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Cryptography

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    $\begingroup$ But you are working in the multiplicative group, which has not prime order. See also this answer $\endgroup$ Commented Oct 31, 2018 at 16:01
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    $\begingroup$ @Ruggero: actually, there do exist finite fields with prime-sized multiplicative groups; for example, $GF(2^{127})$. $\endgroup$ Commented Oct 31, 2018 at 16:32
  • $\begingroup$ @Ruggero The group I'm working with is certainly prime order. $\endgroup$ Commented Oct 31, 2018 at 16:35
  • $\begingroup$ If the multiplicative group you're working in has prime order and n is a power of 2, then the answer is easy; either you're working in $GF(3)$ (and so the answer is 1 and 2), or you're working in a larger group, in which case the only n-th root of 1 is 1, that is $x^n = 1$ only if $x=1$ $\endgroup$ Commented Oct 31, 2018 at 18:29
  • $\begingroup$ @poncho Ops. I confess my mind is often limited to prime fields. $\endgroup$ Commented Nov 5, 2018 at 9:57