How many roots has the equation $x^3+x^2+x+1=0$ modulo $340$?

Question

Question:

How many roots has the equation $x^3+x^2+x+1=0$ modulo $340$?

I am fairly sure this is a very standard problem but I cannot seem to apply Chinese Remainder Theorem.

So far I have gathered there is one-to-one correspondence from $\mathbb{Z}/\langle340\rangle$ to $\mathbb{Z}/\langle4\rangle\times\mathbb{Z}/\langle17\rangle\times\mathbb{Z}/\langle5\rangle$. What can I do next? I know that there is a bijective ring homomorphism $x+\langle 340 \rangle\to(x+\langle 4 \rangle\ ,x+\langle 17 \rangle\ , x+\langle 5 \rangle)$.

Any help is much appreciated!

Answer 1

Hint:

There are $2$ solutions mod $4$, $3$ solutions mod $5$, and $3$ solutions mod $17$.

Answer 2

The equation $x^3+x^2+x+1$ is just $$x^2(x+1)+x+1=(x+1)(x^2+1)$$ which has one root $-1=339\pmod{340}$. For the rest, you need to see if $-1$ is a root of unity$\pmod{340}$.