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We were asked to find primitive root mod $23$, $46$, $529$, $12167$.

My lecturer gave us a hint in finding primitive root mod $23$, but I am confused about his reasoning. My lecturer said $a$ would be a primitive root if $a^2 \not\equiv 1 \bmod 23$ and $a^{11} \not\equiv 1 \bmod 23$ as $\phi(23) = 22$ because the order of a number mod $23$ can only be $2$, $11$, or $22$. Why must $a^2$ and $a^{11}$ not be congruent to $1$ mod $23$, and why is that the order of a number mod $23$ can only be $2$, $11$, or $22$?

I checked $a=2, a=3, a=5$, and found $a=5$ satisfies the condition thus is a primitive root. From this, I found the primitive root mod $529$ using the fact that $23^2=529$ thus $5$ or $5+23$ must be a primitive root. I found $5^{506} \equiv 1 \bmod 529$ using a calculator. (is there a faster way to do this?)

I also can't see any trick in finding primitive root $46$ and $12167$, also I have noted $2 \times 23$ and $12167 = 529 \times 23$.

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hattyhats
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  • See https://math.stackexchange.com/questions/332760/prove-that-a-primitive-root-of-p2-is-also-a-primitive-root-of-pn-for-n1 and https://books.google.co.in/books?id=TtLMrKDsDuIC&pg=PA161&lpg=PA161&dq=primitive+root+2p%5En&source=bl&ots=oQt7C_0Q-R&sig=1bchGzZIH-_oQG763U7WqtguM_c&hl=en&sa=X&ved=0ahUKEwixibuy6ujWAhVJOY8KHZZ6CcwQ6AEIYzAH#v=onepage&q=primitive%20root%202p%5En&f=false – lab bhattacharjee Oct 11 '17 at 14:57
  • That does not answer any part of my question, as I already know how to find primitive root mod 529. My question lies in the confusion on finding primitive root mod 23, 46, and 12167 – hattyhats Oct 11 '17 at 14:59
  • For $(a,23)=1,$ $$a^{23(23-1)}\equiv1\pmod{23^2}$$ – lab bhattacharjee Oct 11 '17 at 15:02
  • https://math.stackexchange.com/questions/127109/order-of-an-element-modulo-n-divides-phin – lab bhattacharjee Oct 11 '17 at 15:03
  • The order of $a$ modulo $n$ is the smallest positive integer $m$ satisfying $a^m\equiv 1\mod n$. So, if $a^2\equiv 1\mod 23$ or $a^{11}\equiv 1\mod 23$, then the order is not $22$, hence $a$ is no primitive root. I we have $a^k\equiv 1\mod n$, we know that the order of $a$ modulo $n$ must divide $k$. – Peter Oct 11 '17 at 15:46
  • If I remember right, there are easy rules to find primitive roots for $p^k$ , $2p$ and $2p^k$, if a primitive root for $p$ is known, but I don't remember these rules. – Peter Oct 11 '17 at 15:50
  • Hint: Wiki writes: No simple general formula to compute primitive roots modulo n is known and continues to describe an algorithm if the factorization of $\phi(p)$ is known. Additionally a method for computing a primitive root for $p^k$ is shown, if a primitive root for $p$ is available. – gammatester Oct 11 '17 at 16:07

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