For my homework, I need to prove that prove that $n^{7}−n$ is divisible by 42 for any integer n.
Here is the part of my proof where I use Fermat's:
$$ \\\textbf{Claim 1: }7\vert n^{7} - n\\ n^{7} - n = n(n^{6} - 1)\\ n^{6}\equiv_71\text{, by Fermat's Little Theorem}\\ n(1-1) \textbf{mod}7 = 0\\ \implies 7\vert n^{7} - n\\ $$
I am assuming that if you have $a\cdot b\cdot c (\textbf{mod} d)$ then you can replace a (WLOG) with (a $\textbf{mod}$ d).
My logic is that you are finding the remainder when you divide $a\cdot b\cdot c $ by d, and you can split that up into $\frac{a}{d}\cdot \frac{b}{d}\cdot \frac{c}{d} $
Is this correct?
I suggest working with
$n^7 - n = n(n^3 - 1)(n^3 + 1) = n(n^3 - 1)(n^2 - n + 1)(n + 1).$