We have the following exercise: "Let $q\in\mathbb C$ and $f:\mathbb N \to \mathbb C$ an arithmetic function such that $$(\star )\space q^n =\sum_{d\mid n}df(d)$$(i) Find the value $f(24)$ and (ii) Prove that, if $q\in \mathbb Z$ then, $f(24)\in\mathbb Z$"
For (i): We apply Mobius's Inverse Formula on $(\star)$ and we have:$$nf(n)=\sum_{d\mid n}q^dμ \left ( \frac{n}{d} \right )$$ where μ is the Mobius function. So,$$24f(24)=\sum_{d\mid n}q^dμ\left ( \frac{24}{d} \right )=q^4-q^8-q^{12}+q^{24} \implies f(24)=\frac{q^4-q^8-q^{12}+q^{24}}{24} $$ For (ii): I made some attempts but only with intuition.
Any help\advise\hint would be very helpful. Thank you in advance.
If $q$ is odd then $q^2\equiv 1 \mod 8$. Then $$q^4-q^8-q^{12}+q^{24}\equiv 1-1-1+1\equiv0 \mod 8$$
If $q$ is even then $q^4\equiv 0 \mod 8$. Then, again, $q^4-q^8-q^{12}+q^{24}\equiv 0 \mod 8$.
Either $q^2\equiv 1 \mod 3$ or $q\equiv 0 \mod 3$ and so $q^4-q^8-q^{12}+q^{24}\equiv 0 \mod 3$.
Therefore $q^4-q^8-q^{12}+q^{24}\equiv 0 \mod 24$, as you require.