6
$\begingroup$

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 $168$?

  • 1
    That's [Carmichael's theorem](http://en.wikipedia.org/wiki/Carmichael_function#Carmichael.27s_theorem): $a^{\lambda(168)}\equiv1\pmod{168}$ and $\lambda(168)=6$.2012-03-07

2 Answers 2

8

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.

  • 1
    I can delete, if a more student-friendly solution appears.2012-03-07
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} \:N = (2,2,6)\\ \rm M = (8,3,7)\end{cases}\ \ \ \ $ This is what Andre does. It's worth emphasis it is of above general form.