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Let \(p\) be a prime number and \(n\) be a positive integer. Prove that \(\phi(p^n-1)\) is divisible by \(n\), where \(\phi\) denotes Euler's totient function.

Note by Finn Hulse 3 years, 10 months ago

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Firstly,

\(p^n\equiv1\mod(p^n-1)\) Note that n is the smallest number with that property, in other words n\(=ord_{p^n-1}(p)\)

Also,

\(p^{\varphi(p^n-1)}\equiv1\mod(p^n-1)\)

But the order must divide every number with that property , so

\(n|\varphi(p^n-1)\)

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Beautiful. :D

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestFirstly,

\(p^n\equiv1\mod(p^n-1)\) Note that n is the smallest number with that property, in other words n\(=ord_{p^n-1}(p)\)

Also,

\(p^{\varphi(p^n-1)}\equiv1\mod(p^n-1)\)

But the order must divide every number with that property , so

\(n|\varphi(p^n-1)\)

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Beautiful. :D

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