+(r+t-rtn)n=1,$$ which shows that $ab$ and $n$ are relatively prime.