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I am struggling to understand a simple fact that should follow from the Chinese remainder theorem. Suppose we have a system of equations.

$$ x^n = 1 (mod\ p) $$ $$ x^n = 1 (mod\ q) $$

Where $p, q$ are prime. I get the idea that each of these equations has a number of solutions $gcd(n, p-1), gcd(n, q-1)$ respectively due to the properties of cyclic fields (explained here). I just don't get how we get that the number of solutions of $x^n = 1 (mod\ pq)$ is the multiple of the number of solutions to the other two equations. Meaning: $$ gcd(n, p-1)*gcd(n, q-1) $$

J. W. Tanner
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user9855939
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  • Welcome to Mathematics Stack Exchange. It's like the multiplication principle, because, by the CRT, each solution mod $pq$ corresponds to a solution mod $p$ and a solution mod $q$ – J. W. Tanner Jul 06 '23 at 23:03
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jul 06 '23 at 23:15
  • The Chinese remainder theorem basically says that a variable $x$ ranging over residue classes modulo $pq$ can be thought of as two independent variables $y$ and $z$, one ranging over residue classes modulo $p$ and the other modulo $q$ such that $x\equiv y\pmod p$ as well as $x\equiv z\pmod q$. – Jyrki Lahtonen Jul 07 '23 at 04:39
  • May be you simply didn't take into account that when looking at the equation modulo $p$ your "universe" (=the set where your variables draw their values from) has size $p$, but in the combined congruence the universe has $pq$ elements, in a sense containing $q$ copies of each residue class modulo $p$, and also $p$ copies of each residue class modulo $q$? – Jyrki Lahtonen Jul 07 '23 at 06:50
  • See the Remarks in the linked dupes. – Bill Dubuque Jul 07 '23 at 16:30

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