Geometric interpretation of Hermite's identity

Question

I came across an elegant interpretation of the well known reciprocity law for the floor function in the USSR problem book and I was wondering if there was a similar sort of interpretation for Hermite's identity;

Reciprocity law interpretation:

(I'm sorry if I'm not supposed to attach images for this kind of thing, but I honestly do not know how to explain it better than the book itself.)

I've seen both an elementary proof for Hermite's identity and one using the periodicity of the function $f(x)=[x]+\left[x+\dfrac{1}{n}\right]...+\left[x+\dfrac{n-1}{n}\right]-[nx]$. Does anyone have a proof that is more geometric in flavor, similar to the one above, for Hermite's identity?

Answer 1

First, let's call our variable $z$ instead of $x$ so we don't get the coordinate confused with the variable. Then we're trying to prove: $$\lfloor z\rfloor + \left\lfloor z + {1\over n}\right\rfloor + ... + \left\lfloor z + {(n-1)\over n}\right\rfloor = \lfloor nz\rfloor$$

Without loss of generality, assume $0 \le z < 1$.

Define a triangle $ABC$ with $A = (0,z)$, $B = (n,z)$, and $C = (n,z+1)$. (See figure.) We shall show that the number of integer-coordinate points inside the triangle is equal to both sides of what we're trying to prove, computed different ways. Note that the $y$-coordinate is 1 for every integer-coordinate point inside the triangle.

The formula for the line $AC$ is $f(x) = z + {x\over n}$, so for an integer $x$ with $0\le x < n$, the number of integer-coordinate points with that $x$ value inside the triangle is $\lfloor z + {x\over n}\rfloor$. Summing over integer-valued $x$ with $0\le x < n$, we get the left-hand side of our identity.

Let $x_0$ be the $x$-coordinate of the point on $AC$ with $y=1$. Then $$1 =f(x_0) = z + {x_0\over n}$$ so $$x_0 = n(1-z)$$ Counting to the right of $x_0$ along the line $y=1$, we see that the number of integer-coordinate points inside the triangle is $$\lfloor n-x_0\rfloor = \lfloor n - n(1-z)\rfloor$$ $$= \lfloor nz\rfloor$$

Answer 2

For the case $z>1$, the following diagram applies;

Counting the integer points vertically inside the trapezium along $x=1,2,3...(n-1)$ yields the sum required on the LHS. Whereas, counting inside the rectangular region horizontally along $y=1,2,3...[z]$ yields $n[z]$ integer points. In the triangular region, there is only one horizontal line $y=[z+1]$ which corresponds to the point on the slanting line $x_0=n[z+1]-z$. Hence, the number of integer points in the triangular region are $[n-x_0]=[n+z-n[z+1]]$.

Hence, the RHS representing the number of integer points in the trapezoid counted horizontally becomes $n[z]+[n+z-n[z+1]]$. How does this simplify into $[nz]$? In fact, when I tried to simplify it, it became $[z]$ alone?