A question about the product of primes

Question

Let $\mathbb{P}$ be the set of all primes in the natural numbers and let $p_i \in \mathbb{P}$ be the $i$th prime, $p_1=2$. Let $m = \prod_{i=1}^n (p_i)$. How many solutions does $x^2 + x \equiv 0 (\bmod m)$ have?

Conjecture: $2^n$.

Answer 1

Your conjecture is correct. Let $e_1,e_2,\dots,e_n$ be any sequence whose entries are $0$ or $-1$. There are $2^n$ such sequences.

By the Chinese Remainder Theorem, the system of congruences $x\equiv e_i\pmod{p_i}$ has a unique solution modulo $m$. Any such $x$ satisfies $x^2+x\equiv 0\pmod{p_i}$, and hence satisfies the congruence $x^2+x\equiv 0\pmod{m}$. So our congruence has at least $2^n$ solutions.

Note that any solution of $x^2+x\equiv 0\pmod{m}$ is a solution of $x^2+x\equiv 0\pmod{p_i}$, and these are $x\equiv 0\pmod{p_i}$ and $x\equiv -1\pmod{p_i}$. So our congruence has exactly $2^n$ solutions.