$a^{13} \equiv a \bmod N$ - proof of maximum $N$ From Fermat's Little Theorem, we know that $a^{13} \equiv a \bmod 13$. Of course $a^{13} \equiv a \bmod p$ is also true for prime $p$ whenever $\phi(p) \mid 12$ - for example, $a^{13} = a^7\cdot a^6 \equiv a\cdot a^6 = a^7 \equiv a \bmod 7$.
So far I have that the largest $N$ for which all $ a^{13} \equiv a \bmod N$, is $N = 2\cdot 3\cdot 5\cdot 7\cdot 13 = 2730$
Can someone either put together an elegant proof of this, or find and prove a different limit?

Suppose that $a^{13}\equiv a\pmod N$, for every $a\in\mathbb Z$, then for every prime $p$ deviding $N$ we have $a^{13}\equiv a\pmod p$, but we have a primitive root $g$ modulo $p$, hence
$$
g^{12}\equiv1\pmod p
$$
which means that $p-1|12$, so $p\in\{2,3,5,7,13\}$. Hence the prime factors of $N$ are from the above set. Now I prove that the number $N$ is squarefree...
Suppose not, then for some prime $p|N$ we have $a^{13}\equiv a\pmod {p^2}$ by choosing $a=p$ we get a contradiction, hence the number $N$ must be squarefree and by above arguments the number $2730$ is the largest possible value for $N$.
