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How can I show that $x^2 + 2x + 3$ is primitive in $GF(5)$?

My idea: $ x^1 = x\\ x^2 = -2x - 3 = 3x + 2\\ x^3 = (3x + 2)x = 3x^2 + 2x = 3(3x + 2) + 2x = x + 1\\ ...\\ x^a = 1\\ $

This would take quite long. Is there a better way? Please only basic Algebra!

Willi Mentzel
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  • My answer to the duplicate is based on the idea that the order is a factor of $24$. The task is to prove that nothing less than $24$ works. To that end it suffices to show that neither $8$ nor $12$ suffices. If my logic there depends on results you are not familiar with (a fairly good possibility), then I suggest that you use repeated squaring and calculate $x^2,x^4,x^8$ and then $x^{12}=x^4\cdot x^8$. Does this make sense to you? – Jyrki Lahtonen Aug 09 '15 at 18:34
  • @JyrkiLathonen It perfectly does make sense! This is exactly what I was searching for. Thank you! Can you still put it in an answer despite the fact that it has been marked a duplicate? – Willi Mentzel Aug 09 '15 at 19:17
  • As I was the mod closing this as a duplicate, it would not look good, if I reopened this in order to answer it myself. OTOH you could post those calculations. Preferrably there. I promise to comment and eventually upvote (if any corrections are due) :-) – Jyrki Lahtonen Aug 09 '15 at 19:21
  • Ok, will do so tomorrow! I only have mobile right now :) I will even put my question and your answer in latex! – Willi Mentzel Aug 09 '15 at 19:31

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