<s>[INST] How to prove that $42|a^7-a$? Suppose we are given a number $a \in \mathbb{Z}$ prove that $42|a^7-a$.
I'm not too sure how to start any ideas?
 [/INST] In fact, $42$ is the largest integer that divides $a^7-a$ for all $a\in\Bbb Z$.
$a^7-a=a\left(a^6-1\right)=a\left(a^3-1\right)\left(a^3+1\right)$
$=a(a-1)\left(a^2+a+1\right)(a+1)\left(a^2-a+1\right)$
$a-1,a,a+1$ are three consecutive integers, so $2,3\mid a^7-a$.
Also by Fermat's Little theorem $7\mid a^7-a$. Therefore $2\cdot 3\cdot 7=42\mid a^7-a$.
$2^7-2=42\cdot 3$ and $3^7-3=42\cdot 52$. But $\gcd(3,52)=1$, 
so it follows that $42$ is the largest integer that divides $a^7-a$ for all $a\in\Bbb Z$.
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