On $\gcd(a,x) = \gcd(b,x)=k \implies gcd(ab,x) = k$ Originally, I was examining $\gcd(a,x) = 1, \gcd(b,x) = 1$ and conjectured $\gcd(ab,x) = 1$. I think this is true, because I thought:
Let $x = p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}\dots$
$a\neq 0\pmod{p_1}$
$a\neq 0\pmod{p_2}$
$a\neq 0\pmod{p_3} \dots$ 
And similarly for $b$, we must have
$ab\neq 0\pmod{p_1}$
$ab\neq 0\pmod{p_2}$
$ab\neq 0\pmod{p_3} \dots$ 
And hence $\gcd(ab,x) = 1$. (Verify? I'm still a bit iffy on this)
If this is true, however, then if $\gcd(a,x) = \gcd(b,x) = k$ for some $k$ a product of distinct prime numbers, then $\gcd(ab,x) = k$ because for each of the primes in $k$'s factorization we have both $a,b = 0 \pmod p$ and for every other prime both $a,b \neq 0 \pmod p$.
Is this generally true for every number $k$, rather than just all $k$ a product of distinct primes? Does $\gcd(a,x) = \gcd(b,x) = 12 \implies \gcd(ab,x) = 12$?

gcd(3,9)=gcd(3,9)=3 but gcd(3*3,9)=9
