prove that $2^{15} - 2^3 $ divides $ a^{15} - a^3$ Prove that $$2^{15} - 2^3 $$ divides $$ a^{15} - a^3$$ for any integer $a$.
Hint: $$ 2^{15} - 2^3 = 5\cdot7\cdot8\cdot9\cdot13$$
Hint: If $13\mid a$, then $13$ divides $a(a^{14}-a^2)=a^{15}-a^3$.  Otherwise, by Euler's theorem, $13$ divides $a^{12}-a^0$, hence also $a^3(a^{12}-a^0)=a^{15}-a^3$.
This is meant mostly as a joke, but partly not:

We clearly have $2^{15}-2^3=32{,}760$ divides $a^{15}-a^3$ when $a=0$, $\pm1$, and $\pm2$, so that leaves only $32{,}755$ other cases to check....

(The part that's not meant as a joke is the implicit question, where do you draw the line between proving something by brute force computation versus spending time looking for a clever theory.)