$n\mid \phi(a^{n}-1)$ for any $a>n$?

Question

I saw the proof which goes as follows:

$a^{n} \equiv 1 \pmod{a^{n}-1} $, and $n$ is the smallest power of a such that this is true.

We also know that by Euler's Identity $a^{\phi(a^{n}-1)}\equiv 1\pmod {a^{n}-1}$

So we get $n\mid \phi(a^{n}-1)$.

Now I am unable to understand where is the fact that $a>n$ is playing role?

It isn't playing a role at all - the statement is still true if $a<n$. Perhaps you mean to ask about the requirement that $a$ and $n$ are relatively prime (which is necessary)? — Zev Chonoles, Jan 03 '13 at 18:54
@ThomasAndrews: Ah, I was thinking that it is a requirement to apply Euler's theorem, but of course in this case the modulus is $a^n-1$, and $a$ and $a^n-1$ are always relatively prime, so I was incorrect. — Zev Chonoles, Jan 03 '13 at 19:01
Only thing one should mention that $a$ is an integer and $a>1$. — user371231, Dec 09 '22 at 10:00

score 4 · Answer 1 · answered Jan 03 '13 at 19:15

4

It's not. Nothing in that proof requires $a>n$, and, indeed, the theorem is true for all $a>1,n\geq 1$.

For example, you gave $n=3,a=2$. But then $a^n-1=7$ and $3|\phi(a^n-1)=6$

answered Jan 03 '13 at 19:15

Thomas Andrews

lab bhattacharjee · Answer 2 · 2013-01-04T03:33:14.817

0

What we can say is $$ord_{(a^n-1)}a\mid gcd(n,\phi(a^{n}-1))$$

As $ord_{(a^n-1)}a=n, n\mid gcd(n,\phi(a^{n}-1))$

If $a=2,n=5, \phi(2^5-1)=31-1=30$ which is divisible by $5(=n)$.

If $a=2,n=7, \phi(2^7-1)=127-1=126$ which is divisible by $7(=n)$.

edited Jan 04 '13 at 03:33

answered Jan 04 '13 at 03:27

lab bhattacharjee

2

Except $a^n\equiv 1\bmod (a^n-1)$ and there can't be any $0<k<n$ such that $a^n-1\mid a^k-1$, so we do know that $\mathrm{ord}{a^n-1}(a)=n$. Because $a$ is relatively prime to $a^n-1$, by Euler's theorem we have $a^{\phi(a^n-1)}\equiv 1\bmod (a^n-1)$, which implies that $n\mid \phi(a^n-1)$ because $\mathrm{ord}{a^n-1}(a)=n$. – Zev Chonoles Jan 04 '13 at 03:29

2 Answers2