$$x^3\equiv x \pmod{105}$$
I'm trying to solve this equation. Here's what I tried so far:
$$x^3\equiv x \pmod{105} \iff x^2\equiv 1 \pmod{105}$$
Then, applying the Chinese remainder theorem, I got the system: $$\cases{x^2 \equiv 1 \pmod{5}\\x^2 \equiv 1 \pmod{7}\\x^2 \equiv 1 \pmod{3}}$$ With the following solutions:
$$\cases{x \equiv \pm1 \pmod{5}\\x \equiv \pm1 \pmod{7}\\x \equiv \pm1 \pmod{3}}$$ At this point, I think I need to pretty much solve these eight systems:
$$\cases{x \equiv 1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv 1 \pmod{3}} \cases{x \equiv 1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv -1 \pmod{3}} \cases{x \equiv 1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv 1 \pmod{3}} \cases{x \equiv -1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv 1 \pmod{3}}$$$$ \cases{x \equiv -1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv -1 \pmod{3}} \cases{x \equiv -1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv 1 \pmod{3}} \cases{x \equiv 1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv -1 \pmod{3}} \cases{x \equiv -1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv -1 \pmod{3}}$$
Here's how I solved the first one: Considering the first two equations, we get: $$x=5k+1=7h+1$$ from which $k = 7+7y, h = 5+5y$, with $y \in \mathbb{Z}$. Therefore, $$x=36+35y\iff x\equiv1\pmod{35}$$ Adding in the third equation, we have that $36+35y = 1+3 w$, from which $x = 1281 + 35w \iff x \equiv1\pmod{105}$.
However, this one seems like a really tedious method as I'd have to do the same calculations for seven more systems. Is there anything I'm missing? Is there a faster way to do this?
As $x^3-x=(x-1)x(x+1)$ is a product of three consecutive integers
$3$ must divide $x^3-x$
So, we need $$x^3\equiv x\pmod{5\cdot7}$$
If $(x-1)x(x+1)\equiv0\pmod 5$
$\implies x\equiv0\ \ \ \ (1), x\equiv-1\ \ \ \ (2), x\equiv1\pmod5\ \ \ \ (3)$
Similarly, $x\equiv0\ \ \ \ (4), x\equiv-1\ \ \ \ (5), x\equiv1\pmod7\ \ \ \ (6)$
Now apply CRT on $(1),(4); (1),(5);(1),(6);(2),(4); (2),(5);(2),(6)$