Number Theory: Prove that $p|m-n$ 
Assume $p$ is an odd prime number, $\displaystyle q = \frac{3p-5}{2}$ and $$ S_q = \frac{1}{2.3.4} + \frac {1}{5.6.7} + ... + \frac{1}{q(q+1)(q+2)}$$.
  If $\displaystyle \frac{1}{p} - 2S_q = \frac{m}{n}$ and $m$ and $n$ are two integers, Prove that $p\  | \ m-n$.

Any hints how to start the proof?

You could start with the identity
$$
\frac{1}{(3n-1)(3n)(3n+1)}=\frac{1}{2}\left(\frac{1}{3n-1}+\frac{1}{3n}+\frac{1}{3n+1}\right)-\frac{1}{2}\frac{1}{n}.
$$
Using this:
$$
\begin{align*}
S_q &= \sum_{n=1}^{(p-1)/2}\frac{1}{2}\bigg[\left(\frac{1}{3n-1}+\frac{1}{3n}+\frac{1}{3n+1}\right)-\frac{1}{n}\bigg]\\
&=\frac{1}{2}\left(\sum_{k=2}^{(3p-1)/2}\frac{1}{k}-\sum_{n=1}^{(p-1)/2}\frac{1}{n}\right)=\frac{1}{2}\left(-1+\sum_{k=(p-1)/2}^{(3p-1)/2}\frac{1}{k}\right).
\end{align*}
$$
Can you take it from there? You might find Wolstenholme's Theorem useful.
