# 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 $$\sum_{k=1}^{m} \left \lfloor{kt}\right \rfloor +\sum_{k=1}^{n} \left \lfloor{\frac{k}{t}}\right \rfloor= mn$$

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I see 3 votes to close, and zero supporting reasons. Anyone care to try to convince me that this question should be closed? – Gerry Myerson 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. – Brian M. Scott 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. – Matemáticos Chibchas 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). – Gerry Myerson Oct 27 '13 at 6:18
@dfeuer: It’s its own context: it’s inherently interesting. – Brian M. Scott 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. ${}$ – Brian M. Scott Oct 28 '13 at 4:06