Peter wrote: Show that for all primes $p$, \[\sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor =\frac{(p+1)(p-1)(p-2)}{4}.\] This is also a problem from the German National Olympiad (DeMO, 4th round) 2002, and I have written a solution for the op02 project. Since op02 does not seem to have survived, I am publishing the solution here: We start by showing something really trivial: Lemma 1. For any integer $n$ and any non-integer real number $a$, we have $\lfloor a \rfloor+\lfloor n-a \rfloor = n-1$. Proof. Since $a$ is not an integer, we have $a=\lfloor a \rfloor+q$ for some real $q$ with $0<q<1$. Hence, $n-a = n-\left(\lfloor a \rfloor+q\right) = \left(n-1-\lfloor a \rfloor\right)+\left(1-q\right)$. Hereby, $0<1-q<1$; thus, $\lfloor n-a \rfloor = n-1-\lfloor a \rfloor$, what yields Lemma 1. For every $k$ such that $1\leq k\leq p-1$, define the number $a = \frac{k^{3}}{p}$. This number $a$ is trivially non-integer, since $k^{3}$ is not divisible by $p$ because $p$ is a prime and $k<p$. Also, set $n = p^{2}-3pk+3k^{2}$. Then, Lemma 1 yields $\left\lfloor \frac{k^{3}}{p}\right\rfloor+\left\lfloor \left(p^{2}-3pk+3k^{2}\right)-\frac{k^{3}}{p}\right\rfloor = \left(p^{2}-3pk+3k^{2}\right)-1$. This simplifies to $\left\lfloor \frac{k^{3}}{p}\right\rfloor+\left\lfloor \frac{\left(p-k\right)^{3}}{p}\right\rfloor = p^{2}-3pk+3k^{2}-1$. Now, the problem is straightforward: $2\sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor = \sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor+\sum^{p-1}_{k=1}\left \lfloor \frac{\left(p-k\right)^{3}}{p}\right \rfloor$ $= \sum^{p-1}_{k=1}\left( \left\lfloor \frac{k^{3}}{p}\right\rfloor+\left\lfloor \frac{\left(p-k\right)^{3}}{p}\right\rfloor \right) = \sum^{p-1}_{k=1}\left(p^{2}-3pk+3k^{2}-1\right)$ $= \left(p-1\right)p^{2}-3p \sum^{p-1}_{k=1}k+3 \sum^{p-1}_{k=1}k^{2}-\left(p-1\right)$ $= \left(p-1\right)p^{2}-3p \frac{\left(p-1\right)p}{2}+3 \frac{\left(p-1\right)p\left(2p-1\right)}{6}-\left(p-1\right) = \frac{\left(p-2\right)\left(p-1\right)\left(p+1\right)}{2}$, so that $\sum^{p-1}_{k=1}\left \lfloor \frac{k^{3}}{p}\right \rfloor = \frac{\left(p-2\right)\left(p-1\right)\left(p+1\right)}{4}$, completing the solution. darij

Peter wrote: Show that for all primes $ p$, \[ \sum^{p-1}_{k=1}\left\lfloor\frac{k^{3}}{p}\right\rfloor =\frac{(p+1)(p-1)(p-2)}{4}.\] The same idea, but with different style. Lemma. Let $ \alpha,\beta\in\mathbb{R}$ with $ \alpha+\beta=n\in\mathbb{Z}$. Then, we have \[ \lfloor\alpha\rfloor+\lfloor\beta\rfloor =\begin{cases}\alpha+\beta &\left(\alpha\in\mathbb{Z}\right),\\ \alpha+\beta-1 &\left(\alpha\not\in\mathbb{Z}\right).\\ \end{cases}\] Proof. In case when $ \alpha\in\mathbb{Z}$, we also have $ \beta\in\mathbb{Z}$. Then, we obtain $ \lfloor\alpha\rfloor =\alpha$ and $ \lfloor\beta\rfloor =\beta$. Hence, $ \lfloor\alpha\rfloor+\lfloor\beta\rfloor =\alpha+\beta$. Now, we consider the case when $ \alpha\not\in\mathbb{Z}$. Since $ \alpha+\beta\in\mathbb{Z}$, we also have $ \beta\not\in\mathbb{Z}$. Since $ \alpha$ is not an integer, one may write $ a = k+q$, where $ k =\lfloor\alpha\rfloor\in\mathbb{Z}$ and $ q\in (0,1)$. Observer that $ \beta = n-\alpha = (n-k-1)+(1-q)$. Since $ n-k-1\in\mathbb{Z}$ and $ 1-q\in (0,1)$, this means that $ \lfloor\beta\rfloor = n-k-1$. It follows that \[ \lfloor\alpha\rfloor+\lfloor\beta\rfloor = k+(n-k-1) = n-1 =\alpha+\beta-1 .\] Lemma. Whenever $ k\in\{ 1,\cdots, p-1\}$, we obtain \[ \left\lfloor\frac{k^{3}}{p}\right\rfloor+\left\lfloor\frac{\left(p-k\right)^{3}}{p}\right\rfloor =\frac{k^{3}}{p}+\frac{\left(p-k\right)^{3}}{p}-1\] Proof. Let $ k\in\{ 1,\cdots, p-1\}$. Since $ p$ is prime, we see that $ k^{3}$ is not divisible by $ p$ so that $ \frac{k^{3}}{p}$ is not an integer. Furthermore, from the congruence $ k^{3}+(p-k)^{3}\equiv k^{3}-k^{3}\equiv 0\; (mod\; p)$, we see that \[ \frac{k^{3}}{p}+\frac{\left(p-k\right)^{3}}{p}\in\mathbb{Z}.\] The above proposition yields the desired result. Now, we compute the summation. By symmetry, we obtain \[ \sum^{p-1}_{k = 1}\frac{k^{3}}{p}=\sum^{p-1}_{k = 1}\frac{(p-k)^{3}}{p}.\] and \[ \sum^{p-1}_{k = 1}\left\lfloor\frac{k^{3}}{p}\right\rfloor =\sum^{p-1}_{k = 1}\left\lfloor\frac{(p-k)^{3}}{p}\right\rfloor.\] Applying the above results we proved, we compute \[ \sum^{p-1}_{k = 1}\left\lfloor\frac{k^{3}}{p}\right\rfloor\\ =\frac{1}{2}\left(\sum^{p-1}_{k = 1}\left\lfloor\frac{k^{3}}{p}\right\rfloor+\sum^{p-1}_{k = 1}\left\lfloor\frac{(p-k)^{3}}{p}\right\rfloor\right)\\ =\frac{1}{2}\sum^{p-1}_{k = 1}\left(\left\lfloor\frac{k^{3}}{p}\right\rfloor+\left\lfloor\frac{(p-k)^{3}}{p}\right\rfloor\right)\\ =\frac{1}{2}\sum^{p-1}_{k = 1}\left(\frac{k^{3}}{p}+\frac{\left(p-k\right)^{3}}{p}-1\right)\\ =-\frac{p-1}{2}+\frac{1}{2}\sum^{p-1}_{k = 1}\left(\frac{k^{3}}{p}+\frac{\left(p-k\right)^{3}}{p}\right)\\ =-\frac{p-1}{2}+\sum^{p-1}_{k = 1}\frac{k^{3}}{p}\\ =-\frac{p-1}{2}+\frac{1}{p}\left(\frac{(p-1)p}{2}\right)^{2}\\ =\frac{(p+1)(p-1)(p-2)}{4}.\\ \]