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

Question

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 $168$?

Answer 1

By Fermat's Theorem, $a^6\equiv 1 \pmod 7$. Also by Fermat's Theorem, or otherwise, $a^2\equiv 1 \pmod 3$. Thus $a^6\equiv 1 \pmod 3$. So far, we have that $$a^6\equiv 1 \pmod {21}.$$ But $a$ is odd, so $a^2\equiv 1 \pmod 8$. It follows that $$a^6 \equiv 1 \pmod {8}.$$ Now it's over.

Answer 2

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.