If $\gcd(A,B,C)=1$, can we find $h$ s.t. $\gcd(A,B+hC)=1$? If $\gcd(A,B,C)=1$, can we find $h$ s.t. $\gcd(A,B+hC)=1$?
I have tried but I find I am not able to prove this. Maybe I do not know some important thing? Could someone help? Thanks!
 A: We must check prime divisors $p$ of $A$:
for $p|A,C$ (and not $B$) every $h$ is fine,
for $p|A,B$ (and not $C$) we must have $h \equiv 1 \pmod p$,
for $p|A$ (and not B and C) we must have $h \equiv 1 -BC^{-1} \pmod p$, that is fine since $C$ is invertible in this case.
Put everything in a big Chinese Remainder System, you know you can solve it!
E.G.:
$(A,B,C)=(30,14,35)$
we have $2,3,5|30$.
$5|A,C$ so no problem.
$2|A,B$ thus we can take $h \equiv 1 \pmod 2$
$3|A$ we can take $h \equiv 1 - 14 \cdot (-1) \equiv 0 \pmod 3$.
$h=3k$ with $k$ odd verifies all the $3$ requests, you can easily check that $gcd(A,B+hC)=1$ for every $h$ of this form. (They are not the only one).
A: This isn't an elementary proof, but it's all I have.
$1 = \gcd(A,B,C) = \gcd(A,\gcd(B,C))$.
Let $g = \gcd(B,C)$. Then $\dfrac Bg$ and $\dfrac Cg$ are integers and
$\gcd \left( \dfrac Bg, \dfrac Cg \right) = 1$.
$\underline{\text{Dirichlet's Theorem on Primes in Arithmetic Progressions:}}$
If a and b are relatively prime positive integers, then the 
      arithmetic progression a, a+b, a+2b, a+3b, ... contains
      infinitely many primes.
So the arithmetic progression $\dfrac Bg + n\dfrac Cg$ contains infinitely many prime numbers. So, for some $n, \dfrac Bg + n\dfrac Cg = p$ where $p$ is a prime number such that $\gcd(A,p) = 1$. Since, also, $\gcd(A,g) = 1$, then we must have
\begin{align}
  1 &= \gcd(A,gp) \\
    &= \gcd\left(A, g\left(\dfrac Bg + n\dfrac Cg\right) \right) \\
   &= \gcd(A, B + nC)
\end{align}
