Finding $\sum_{k=1}^{p-1}\left\lfloor\frac {k^3}{p} \right\rfloor $

Question

Let, $p$ be an odd prime. Find $$\sum_{k=1}^{p-1}\left\lfloor\frac {k^3}{p} \right\rfloor $$

Any hints will be helpful. Thanks.

Answer 1

Hint: Show that (for $ 1\leq k \leq p-1$, $p$ prime), $$ \lfloor \frac{ k^3 } {p} \rfloor + \lfloor \frac{ (p-k)^3 } { p} \rfloor = p^2 - 3kp + 3k^2 -1 = \frac{k^3}{p} + \frac{ (p-k)^3} {p} - 1$$

Corollary: The problem follows by taking the summation and dividing by 2.

The answer is $\frac{ (p-1)^2p}{4} - \frac{(p-1)}{2} = \frac{1}{4} (p-2)(p-1)(p+1)$