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I like to find those finite field $GF(p^n)$ which contains primitive $3$rd root of unity. One thing is clear that $GF(p^n)^*$ is cyclic group of size $p^n-1. $ So for $3$rd primitive root we must have $3/(p^n-1).$ i.e. $$p^n-1\equiv 0\mod3$$ so for primes of the form $p=3n+1$ finite field $GF(p^n)$ must have $3$rd primitive root of unity. Also i noticed that for any prime $p$ and even $n$ $p^n\equiv 1\mod3$ i.e. finite field $GF(p^n)$ always has $3$rd primitive root of unity if $n$ is even. Please help me to find all prime $p$ so that $GF(p^n)$ has $3$rd primitive root of unity. Thanks.

neelkanth
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    For $p=2$ and $n>1$ with $n$ even, the field $GF(p^n)$ contains a primitive $3$rd root of unity since $x^2+x+1$ is reducible (in $GF(4)$). It is also clearly true that for $p=3$ any $GF(p^n)$ contains a primitive $3$rd root of unity. I hope someone gives a general answer. – Levent Mar 05 '18 at 12:32
  • Yes bhut it is already noticed in question statement – neelkanth Mar 05 '18 at 12:43
  • Oh, I am sorry, I did not notice that. – Levent Mar 05 '18 at 12:46
  • There is nothing essential missing from the list of key observations in your post. The 3rd primitive root is there if and only if $p\equiv1\pmod3$ or $n$ is even (of course $p\neq3$ is a must). So: WHAT IS THE QUESTION? – Jyrki Lahtonen Mar 05 '18 at 21:42

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