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: $$ 2^{15} - 2^3 = 5\cdot7\cdot8\cdot9\cdot13$$

Apply  Carmichael function,
to find $\lambda(8)=2,\lambda(5)=\phi(5)=4, \lambda(9)=\phi(9)=6$ etc.
Then, for $9, F=a^{15}-a^3=a^3(a^{12}-1)=a^3(a^6-1)(a^6+1)$
As $3$ is prime, ether $(i)\ 3|a\implies 3^3|F$ or $(ii)\ (3,a)=1\implies 9|(a^6-1)$
For $8, F=a^{15}-a^3=a^3(a^{12}-1)$
As $2$ is prime, ether $(i)\ 2|a\implies 2^3|F$ or $(ii)\ (2,a)=1\implies 8|(a^2-1)$
For $5, F=a^{15}-a^3=a^3(a^{12}-1)=a^3\{(a^4)^3-1\}=a^3(a^4-1)(a^8+a^4+1)$
As $5$ is prime, ether $(i)\ 5|a\implies 5|F$ or $(ii)\ (5,a)=1\implies 5|(a^4-1)$
Similarly for $7,13$
