How to find $n$, such that $\gcd(a, b+nc)\ne1$

Question

Given three polynomials $a(x)$, $b(x)$, and $c(x)$ over the integers, one needs to find an integer $n$, such that $\gcd(a(x), b(x) + n\cdot c(x)) \ne 1$.

Is there a general way to find such an $n$ or to determine that no such integer exists?

Answer 1

Hint If $gcd(a,b,c)=d \neq 1$ then show that $d| \gcd(a(x), b(x) + n\cdot c(x))$ for all $n$.

Hint 2 If $gcd(a,b,c)= 1$, let $d_1,..., d_k$ be the irreducible factors of $a$.

Show that for each $d_j$ there exists at most one $n \in \mathbb Z$ such that $$ d_j |b+nc $$

This shows that in this case there are at most $k$ values of $n \in \mathbb Z$ which are bad.

Addedd

For the edited version.

Let $d_1,..., d_k$ be the irreducible factors of $a$. Note that $gcd(a,b+nc) \neq 1$ if and only if there exists some $1 \leq j \leq k$ such that $d_j |b+nc$.

The problem can be restated as:

Given an irreducible polynomial $d_j$ when does there exists some $n$ so that $d_j | b+nc$?

This is equivalent to $$b=-nc \pmod{d_j}$$ i.e. the reminders when $b$ and $c$ are divided by $d_j$ being proportional.