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How many 19th power residues modulo 229 are there?

My approach was that as gcd(19,228) = 19, then by a proposition that I am allowed to use from lecture, any b is a 19th power residue mod 229 if and only if $b = x^{19}$ for some x an element of unit group of (Z/229). This group has 228 elements, so I conclude that there are 228 19th power residues modulo 229.

Is my logic flawed in any sense? I am not sure if the fact I am using necessarily implies a bijection between the number of power residues and elements in the unit group.

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  • $\begingroup$ Careful...$19\,|\,228$ so there are non-trivial solutions to $a^{19}\equiv 1 \pmod {229}$. $\endgroup$
    – lulu
    Oct 6, 2023 at 17:38
  • $\begingroup$ Interesting, how would one go about finding the non-trivial 19th power residues? $\endgroup$
    – Anon
    Oct 6, 2023 at 17:42
  • $\begingroup$ You could just search, but there's no need. The goal is to count them, not list them. $\endgroup$
    – lulu
    Oct 6, 2023 at 17:42
  • $\begingroup$ I suggest: start with a smaller example. How many cubes are there $\pmod {13}$? Try to generalize that. $\endgroup$
    – lulu
    Oct 6, 2023 at 17:44
  • $\begingroup$ Here is the proposition shown in lecture: If (n,p-1) = e, then some b is nth power residue mod p if and only if b = x^e for some x in unit group of Z/p. Doesn't this imply that the solutions MUST be elements of the unit group? $\endgroup$
    – Anon
    Oct 6, 2023 at 17:47

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$229$ is prime so $(\mathbb Z/229)^×$ is cyclic group and the collection of all $19$th powers forms subgroup, just find its order. I don't know if such simple answer exists if it wasn't a prime.

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