Show that $2^{15}-2^3$ divides $a^{15}-a^3$ for all $a$ Show that for all $a$, $2^{15}-2^3$ divides $a^{15}-a^3$.
I was able to prove that this is true for all $a$, such that $\gcd(a,2^{15}-2^3)=1$, by using Euler's theorem, where I concluded that $a^{12}\equiv1 \pmod{2^{15}-2^3}$, since $a^{12}\equiv1$, mod all of $2^{15}-2^3$ divisors.
For $a$, such that $\gcd(a, 2^{15}-2^3)\neq1$, I wasn't able to solve this. I thought of going over all of the cases of $\gcd(a,2^{15}-2^3)$, but this seems tedious.

Similar to Ian's answer san the brute force: since 
$$2^{15}-2^3=2^3\times 3^2 \times 5 \times 7 \times 13$$
we only need to show $a^{15}-a^3=a^3(a^{12}-1)$ is divisible by each of the  above factors.
By Fermat's theorem


*

*$a(a^{12}-1)$ is divisible by $13$.

*$a(a^6-1)$ is divisible by $7$.

*$a(a^4-1)$ is divisible by $5$.


On the other hand, it is well-known that $a^2-1$ is divisible by $8$ if $a$ is odd. So $a^3(a^2-1)$ is divisible by $8$.
Lastly, to show divisibility by $3^2$, note that if $a$ is divisible by $3$, then we are done. If $a$ is not divisible by $3$, then $a^4=3k+1$, so $a^{12}-1=(3k+1)^3-1$ is divisible by $3^2$.
