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Suppose we have set $Z = \{0, 1, \dots, N-1\}$ with arithmetic operations modulo $N$; $a > 0$ is an element in $Z$.

Is it possible that $a^{-1}$ does not exist but $a^{-n}$ exists for some $n$, $1 < n < N$?

If this is impossible how to prove it?

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  • $\begingroup$ What do you mean by $a^{-n}$. Is it $(a^n)^{-1}$? $\endgroup$ – Prahlad Vaidyanathan Dec 27 '13 at 5:30
  • $\begingroup$ yes, inverse of $a^n$ $\endgroup$ – kludg Dec 27 '13 at 5:32
  • $\begingroup$ Minor note: since the ordering of $Z$ is not significant, you should probably have written $a\ne 0$ rather than $a>0$. $\endgroup$ – dfeuer Dec 27 '13 at 5:32
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Let $a^nb=1$. Then $a(a^{n-1}b)=1$.

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