Solve the following congruence: $x(x+1)(x+2) \equiv 0 \pmod{221}$

Question

Find the first five solutions for, $$x(x+1)(x+2) \equiv 0 \pmod{221}$$

I am very confused. By CRT,

$x(x+1)(x+2) \equiv 0 \pmod{13}$ and $x(x+1)(x+2) \equiv 0 \pmod{17}$

But these two congruences are also reckless.

Answer 1

$$x(x+1)(x+2)\equiv 0\pmod{221}\iff \begin{cases}x(x+1)(x+2)\equiv 0\pmod{13}\\x(x+1)(x+2)\equiv 0\pmod{17}\end{cases}$$

By Euclid's Lemma:

$$\iff \begin{cases}x\equiv \{0,-1,-2\}\pmod{13}\\x\equiv \{0,-1,-2\}\pmod{17}\end{cases}$$

By the Chinese Remainder Theorem (CRT) there are $9$ solutions mod $221$.

E.g., if $x\equiv 0\pmod{13}$ and $x\equiv -1\pmod{17}$, then $x\equiv 169 \pmod{221}$.

You should already know how to use CRT. You'll get:

$$x\equiv \{-2,-1,0,50,51,102,117,168,169\}\pmod{221}$$

Answer 2

Since $13$ ans $17$ are primes, we can infer that one of the factors $x,x+1,x+2$ is a multiple of $13$ and one (possibly the same or not) is a multiple of $17$. All in all, you end up with nine cases ...