4

I have a proof for the case of $a$ being prime I believe, I think this is also true for $a$ composite since I ran a test for the first $100$ numbers over the first $100$ values of $n$ and it seems to be true.

Anyway the proof for 'a' prime is indirect and through finite field theory. $ \phi(a^n - 1)$ is the number of primitive roots in $GF(a^n)$ and they have irreducible polynomials of degree n over $GF(a)$ and all the roots of that irreducible equation have to be primitive roots. So the irreducible equations partition the set of primitive roots into sets of $n$ elements which means $ \phi(a^n - 1)$ is divible by $n$.

Any help would be appreciated.

Jared
  • 31,451
Asvin
  • 8,229

3 Answers3

7

If $a > 1$ is an integer, then the order of $a$ in $\mathbb{Z}_{a^n-1}^*$ is $n$.

spin
  • 11,956
3

maybe this helps: $$ (a , a^n - 1) = 1 $$ so

$$ a^{\Phi (a^n - 1)} = 1 \pmod {a^n-1} $$ on the other hand $$ a^n = 1 \pmod {a^n-1} $$

MITjanitor
  • 2,690
Mehdi
  • 604
2

We'll assume $a > 1$. Note that the order of $a$ modulo $(a^n-1)$ is exactly $n$:

  • $a^n\equiv 1 \mod (a^n-1)$ (the order divides $n$)

  • $|a^k - 1| < |a^n - 1|$ for $k<n$ (the order is at least $n$).

By Lagrange's Theorem, the order of an element divides the order of the group, thus: $$n | \phi(a^n-1)$$

Ofir
  • 6,245