How to prove :If $p$ is prime greater than $3$ and $\gcd(a,24\cdot p)=1$ then $a^{p-1} \equiv 1 \pmod {24\cdot p}$?

Hint: The claim follows from proving the following facts separately: $a^{p-1}\equiv 1\pmod 8$, $a^{p-1}\equiv 1\pmod 3$, and $a^{p-1}\equiv 1\pmod p$.

Since $(a,24\cdot p)=1$, it also follows that $(a,p)=(a,3)=(a,8)=1$.

By the generalization of Fermat's little theorem, $a^{p-1}\equiv 1\pmod{p}$, $a^2\equiv 1\pmod{3}$, and $a^4\equiv 1\pmod{8}$. But $a^4\equiv 1\pmod{8}$, implies $a\equiv 1,3,5,7\pmod{8}$, and in all cases $a^2\equiv 1\pmod{8}$.

Since $p-1$ is even, $a^{p-1}\equiv 1$ in all cases, since $a^{p-1}\equiv (a^2)^{(p-1)/2}\equiv 1^{(p-1)/2}\pmod{3,8}$.

Then $a^{p-1}-1$ is divisible by $3$, $8$, and $p$, and thus as a common multiple, is divisible by $\mathrm{lcm}(3,8,p)=24\cdot p$, so $a^{p-1}\equiv 1\pmod{24\cdot p}$.

HINT $\ $ By Carmichael's simple generalization of Euler $\phi$, since prime $\rm\:p\:$ is coprime to $2,3$

$\rm\ \ \lambda(8\cdot3\cdot p) = lcm(\lambda(8),\lambda(3),\lambda(p)) = lcm(\phi(8)/2,\phi(3),\phi(p)) = lcm(2,2,p-1) = p-1\ $

therefore $\rm\quad gcd(n,24\:p) = 1\ \ \Rightarrow\ \ 1\ \equiv\ n^{\:\lambda(24\:p)}\: \equiv \ n^{p-1}$