Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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
share|improve this question

marked as duplicate by Dilip Sarwate, Henry T. Horton, Old John, Micah, rschwieb Dec 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.

    
I totally forgot that this is a dup. Good job spotting that, guys. –  Jyrki Lahtonen Dec 19 '12 at 8:09
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
1  
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
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.