# Hard floor function problem

Let $\left \lfloor{x}\right \rfloor$ denote the floor of $x$. Supose $m\in \mathbb{N}$, and that $t$ is a positive irrational number. Put $n=\left \lfloor{mt}\right \rfloor$. Prove that $$\sum_{k=1}^{m} \left \lfloor{kt}\right \rfloor +\sum_{k=1}^{n} \left \lfloor{\frac{k}{t}}\right \rfloor= mn$$

• I see 3 votes to close, and zero supporting reasons. Anyone care to try to convince me that this question should be closed? Oct 27 '13 at 6:10
• It looks like a fairly non-trivial problem; I’d certainly be interested in seeing a solution, if I don’t find one myself. Oct 27 '13 at 6:15
• @dfeuer I think that "lack of context" is not enough reason to close this specific question, which is interesting per se. Oct 27 '13 at 6:18
• OK, James, you've been warned: your problem lacks context. That means people want to know where it comes from, and why you are interested in it (and they'll close it if you don't tell them). Oct 27 '13 at 6:18
• @dfeuer: It’s its own context: it’s inherently interesting. Oct 27 '13 at 6:23

The rectangle $R=[0,m] \times [0,n]$ has $mn$ lattice points in it with positive coordinates. The line $L: y=tx$ does not pass through any of these lattice points since $t$ is irrational. The first sum counts the lattice points below $L$, and the second counts lattice points above $L$, since a point above $L$ is also to the left of $L$ and $L$ may also be written as $x=y/t$
Note: one detail is that from $n=\left \lfloor{mt}\right \rfloor$ we have the highest lattice point of the form $(m,k)$ is the upper right corner $(m,n)$ --actually the important feature is that the largest thing counted in the first sum is this upper right point in the rectangular lattice $R.$ Then the line $L$ cuts through the top edge of $R$ in such a way that nothing counted in the two sums happens to lie outside the rectangle.
• Very nice. ${}$ Oct 28 '13 at 4:06