Given a prime $p$ and number $a$, how do I find the smallest $n$ such that $n^a\equiv 1 \pmod p$? Is there a non trial-and-error method?
Edit: obviously removing $n=1$!
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2$n=1$ works for all and it's the smallest – Stefan4024 Mar 06 '15 at 00:41
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I've edited your notation since $\mod{ }$ when used without equivalence signs is an operation and trivially $1\mod p = 1$ so that you're asking for the lowest $n$ s.t. $n^a=1$, which of course is not what you want. – user26486 Mar 06 '15 at 00:41
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Do you mean $,a^n,$ vs. $,n^a?\ \ $ – Bill Dubuque Mar 06 '15 at 00:49
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No. I am not talking about the order. I mean what I said for $n$ not $1$. – math_lover Mar 06 '15 at 01:12
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Sometimes $n=1$ is the only answer, such as when $p=5$ and $a=3$. – NovaDenizen Apr 10 '19 at 15:36
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The Tonelli-Shanks and Cipolla algorithms for square-roots can easily be generalized to compute d'th roots in finite fields, e.g. see Adleman; Manders; Miller: On taking roots in finite fields, and Bach; Shallit: Algorithmic number theory, section 7.3. See also this answer.
Bill Dubuque
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I haven't yet found an example where $n \not\equiv 1$, $a=3$, and $\gcd(p-1,3) = 1$. So I suspect nontrivial answers only exist when $\gcd(p-1,3) \ne 1$.
When $\gcd(a,p-1) = k \ne 1$, we have $\forall n \not\equiv 0, n^{\frac{a(p-1)}{k}} = \left(n^{\frac{a}{k}}\right)^{p-1} \equiv 1$, so $\forall n \not\equiv 0, \left(n^{\frac{p-1}{k}}\right)^a \equiv 1$.
This suggests an algorithm where you search for the smallest $\frac{p-1}{k}$th power, but I'm not sure how you'd go about that.
NovaDenizen
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Note that for $p>3$, $p$ has the form $6m\pm 1 \Rightarrow p-1=6m,6m-2$. gcd$(p-1,3)\ne 1 \Rightarrow p=6m+1$ – Keith Backman Apr 10 '19 at 17:17