Solve $x^{18} \equiv 7^{99} - 7, \mod 592$

Question

What I tried:

$x^{18} \equiv 7^{99} - 7, \mod 592 \iff \begin{cases} x^{18} \equiv 7^{99}-7 & \mod 7 \\ x^{18} \equiv 7^{99}-7 & \mod 2 \\ x^{18} \equiv 7^{99}-7 & \mod 3\end{cases} \iff x^{18} \equiv 0, \mod 7,2,3. $

I'm not sure how to proceed: is the last step equivalent to saying $x^{18} \equiv 0, \mod 42 (=7 \cdot 2 \cdot 3)$ or $x^{18} \equiv 0, \mod 592$?

Answer 1

Hint $\bmod 37\!:\,\ x^{\large 18}\equiv \color{#c00}{7^{\large 99}}\!-7\equiv -6\,\overset{\rm square}\Longrightarrow\,x^{\large 36}\equiv -1\,$ contra little Fermat

because: $\ \ 7 \equiv 3^{\large 4}\,\Rightarrow\, \color{#c00}{7^{\large 99}}\equiv (3^{\large 4})^{\large 99}\equiv (3^{\large 36})^{\large 11}\equiv 1^{\large 11}\equiv\color{#c00}{\bf 1}$