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I would like to prove the following: Let b > 1 a natural. Let p a prime number where p ^ b = 1. The decimals of 1/p in base b are infinite and periodic from the first decimal (I've proven that already), let t be that period.

Prove that t | p-1

kheorus 7
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    What do you mean by b ^ p = 1? The greatest common divisor? – 5xum Oct 04 '22 at 09:06
  • I mean that b is prime with p, aka p doesn't divide b since p is prime – kheorus 7 Oct 04 '22 at 09:14
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    So it seems that your answer to 5xum is "yes". To clarify, you can edit your question to replace b ^ p = 1 by $b\land p=1$, or even better: "$p$ does not divide $b$". – Anne Bauval Oct 04 '22 at 09:34

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By the same proof as when $b=10$, the period in base $b$ of $\frac1p$ is the order of $b$ in the multiplicative group $(\mathbb Z/p\mathbb Z)^*$. By Lagrange's theorem, this order divides the order $p-1$ of that group.

Anne Bauval
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