How to find the amount of solutions of polynomial congruence?

The given congruence is $ x^4-5x-6 ≡ 0 $ mod $(100^{100})$. Find the amount of it’s solutions.

I’ve factorised it: $ (x+1)(x-2)(x^2+x+3) ≡ 0$ mod $(100^{100})$.

From here I get solutions for the first two brackets: $ x ≡ -1 $ mod $(5^{200}2^{200}) $ and $ x ≡ 2 $ mod $(5^{200}2^{200}) $.

The left one bracket is: $ x^2+x+3 ≡ 0$ mod $(5^{200}2^{200})$.

Using the Chinese remainder theorem: $ \begin{cases} x^2+x+3 ≡ 0(5^{200}) \\ x^2+x+3 ≡ 0(2^{200}) \end{cases} $

Using the Legendre’s(Jacobi’s) symbol and the following facts:

1. if we have $ax+by+c ≡ 0$ mod(m) and $a,b,c$ are integers, m – a positive integer and $GCD(2a,m)=1$, than the given expression and $y^2 ≡ b^2-4ac$ mod(m) have the same amount of solutions.
2. if $p$ is prime and do not divide $a$, then $x^2 ≡ a$ mod $p^n$ is solvable(and has two solutions), if a is quadratic residue modulo p.

With given information, I get that the first expression in my system has 2 solutions. What about the second one? What should I do with 2 exponent?

The answer in my textbook for the amount of solutions for the firstly given expression is 8.

Solution 1:

Show the following

1. $x^2 + x + 3 \equiv 0 \pmod{2^{200}}$ has 0 solutions.
2. $ x^4 - 4x -6 \equiv 0 \pmod{2^{200}}$ has 2 solutions $\pmod{2^{200}}$.
3. $x^2 + x + 3 \equiv 0 \pmod{5^{200}}$ has 2 solutions $\pmod{5^{200}}$. (There is no need to determine these solutions.)
4. $ x^4 - 4x -6 \equiv 0 \pmod{5^{200}}$ has 4 solutions $\pmod{5^{200}}$.
5. $ x^4 - 4x -6 \equiv 0 \pmod{10^{200}}$ has $2\times 4 = 8 $ solutions $\pmod{10^{200}}$.