For a prime $p$, $6p\mid a^p+1$ for no $a$ or infinitely many $a$ BdMO Nationals Secondary:

Show that for any prime $p$, there are either infinitely many or no positive integer $a$, so that $6p$ divides $a^p+1$
  . Find all those primes for which there exists no solution.

At first,we consider $p=2$. Then certainly, $6\cdot2\nmid a^2+1$ for all $a$ since looking at the equation mod $3$ and $4$(we pick $3$ and $4$ because $3\cdot4=6\cdot2$ and $\gcd(3,4)=1$) shows that there is no solution mod $3$.Any prime number greater than $2$ must be odd.Then,we know that for all odd $p$,$$a^p+1=(a+1)(a^{p-1}-a^{p-2}......+1)$$
We can pick $a+1$ to be any multiple of $6p$ and that clearly gives us an infinite number of solutions.
Q.E.D
Am I right?Are there are other ways(not necessarily shorter or easier)?

For the sake of eliminating an unanswered question, I repeat: Your proof is nice.
