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While 'playing' with the multiplicative group of integers mod $n$, I noticed that $n\mid \phi(a^n-1)$. The proof is straightforward:

  • $a \in \left ( \mathbb{Z}/(a^n-1)\mathbb{Z} \right )^\times$ because $a \equiv 0 \pmod{p} \Leftrightarrow a^n-1 \equiv -1 \pmod{p}$ for a prime $p$, and hence $\gcd(a,a^n-1)=1$
  • $\left | a \right |=n $ because $ a^n = (a^n-1)+1 \equiv 1 \pmod{a^n-1} $ and if $ 1 \leq k \leq n-1 $ we have $ a<a^k<a^n-1 \Rightarrow a^k\not\equiv 1 \pmod{a^n-1} $ so that $n $ is the smallest positive power for which $a^k \equiv 1$
  • Finally, by Lagrange's theorem, we get: $\left| a \right| \mid \left|\left ( \mathbb{Z}/(a^n-1)\mathbb{Z} \right )^\times\right| \Rightarrow n\mid\phi(a^n-1)$

I assumed that such a simple looking identity had to have a simple direct number theoretic explanation, and with my limited knowledge of number theory, I tried to tackle the problem but without success.

I still have a feeling that there is a simple explanation and that I'm just missing the elephant in the room that easily illuminates this fact.

Off Topic: I had difficulty formatting the 'proof' because I am new to $\LaTeX$ and I don't know how things should really be done on this website, so feel free to edit anything to make it "more right", I will certainly appreciate it.

Michael Hardy
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Fujoyaki
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1 Answers1

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The standard "classical" number-theoretic proof is the same. Let $a\gt 1$. We have $a^n\equiv 1\pmod{a^n-1}$, and obviously $a^k\not\equiv 1\pmod{a^n-1}$ if $1\le k\lt n$. So $a$ has order $n$ modulo $a^n-1$, and therefore $n$ divides $\varphi(a^n-1)$.

At the end one does not appeal to Lagrange's Theorem explicitly. But a standard easy early number-theoretic lemma is that if $a$ has order $k$ modulo $m$, then $k$ divides $\varphi(m)$.

André Nicolas
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  • Thank you for your contribution, but what I meant by "number theoretic" is more like "non-algebraic", i.e. no Lagrange's theorem for example. – Fujoyaki Sep 14 '14 at 00:19
  • The above proof has been around since before the first glimmer of groups. The order of an element $a$ modulo $m$ was defined before order of a group was defined. – André Nicolas Sep 14 '14 at 17:15