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$a$ and $m$ are natural numbers and $a>1$. Prove that $m | \phi(a^m - 1)$.

Any hints how can I prove this statement?

Martin Sleziak
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Ju Bc
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    do you heart about Lehmer Conjecture ?, try to let a=2 , it is the only part for L.Conjecture which it is proved – zeraoulia rafik May 05 '17 at 17:54
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    https://math.stackexchange.com/questions/594233/prove-that-n-divides-phian-1-where-a-n-are-positive-integer-without OR https://math.stackexchange.com/questions/774814/prove-that-n-divides-phian-1-where-phi-is-eulers-phi-function – lab bhattacharjee May 05 '17 at 18:03
  • The goal of this question is to show that m is prime iff m | phi(a^m-1) , and this happen if a^m-1 is prime , just we look for a=2 . – zeraoulia rafik May 05 '17 at 18:24
  • @zeraouliarafik Lehmer's Totient Problem asks whether there is any composite number $n$ such that Euler's totient function $φ(n)$ divides $n − 1$. – Dietrich Burde May 05 '17 at 18:40
  • it could be proved easily for any number $m=(p_1-1)^{e_1} (p_2-1)^{e_2} \cdots $ where $p_i$ are prime numbers (not necessarily in order). – Ahmad May 05 '17 at 20:40

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From $\gcd(a,a^m-1)=1$ we have $$a^{\varphi(a^m-1)}\equiv 1 \pmod{a^m-1}$$ also $$a^{m}\equiv 1 \pmod{a^m-1}$$ If we assume $\varphi(a^m-1)=mq+r, 0<r<m$ then $$a^{\varphi(a^m-1)}\equiv a^{mq+r} \equiv (a^m)^qa^r \equiv (1)^qa^r \equiv a^r\equiv 1 \pmod{a^m-1}$$ which is a contradiction (because of $0<r<m$), so $r=0$.

rtybase
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