Prove that $a^5 ≡ a$ (mod 15) for every integer $a$
- 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.
Best Answer
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 $a\in{\mathbb Z}$.