For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to $f(x)\equiv 0 \mod(n)$ where $n$ is composite? For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to $f(x)\equiv 0 \mod(n)$ where $n$ is composite? 

Is there a general way to determine the number of incongruent solutions modulo $n$? 

My first idea is that we can of course break $n$ into its prime power factorization and look at $f(x)\equiv 0 \mod(p_{i}^{e_{i}})$ where $(p_{i}^{e_{i}})$ appears as a prime power factor in $n$.

Here's where I start to become confused, if $f(x)=x$ then the Chinese remainder theorem tells us that the solution is unique modulo $n$, but if $f(x)$ is non-constant and non-linear then we need to use the lifting method to solve $f(x)\equiv 0$ for each $\mod(p_{i}^{e_{i}})$ - but so far the method tells us nothing about the number of solutions. 

I presume I am not incorrect in saying that the number of incongruent solutions to $f(x)\equiv \mod(p_{i}^{e_{i}})$ is at most $min(deg(f), p_{i}^{e_{i}})$, but is there a general way to determine precisely how many solutions are there?
First factor $m$ to powers of primes. $$N(m) = PI\ N(p^a)$$ $N$ is number of roots. This is by CRT.

Then for each prime factor p of m, a polynomial $P(x)$ can be reduced to $P'(x)$ with degree $&lt;= p$. If $P'(x)$ is a factor of $(x^p - x) \mod p$ (after dividing, the remainder is multiple of $p$), then it has exactly $n$ roots where $n$ is degree of $P'(x)$.

From $p$ to $p^a$, it's Hensel's Lemma.
First factor $m$ to powers of primes. $$N(m) = PI\ N(p^a)$$ $N$ is number of roots. This is by CRT.

Then for each prime factor p of m, a polynomial $P(x)$ can be reduced to $P'(x)$ with degree $&lt;= p$. If $P'(x)$ is a factor of $(x^p - x) \mod p$ (after dividing, the remainder is multiple of $p$), then it has exactly $n$ roots where $n$ is degree of $P'(x)$.

From $p$ to $p^a$, it's Hensel's Lemma.