Sean $p, q$ enteros y $p, q > 1$ , $gcd(p, \,6q)=1$ . Demuestra que: $$\sum_{k=1}^{q-1}\left \lfloor \frac{pk}{q}\right\rfloor^2 \equiv 2p \sum_{k=1}^{q-1}k\left\lfloor \frac{pk}{q} \right\rfloor (mod \, q-1)$$