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Q: Solve $x^{14}\equiv 16 (mod\ 23)$.

If $23$ has replaced with a much smaller number, say $m$ I would be inclined to simply test each $0\le k< m$. Moreover, $23$ itself is prime so I can't split it into prime factors, apply Hensel's lemma or similar and then use the chinese remainder theorem successively.

I think this will involve the use of order and primitive roots, but I can't immediately see how that would be a viabe option here.

Bill Dubuque
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  • $23$ is really quite small. But perhaps it helps to note that the solutions to $n^2\equiv {16}\pmod {23}$ are $4$ and $19$. Still, given how small $23$ is, I'd just brute force the computation. – lulu Mar 15 '22 at 18:26
  • It's equivalent to solving $,x^7\equiv 4,,$ for which see the linked dupe (and its links). The sought roots are then $\equiv \pm x\ \ $ – Bill Dubuque Mar 15 '22 at 18:34

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