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

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 $$x^{18} \equiv 0, \mod 592$$?

• The prime factorisation of $592$ is $2^4\,37$. – Bernard Jan 27 at 13:28
• by the chinese remainder theorem, the last $a \equiv 0, \mod 7,2,3 \iff a\equiv 0, \mod 42$ but the first step $a \equiv b, \mod 592 \iff a\equiv b, \mod, 7,2,3$ is completely wrong and weird and utterly out of the blue and has no justification at all. By CRT we can get $x^{18} \equiv 7^{99}-7, \mod 592=2^4*37 \iff x^{18}\equiv 7^{99} - 7, \mod 16,37$ – fleablood Jan 27 at 16:49
• @fleablood Perplexing indeed. Possibly $\bmod 37\,$ was misread as $\bmod 3,\!7\ \$ – Bill Dubuque Jan 27 at 16:56

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: $$\ \ 7 \equiv 3^{\large 4}\,\Rightarrow\, \color{#c00}{7^{\large 99}}\equiv (3^{\large 4})^{\large 99}\equiv (3^{\large 36})^{\large 11}\equiv 1^{\large 11}\equiv\color{#c00}{\bf 1}$$
• So the contradiction implies that we don't have to consider the congruence modulo 37 and that we can just look for solutions modulo 16, right? And since $16 = 2^4$, is it enough to only consider $x^{18} \equiv 7^{99}-7 \, (\operatorname{mod} 2)$? – Zachary Jan 27 at 17:05
• It implies that there is no solution mod $592,\,$ since such s solution remains a solution of reduced mod $37.\$ – Bill Dubuque Jan 27 at 17:26