Prove that $36|n^{12k-6}-n^6$ Prove that if gcd(n,6)=1 and k>0, then  $36|n^{12k-6}-n^6$.
Idea: I want to show that  $2|n^{12k-6}-n^6$ and  $3|n^{12k-6}-n^6$.
For the first part, since gcd(n,6)=1, n must be an odd number, so it is easy to show that $n^{12k-6}-n^6$ is divisible by 2. I am stuck by the second part and i think that I need to use Fermat's little theorem somehow. Anyone can give some hint?

Let $n$ and $k>0$ integers. Notice that
\begin{align*}
n^{12k-6}-n^6&=\left(n^{6k-3}-n^3\right)\left(n^{6k-3}+n^3\right)\\
&=\left(n^{2k-1}-n\right)\left(n^{4k-2}+n^{2k}+n^2\right)\left(n^{2k-1}+n\right)\left(n^{4k-2}-n^{2k}+n^2\right)\\
&=n^6\left(n^{2k-2}-1\right)\left(n^{4k-4}+n^{2k-2}+1\right)\left(n^{2k-2}+1\right)\left(n^{4k-4}-n^{2k-2}+1\right)
\end{align*}


*

*If $n$ is even, then $2^6=64$ divides $n^{12k-6}-n^6$. 

*For $n$ odd we have $4|(n^{2k-2}-1)$ and $2|(n^{2k-2}+1)$, so $8$ divides $n^{12k-6}-n^6$.

*If $3|n$, then $729$ divides $n^{12k-6}-n^6$.

*If $3$ does not divide $n$, then the squares $n^{4k-4}$ and $n^{2k-2}$ gives remainder $1$ under division by $3$, then $3$ divides $n^{2k-2}-1$ and $n^{4k-4}+n^{2k-2}+1$, so $9$ divides $n^{12k-6}-n^6$.


Then, $72$ divides $n^{12k-6}-n^6$ for any integer $n$.
