I don't know if I can only give a hint: Lemma 1:${p^{\alpha}dq-1 \choose p^{\alpha}d-1} \equiv {p^{\alpha-1}dq-1 \choose p^{\alpha-1}d-1}(\mod p^{\alpha+1}),\alpha>=1 $ Lemma 2:$f(n,q)=q\sum_{d|n}{{dq-1\choose d}\phi(\frac{n}{d})}$ Lemma 3:$pf(\frac{n}{p},q)\equiv f(n,q)$(We prove the Lemma 3with Lemma2 and Lemma1) then by Lemma 3,we can proved by induction $nq|f(n,q)$

Thank you for your advice,I need to be patient and careful in CMO My friend,if you still have any patience,please read the following solution Lemma$ X_{\alpha}\equiv X_{\alpha + 1},(modp^{\alpha + 1})$, where$ p$ is a prime $ X_{\alpha} = \binom{dp^{\alpha}q - 1}{dp^{\alpha} - 1}$ OK.set $ f(x) = (x - 1)(x - 1)...(x - p + 1)$ where $ p$ is a prime. use definition of $ \binom{m}{n}$ to expand $ X_{\alpha},X_{\alpha + 1}$ We get $ X_{\alpha + 1} - X_{\alpha} = X_{\alpha}(\frac {f(dp^{\alpha + 1}q)f(dp^{\alpha + 1}q - p)...f(dp^{\alpha + 1}q - dp^{\alpha + 1} + p)}{f(dp^{\alpha + 1})f(dp^{\alpha + 1} - p)...f(p)} - 1)$ Note that $ f(kp)$ is not a multiple of $ p$ We're enough to show $ f(dp^{\alpha + 1}q)f(dp^{\alpha + 1}q - p)...f(dp^{\alpha + 1}q - dp^{\alpha + 1} + p)$ $ \equiv f(dp^{\alpha + 1})f(dp^{\alpha + 1} - p)...f(p),modp^{\alpha + 1}$ It is obvious because $ f(dp^{\alpha + 1}q - kp)\equiv f(dp^{\alpha + 1} - kp),modp^{\alpha + 1}$ for any integer $ k$ which completes the proof for our lemma Let's come back to our initial problem: It is equivalent to $ \leftrightarrow n|\sum_{d|n}\binom{dq - 1}{d - 1}\phi(\frac {n}{d})$ If we denote $ n$ by $ n = p^{\alpha}m$ where $ (m,p) = 1$ we are sufficient to show $ \leftrightarrow p^{\alpha}|\sum_{d|m}\sum_{k = 0}^{\alpha}\binom{dp^{k}q - 1}{dp^{k} - 1}\phi(\frac {p^{\alpha - k}m}{d})$ Moreover we are sufficient to show $ p^{\alpha}|A_{\alpha}$ where$ A_{\alpha} = \sum_{k = 0}^{\alpha}\binom{dp^{k}q - 1}{dp^{k} - 1}\phi(\frac {p^{\alpha - k}m}{d})$ Let's induction on $ \alpha$ Note that $ A_{\alpha + 1} = \sum_{k = 0}^{\alpha + 1}\binom{dp^{k}q - 1}{dp^{k} - 1}\phi(\frac {p^{\alpha + 1 - k}m}{d})$ and $ \phi(\frac {mp}{d}) = (p - 1)\phi(\frac {m}{d})$,$ \phi(\frac {mp^a}{d}) = p\phi(\frac {m^{a - 1}}{d})$for $ a > 1$ We get $ A_{\alpha + 1} = pA_{\alpha} + \binom{dqp^{\alpha + 1} - 1}{dp^{\alpha + 1} - 1}\phi(\frac {m}{d}) - \binom{dqp^{\alpha} - 1}{dp^{\alpha} - 1}\phi(\frac {m}{d})$ Thus $ A_{\alpha + 1}\equiv \phi(\frac {m}{d}) (\binom{dqp^{\alpha + 1} - 1}{dp^{\alpha + 1} - 1} - \binom{dqp^{\alpha} - 1}{dp^{\alpha} - 1}) \equiv 0,(mod p^{\alpha + 1})$ by the lemma which completes our induction QED Is it OK now? and dear moderator Peter,could you please delete all of my previous posts(including the attachments)Thank you in advance!!!

hxy09 wrote: and dear moderator Peter,could you please delete all of my previous posts(including the attachments)Thank you in advance!!! Done. I hope this problem has finally found its solution now.

There is a, in my opinion, much nicer solution using Burnside's Lemma. Consider the set $ X $ of regular $ nq $-gons where exactly $ n $ of the vertices are colored red, and the other $ n(q - 1) $ of the vertices are colored blue. Let $ G $ be the set of rotations of a regular $ nq $-gon by $ \frac{2k\pi}{nq} $ radians for $ k \in \{0, 1, 2, \dots, nq - 1\}. $ Consider the group action of $ G $ on $ X. $ Because no element of $ X $ is fixed by a rotation of $ \frac{2k\pi}{nq} $ radians if $ q \nmid k $ we have by Burnside's Lemma that the number of orbits of this group action is given by: \[ \frac{1}{nq}\sum_{k = 0}^{n - 1}\binom{\text{gcd}(n, k)q}{\text{gcd}(n, k)} \] Which must be an integer so we are done.