Polynomials' multiplicative inverse

Question

Let $A(x)$ be a polynomial with integer coefficients. Is there always a polynomial $B(x)$ for which

$$A(x)\cdot B(x)\equiv 1\pmod n$$

(for a given integer $n$). If the answer isn't yes, an answer "yes if $n$ is ____ (fill in with a characteristic the number has to have" would be interesting as well. Of course, no is also an answer.

This question is on the track of what I'm interested in, it just doesn't have the key ingredient, the mod (that's why it turned out trivial).

Answer 1

If we take $A(x)=xp(x)$, then naturally no such $B$ exists for any $n>1$.

Answer 2

If $n$ were a prime, then the answer is "no" (unless $A(x)$ is a non zero constant), because the residue classes modulo a prime form a field, and the argument in an answer to the linked question covers that case. A consequence of this is that for $A(x)$ to have an inverse it is necessary that $A(x)$ is a non-zero constant modulo all the prime factors of $n$. This is actually also sufficient. To cut a longish story short:

1. Chinese remainder theorem.
2. In a commutative ring elements of the form $u+y$, with $u$ a unit and $y$ nilpotent, are always invertible. See this question for examples in the case $n=4$.