How many solutions are there to $x^3\equiv-1\pmod{365}$?

Question

How many solutions are there to $x^3\equiv-1\pmod{365}$?

I found that there is only one solution mod $5$, but mod $73$ I'm a little confused. Factoring out $x^3 + 1=(x+1)(x^2-x+1)$, the second equation has no real solutions, hence $x\equiv-1\pmod{73}$ is the only solution mod $73$. So by the Chinese remainder theorem, there is only $1*1=1$ solution. Is this correct?

Answer 1

Note that $x^3+1$ factors as $x^3+1=(x+1)(x^2-x+1)$, where $x^2-x+1$ is irreducible in $\Bbb{F}_5[x]$. So indeed there is only one solution $\mod5$. But in $\Bbb{F}_{73}[x]$ we have $$x^3+1=(x+1)(x+8)(x-9),$$ so there are three solutions $\mod{73}$, which are $-1$, $-8$ and $9$. By the Chinese remainder theorem we then have $1\times3=3$ solutions.

Answer 2

See @Servaes for a very nice answer.

For something a bit more systematic, $73$ is prime and so there is a primitive root $g$ modulo $73$. Every element of $\Bbb Z_{73}$ except for $0$ (which is not a solution) can be written as $x=g^k$ for some $k=0,1,2,\ldots,71$. Since $-1\equiv g^{36}$ we have $$\eqalign{x^3\equiv-1\pmod{73}\quad &\Leftrightarrow\quad g^{3k}\equiv g^{36}\pmod{73}\cr &\Leftrightarrow\quad 3k\equiv36\pmod{72}\ ;\cr}$$ since $\gcd(3,72)=3$ and this is a factor of $36$, there are three solutions.