Prove that $a^5 ≡ a$ (mod 15) for every integer $a$ Prove that $a^5 ≡ a$ (mod 15) for every integer $a$


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*We are currently studying modular arithmetic and congruence and I came across this proof on my study set that I'm really not sure how to approach. We have gone over equivalence relations in mod arithmetic, equivalence classes (complete set of residues), and have gone through several mod arithmetic examples w/o variables. Looking through my notes and book I'm still not sure how to proceed with this problem, any help is appreciated. 

 A: If $a\in{\mathbb Z}$ then
$$c:==a^5-a=a(a-1)(a+1)(a^2+1)$$
is obviously divisible by $3$, and is obviously divisible by $5$ if $a\in\{-1,0,1\}\ {\rm mod}\ 5$. If $a=\pm2\ {\rm mod}\ 5$ then $a^2+1=0\ {\rm mod}\ 5$. It follows that $c=0\ {\rm mod}\ 15$ for all $a\in{\mathbb Z}$.
A: Observe that an integer $n$ satisfies $n \equiv 0 \pmod{15}$ if and only if $n \equiv 0 \pmod 3$ and $n \equiv 0 \pmod 5$. Therefore, it suffices to prove that 
$a^5 \equiv a \pmod 5$ and $a^5 \equiv a \pmod 3$. The first formula follows from  Fermat's little theorem.
For the second formula, it con for instance consider separately the three cases: 


*

*if $a \equiv 0 \pmod 3$, then $a^5 \equiv 0 \pmod 3$, 

*if $a \equiv 1 \pmod 3$, then $a^5 \equiv 1 \pmod 3$

*if $a \equiv -1 \pmod 3$, then $a^5 \equiv -1 \pmod 3$.

A: Fermat little theorem:
$a^5 \equiv a \mod 5$ so $a^5 -a \equiv 0 \mod 5$.
If $3\not |a$, $a^2 \equiv 1 \mod 3$ so $a^5 \equiv a \mod 3$ so $a^5 -a \equiv 0 \mod 3$
So, $a^5 -a \equiv 0 \mod 15$.
Or... we could factor $a^5 - a = (a-1)a (a+1)(a^2+1) $.  $3|(a-1)a (a+1)$ and $5|a^2 +1 \iff 5|a^2+5a+6=(a+2)(a+3) $ and $5|(a-1)a (a+1)(a+2)(a+3) $ 
