Show that for any a ∈ ℤ, 42 | $(a^7 − a).$ Show that for any $a \in \mathbb{Z}, 42 \mid (a^{7} − a)$.
I saw this question on Rosen textbook and it doesn't have answer key so I am wondering can you guide me how to do it? 
What I have tried is that since I don't know the number a so I substitute a number for a.
 A: $$a^7-a=a(a-1)(a+1)(a^2-a+1)(a^2+a+1).$$ 
Since it has a product of three consecutive numbers as its factor, it is divisible by $3!$.
By Fermat's little theorem, it is divisible by $7$.
Hence by $6\times7=42.$
A: (1) If $a$ is multiple of $7$ then $a^7-a$ is a multiple of $7$; if $\gcd(a, 7)=1$ then by FLT $a^6\equiv1\pmod{7}$ hence $a^7-a=a(a^6-1)$ is a multiple of 7.
(2) If $a$ is multiple of $3$ then $a^7-a$ is a multiple of $3$; if $\gcd(a, 3)=1$ then by FLT $a^2\equiv1\pmod{3}$ hence $a^7-a=a(a^2-1)(a^4+a^2+1)$ is a multiple of 3.
(3) If $a$ is multiple of $2$ then $a^7-a$ is a multiple of $2$; if $\gcd(a, 2)=1$ then by FLT $a\equiv1\pmod{2}$ hence $a^7-a=a(a-1)(a+1)(a^4+a^2+1)$ is a multiple of 2.
A: Hint : $a = b \mod p$ implies that $  a^n-a = b^n-b \mod p$
Use it with $p=2,3,7$ (as $42=2*3*7$) and all the possible $b$ (ie $0,1,..,p-1$)
A: $42 = 2\cdot3\cdot7$, so we can check individually if 
$2|a^7-a$ and  $3|a^7-a$ and $7|a^7-a$ 
simultaneously hold. For the first one, we consider two cases: either $a$ is odd or is even. This means that either $a\equiv0\pmod 2$ or $a\equiv1\pmod2$. 
Thus, we either have $1^7-1\equiv 0\pmod2$ or $0^7-0\equiv0\pmod2$. In both case, we can conclude that $2|a^7-a$. 
Next we deal with 3 cases: either $a\equiv0\pmod3$ or $a\equiv1\pmod3$ or $a\equiv2\pmod3$. Plug the $a$ in any of the three cases and you should get that $a^7-a\equiv0\pmod3$
Finally you deal with 7 cases the same way around.
