I showed that $a^{15}-a^3=(a^5-a)$ so by Fermat's, $5|a^5-a$ so $5|a^{15}-a^3$. I also showed that $a^{15}-a^3=a^3(a^{12}-1)$ so by Euler, $a^{12}-1 \equiv 0$ mod 13.
I am having a hard time showing that 7 divides $a^{15}-a^3=(a^5-a)$. They talk about it some in this post but I am still not seeing it. Thank you.
The $5$ divisibility won't be relevant here. Just note that $a^{15}-a^3=a^{3}(a^{12}-1)$ as you did and if $7|a$ you are done, otherwise $a^6\equiv 1\mod 7$ by Fermat and so $a^{12}=(a^6)^2\equiv 1\mod 7$ so $a^{12}-1\equiv 0\mod 7$ as desired.
You can prove it uniformly for all three primes as in the following Theorem, which implies that $\,5\cdot 7\cdot 13\mid a^{13}-a\,$ so also $\,a^2(a^{13}-a) = a^{15}-a^3$
Theorem $\ $ For natural numbers $\rm\:a,e,n\:$ with $\rm\:e,n>1$
$\qquad\rm n\:|\:a^{\large e}-a\:$ for all $\rm\:a\:\iff n\:$ is squarefree, and prime $\rm\:p\:|\:n\:\Rightarrow\: p\!-\!1\:|\:e\!-\!1$
Proof $\ (\Leftarrow)\ \ $ Since a squarefree natural divides another iff all its
prime factors do, we need only show $\rm\:p\:|\:a^{\large e}\!-\!a\:$ for each prime $\rm\:p\:|\:n,\:$ or,
that $\rm\:a \not\equiv 0\:\Rightarrow\: a^{\large e-1} \equiv 1\pmod p,\:$ which, since $\rm\:p\!-\!1\:|\:e\!-\!1,\:$ follows
from $\rm\:a \not\equiv 0\:$ $\Rightarrow$ $\rm\: a^{\large p-1} \equiv 1 \pmod p,\:$ by little Fermat.
$(\Rightarrow)\ \ $ See this answer (not required here).
$a^{15}-a^3 = a^3(a^{12}-1) = a^3(a^6-1)(a^6+1)= (a^7-a)(a^9+a^3)$
$7|a^7-a$ by LFT
If $a\equiv 0\mod7$, it's obvious. If not, Lil' Fermat ensures $\;a^{15}\equiv a^3\mod7$.
In $mod \space 7$ $$\large{a^{\phi(7)}}\equiv 1 \\a^6 \equiv 1 \to \\{(a^6)}^2\equiv 1^2 $$now,multiply by $a^3$ $$a^{12} \equiv 1 \\\times a^3\\a^{12+3}\equiv1.a^3$$
