Q:

Three relatively prime numbers

Is it true that if $\gcd(a,b,c)=1$ then there exists $x,y\in\mathbb{Z}$ such that $\gcd(a+xc,b+yc)=1$?  
I came upon this while trying to prove that the natural homomorphism $r_m:\operatorname{SL}_n(\mathbb{Z})\to\operatorname{SL}_n(\mathbb{Z}/m\mathbb{Z})$ is surjective.  I was trying to show that for $n=2$, if $A\in\operatorname{SL}_2(\mathbb{Z}/m\mathbb{Z})$ then it suffices to show that there exists $B\in M_2(\mathbb{Z})$ such that $r_m(B)=A$ and $\gcd(b_{11},b_{12})=1$.

A:

If the highest common factor of $a$ and $c$ is $d$, so that $a=pd$ and $c=qd$ with $p$ and $q$ co-prime, then $a+xc=d(p+xq)$.
We know that $d$ is co-prime to $b$, and Dirichlet's theorem on primes in arithmetic progression tells us that $p+qx$ is a prime infinitely often.  But $b$ only has a finite number of prime factors. 
So in fact we can do this with $y=0$.
Whether Dirichlet's theorem is necessary for this, I don't know off the top of my head.  It feels like there ought to be something simpler.  But this at least answers the question.