$n^2$ divides $\phi(a^n-1)$ whenever $n$ is even and $a>2$ My problem is to show that 
$n^2$ divides $\phi(a^n-1)$ whenever $n$ is even and $a>2$.
I have thought a solution but it is quite long and tedious.
I wonder if anyone has a nice and clear solution.
(May be consider to find an element having order $n^2$ in the cyclic group? I tried but failed.)
Welcome for discussion =)
 A: My method followed by a several claims:
Let $n$ be a positive integer (instead of an even integer).


*

*$n \mid \phi(a^n-1)$ and $2n \mid \phi(a^n+1)$,
by considering order of $a$ is $n$ modulo $a^n-1$ and order of $a$ is $2n$ modulo $a^n+1$ 

*$2n \mid \phi(a^n-1)$, whenever $a>1$, $n>1$.
I prove it by induction on $k$ in the following statement:
"For any nonegative integer $k$, $2n \mid \phi(a^n-1)$, whenever $a>1$, $n=2^km$, where $m$ is positive odd integer."
The proof of (2) is here:
I: $k=0$, $n$ is odd and $n \mid \phi(a^n-1)$ (by(1)). As $a>2$ and $n>1$, so $a^n-1>1$ and $\phi(a^n-1)$ is even.
Hence $2n \mid \phi(a^n-1)$.
II: Assume $2n \mid \phi(a^n-1)$ if $n=2^km$.
Then $2^{k+1}m \mid \phi(a^{2^{k}m}-1)$ by assumption.
$2(2^km) \mid \phi(a^{2^{k}m}+1)$ (by(1)) so we have $2^{k+1}m \mid \phi(a^{2^{k}m}+1)$.
III: $\gcd(a^{2^{k}m}-1,a^{2^{k}m}+1)=1$ or $2$ depends on whether $a$ is odd or even (using Bézout's lemma).
By $\gcd(c,d)\phi(c)\phi(d)=\phi(cd)\phi(\gcd(c,d))$ and taking $c=a^{2^{k}m}-1, d=a^{2^{k}m}+1$, we obtain $\phi(a^{2^{k}m}-1)\phi(a^{2^{k}m}+1) \mid \phi(a^{2^{k+1}m}-1)$ no matter $a$ is odd or even.
IV. Mutiply the two result in (II) together, we have $(2^{k+1}m)^2 \mid \phi(a^{2^{k+1}m}-1)$.
Hence $2(2^{k+1}m) \mid \phi(a^{2^{k+1}m}-1)$, our induction is done here.
Finally, we multiply the result in (1) and (2) to obtain
$(2n)^2 \mid \phi(a^n+1)\phi(a^n-1) \mid \phi(a^{2n}-1)$, whenever $a>1, n>1$,
and the last "$\mid$" follows by the same argument in (III).
