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.
A: We assume $c\neq 0$, else there is nothing to do.  
Let $d=\gcd(a,b)$ and write $a=md,b=nd$. There exists integers $u,v$ such that $\gcd(u,v)=1$ and 
$$
umd + vnd = d
$$
Therefore
$$
\begin{align*}
(umd + uvc) + (vnd - uvc) &= d\\
u(a+vc) + v(b-uc) &= d
\end{align*}
$$
First assume $u,v$ are non-zero.  
We claim that $D=\gcd(a+vc,b-uc)=1$.   
Suppose otherwise, let $p$ be a prime dividing $D$. Then $p$ divides $d$, hence $p$ divides $a$ and $b$.  
Since $p$ also divides $a+vc,b-uc$, this shows that $p$ divides $vc,uc$. By assumption $\gcd(a,b,c)=1$, hence $p$ cannot also divide $c$. This means that $p$ divides $u,v$, contradicting $\gcd(u,v)=1$.
Therefore $D$ must be equal to 1.  

If $u=0$, then $v=n=1$ and $a=md, b =d$. We may set $u,v$ to $u=-1, v=1+m$ so that $u,v$ are nonzero and
$$
ua +vb = -md + (1+m)d = d
$$
Similarly for $v=0$.
