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

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    $\begingroup$ The prime factorisation of $592$ is $2^4\,37$. $\endgroup$ – Bernard Jan 27 at 13:28
  • $\begingroup$ 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 $x^{18} \equiv 7^{99}-7, \mod 592=2^4*37 \iff x^{18}\equiv 7^{99} - 7, \mod 16,37$ $\endgroup$ – fleablood Jan 27 at 16:49
  • $\begingroup$ @fleablood Perplexing indeed. Possibly $\bmod 37\,$ was misread as $\bmod 3,\!7\ \ $ $\endgroup$ – Bill Dubuque Jan 27 at 16:56
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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: $\ \ 7 \equiv 3^{\large 4}\,\Rightarrow\, \color{#c00}{7^{\large 99}}\equiv (3^{\large 4})^{\large 99}\equiv (3^{\large 36})^{\large 11}\equiv 1^{\large 11}\equiv\color{#c00}{\bf 1}$

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  • $\begingroup$ 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 $x^{18} \equiv 7^{99}-7 \, (\operatorname{mod} 2)$? $\endgroup$ – Zachary Jan 27 at 17:05
  • $\begingroup$ It implies that there is no solution mod $592,\,$ since such s solution remains a solution of reduced mod $37.\ $ $\endgroup$ – Bill Dubuque Jan 27 at 17:26
  • $\begingroup$ "So the contradiction implies that we don't have to consider the congruence modulo 37" A contradiction means everything is impossible. We have to quit. We can not solve this. There is no solution. $\endgroup$ – fleablood Jan 27 at 18:11

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