How many solutions does this equation have? $$ x^4+2x^3+x+1 \equiv 0 \pmod{5} $$
I know that I can rewrite it was $$ x \left( x^3+2x^2+1 \right) \equiv -1 \pmod{5} $$ but I'm not sure if anything else could be done here. I know how to do the solving and the Chinese Remainder Theorem, but I'm not sure how I can just tell how many solutions there are without actually going through the process.
Inspection shows the sum of the coefficients to be $5$, the modulus in question, so we have $x=1$ as our first solution.
$$x^4+2x^3+x+1 \equiv (x-1)(x^3 -2x^2 -2x -1) \pmod{5}$$
$0, 1$ and $-1$ all fail, but $$2^3 - 2 \cdot 2^2 - 2\cdot 2 - 1 = -5 \equiv 0 \pmod{5}$$ so $$(x-1)(x^3 -2x^2 -2x -1) \equiv (x-1)(x-2)(x^2 -2) \pmod{5}$$
and finally, note that $2$ is not a square modulo $5$.
The only way to do these is really to do some intelligent guess and check.
Clearly, $~x~$ is a solution if and only if $~(x + 5)~$ is a solution. This is because
$$\left[ ~x \in \Bbb{Z}, ~n \in \Bbb{Z^+} ~\right] \implies (x + 5)^n \equiv x^n \pmod{5}. \tag1 $$
The identity in (1) above is easily demonstrated by examining the binomial expansion of $~(x + 5)^n.$
So, the problem reduces to determining which elements in $~\{0,1,2,3,4\}~$ solve the problem. Let $~f(x) = x^4 + 2x^3 + x + 1~$ and consider the following table:
\begin{array}{| r | r | r | r | r |} \hline x & x \pmod{5} & x^4 \pmod{5} & 2x^3 \pmod{5} & f(x) \pmod{5} \\ \hline 0 & 0 & 0 & 0 & 1 \\ \hline 1 & 1 & 1 & 2 & 0 \\ \hline 2 & 2 & 1 & 1 & 0 \\ \hline 3 & -2 & 1 & -1 & -1 \\ \hline 4 & -1 & 1 & -2 & -1 \\ \hline \end{array}
So, the only solutions are $~x \in \{1,2\}.~$
That is, all elements of the set $~\{ ~a + 5b ~: ~[ ~a = 1 ~\text{or} ~a = 2 ~], ~b \in \Bbb{Z} ~\}~$ are solutions.
I'm not sure how I can just tell how many solutions there are without actually going through the process.
It depends on the polynomial. With a small modulus, a direct evaluation of the polynomial at each number in that modulus is reasonable. But since you are interested in other methods, here is a way to count the number of distinct roots mod $p$ without evaluating the polynomial at all.
Theorem. When $p$ is prime and $f(x)$ is a mod $p$ polynomial, the roots of $f(x)$ in the integers mod $p$ are the same as the roots of $\gcd(f(x),x^p-x)$.
Proof. As a mod $p$ polynomial, $x^p - x \equiv x(x-1)\cdots(x-(p-1)) \bmod p$, so its roots are all integers mod $p$ exactly once. Therefore $\gcd(f(x),x^p-x)$ is, up to a nonzero constant factor, the product of $x-r$ as $r$ runs over the distinct roots of $f(x) \bmod p$. QED
Corollary. When $p$ is prime and $f(x)$ is a mod $p$ polynomial, the number of roots of $f(x)$ in the integers mod $p$ is the degree of $\gcd(f(x),x^p-x)$.
Proof. This gcd is the product of $x-r$ as $r$ runs over the distinct roots of $f(x) \bmod p$, so the degree of $\gcd(f(x),x^p-x)$ counts the number of those roots.
Example. Let $f(x) = x^4 + 2x^3 + x + 1 \bmod 5$. By Euclid's algorithm, $\gcd(f(x),x^5-x) = x^2 + 2x + 2$, so $f(x) \bmod 5$ has two roots in the integers mod $5$ and they are the mod $5$ roots of $x^2 + 2x + 2$.
As others suggested, you can use Fermat's Little Theorem. You can iterate it since $x\ne 0$. Essentially, divide and conquer to reduce the 4th degree polynomial to a two degree polynomial.
$x^4+2x^3+x+1 \equiv 0 \pmod{5}$
$2x^3+x+2 \equiv 0 \pmod{5}$
$2x^4+x^2+2x\equiv 0 \pmod{5}$
$x^2+2x +2\equiv 0 \pmod{5}$ // here you might legitimately claim two or fewer solutions. The constant term is not a perfect square modulo 5, so there is no root of multiplicity 2. That means there's either two roots or no roots. By inspection we know 1 is a root. A variation of the rational root theorem might have told us that at step 1. We thus eliminate the no-root possibility. So we have two roots without knowing both. It is easy to find the other though.
$(x+1)^2\equiv 4 \pmod{5}$
$(x-1)(x+3) \equiv 0 \pmod 5$
$\implies x \in \{1,2\}$
