TITLE: Show that $a^6-1$ is divisible by $168$ whenever $(a,42)=1$.
QUESTION [6 upvotes]: I have been running into this type of problem a lot:

Show that $a^6-1$ is divisible by $168$ whenever $(a,42)=1$.

First of all, by Euler's theorem, we have that
$$a^{\phi(42)}\equiv a^{12}\equiv1\pmod{42}.$$
Notice that
$$a^6a^6\equiv1\pmod{42}\text{ and }168=4\cdot42.$$
It follows that
$$a^{12}\equiv1\pmod{168},$$
$$a^{12}-a^6\equiv1-a^6\pmod{168},$$
$$a^6(a^6-1)\equiv1-a^6\pmod{168},$$
$$a^6-1\equiv a^6(1-a^6)\pmod{168},$$
$$a^6-1\equiv a^6-a^{12}\pmod{168}.$$
I get stuck here: How can I show that the RHS is congruent to zero modulo $168$?

REPLY [2 votes]: Hint $\ $ Either apply Carmichael's generalization of Euler-Fermat, or proceed directly via
$$\rm A^{N_j}\equiv 1\ \ (mod\ M_j)\ \Rightarrow\ A^{lcm\ N_j}\equiv 1\ \ (mod\ lcm\ M_j)$$
for $\rm \begin{cases}\rm \:N = (2,2,6)\\ \rm M = (8,3,7)\end{cases}\, $ by CCRT. That's what Andre does. It's worth emphasis it's general form.