# Non-trivial 93rd roots of unity in finite fields [duplicate]

Possible Duplicate:
Finding the values of $n$ for which $\mathbb{F}_{5^{n}}$, the finite field with $5^{n}$ elements, contains a non-trivial $93$rd root of unity

For which of the following value of $n$, does the finite field $\Bbb F$, with $5^n$ elements contain a non trivial $93$-rd root of unity?

1. 92
2. 30
3. 15
4. 6
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I totally forgot that this is a dup. Good job spotting that, guys. – Jyrki Lahtonen Dec 19 '12 at 8:09

## marked as duplicate by Dilip Sarwate, Henry T. Horton, John Wordsworth, Micah, rschwiebDec 19 '12 at 1:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

The multiplicative group of a finite field $\Bbb F$ with $\mid \Bbb F\mid=q$ is cyclic and so, by the theory of cyclic groups, it contains a unique subgroup of order $d$ for each divisor $d$ of $q-1$. Thus, the question becomes: for which values of $n$ we have that $93$ divides $5^n-1$?

Now:

• $5^{92}\equiv67^{23}\equiv40\bmod93$,

• $5^{30}\equiv(5^6)^5\equiv1\bmod93$,

• $5^{15}\equiv56^3=32\not\equiv1\bmod93$,

• $5^6\equiv1\bmod93$.

So, of the $n$ listed, the answer is 6 and 30.

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What's with the intermediate steps? Why those particular factorisations? – Ben Millwood Dec 18 '12 at 14:00
thanks for ur precious advice. – Alka Goyal Dec 18 '12 at 16:15
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