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I'm trying to show 7 is a primitive root modulo 11, 22, 121, and 14641.

I think I need to show

  1. $7^{10} \equiv 1 \pmod{11}$
  2. $7^{10} \equiv 1 \pmod{22}$
  3. $7^{110} \equiv 1 \pmod{121}$
  4. $7^{13310} \equiv 1 \pmod{14641}$

Since $gcd(11,22)\ne 1$, it seems I can't use CRT?

Attempt

As been pointed out by @Arthur. I need to show that the smallest positive solution to $7^≡1 \pmod{11}$ is =10. I can calculate $7^ \pmod{11}$ for integer 1<=n<=10 to show this. But can I show that the smallest positive solution to $7^n \equiv 1 \pmod{14641}$ is n=13310 without calculating all?

Mzq
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    There are several possibilities , the easiest one is Euler's therem : If $\gcd(a,n)=1$ , then we have $a^{\varphi(n)}\equiv 1\mod n$ – Peter Oct 11 '23 at 12:31
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    None of these address $7$ being a primitive root. As @Peter says, we can know that all your congruences are true without really calculating any powers. But they aren't what you need to prove. – Arthur Oct 11 '23 at 12:39
  • ah, so how should I show 7 is a primitive root without calculating all orders of 14641? – Mzq Oct 11 '23 at 12:40
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    To show that $7$ is a primitive root modulo 11, you need to show that $7^{10}\equiv_{11}1$, yes. That's what makes $7$ a root (of unity). But that's rather easy, as it's true for any number not divisible by $11$ (Fermat's little theorem in this case, but also Euler's theorem, or Lagrange's theorem, for more advanced versions). What you need to also show, however, and where most of the work lies, is that the smallest positive solution to $7^n\equiv_{11}1$ is $n = 10$. That's what makes $7$ a primitive root. – Arthur Oct 11 '23 at 12:42
  • Is there a "math way" to show that? As a programmer, I know I can write code to verify n=13310 is the smallest number to get $7^n \equiv 1 \pmod{14641}$ – Mzq Oct 11 '23 at 13:03
  • You could start with reading some materials regarding primitive roots. For example, the Wikipedia page tells you how to show that a number is a primitive root modulo a given prime $p$, and in which case is a primitive root modulo $p$ also a primitive root modulo $p^k$. – Jianing Song Oct 11 '23 at 22:28
  • The Order Test easily verifies the first two. The rest follow by the first dupe, since the primitive root $7$ lifts from $!\bmod 11$ to $\bmod 11^n,,$ since $7^{10}\not\equiv 1\pmod{!11^2}.\ \ $ – Bill Dubuque Oct 11 '23 at 22:58

1 Answers1

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As others have noted, you know $7^{\varphi(n)} \equiv 1 \mod n$ by Euler, and you want to show $7^k \not\equiv 1 \mod n$ for all $k \in \{1, \ldots \varphi(n)-1\}$. What makes this easily doable is that you only need to check those $k$ that divide $\varphi(n)$, since those are the only possibilities for the period of $7$ in the multiplicative group mod $n$.

Robert Israel
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