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I tried induction but got stuck. For $n=1$ it is true. Suppose it holds for $n$, i.e, the order of 3 is $\phi(7^n)$. Now I should prove it for $\phi(7^{n+1})=6\cdot 7^n$.

$7^n\mid3^{6\cdot 7^n}-1$. How do I derive that $7^{n+1}\mid 3^{6\cdot 7^{n+1}} - 1$. I would appreciate your help.

Michael Hardy
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Meitar
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    Well, that's not all you need. Indeed $m\mid 3^{\phi(m)}-1$ for any $m$ not divisible by $3$. What you need is that this $\phi(7^{n+1})$ is the least number $d$ such that $7^{n+1}\mid 3^d-1$. – Thomas Andrews May 16 '15 at 16:45
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    I wouldn't say "exact," because people do actually post truly exact duplicates, where they copy and paste questions. But it is a duplicate. @EthanBolker – Thomas Andrews May 16 '15 at 16:50
  • Well the main solution there relies on a theorem of which I haven't heard. – Meitar May 16 '15 at 16:51
  • Your actual request holds for all numbers coprime to 7, not just primitive roots - what you have there is not adequate to show that 3 is a primitive root. – Joffan May 16 '15 at 17:24

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