Given $A,B\in {\Bbb Z}^2$, write $A \leftrightarrows B$ if the interior of the line segment AB misses ${\Bbb Z}^2$.

For $r>0$, define $S_r:=\{ \{A, B\} | A,B\in {\Bbb Z}^2,||A||<r,||B||<r, | ||A||-||B|| |<1 {\rm\ and\ } A \leftrightarrows B \}\ .$

A little calculus gives the equivalence of $\zeta(2)=\pi^2/6$ and $$\lim_{r\to\infty} \frac{|S_r|}{(2r)^3} = 1\ .$$

Of course $(2r)^3$ counts lattice points in a cube $C_{r}:=[-r,r)^3$.

Question: Does there exist an approximately bijective proof of this limit (or some variant), one that matches most of $S_r$ with most of $C_r$?

Given $A,B\in {\Bbb Z}^2$, write $A \leftrightarrows B$ if the interior of the line segment $AB$ misses ${\Bbb Z}^2$.

For $r>0$, define $S_r:=\{ \{A, B\} \mid A,B\in {\Bbb Z}^2,\|A\|<r,\|B\|<r, \left| \|A\|-\|B\| \right|<1 \text{ and } A \leftrightarrows B \}\ .$

A little calculus gives the equivalence of $\zeta(2)=\pi^2/6$ and $$\lim_{r\to\infty} \frac{|S_r|}{(2r)^3} = 1\ .$$

Of course $(2r)^3$ counts lattice points in a cube $C_r:=[-r,r)^3$.

Question: Does there exist an approximately bijective proof of this limit (or some variant), one that matches most of $S_r$ with most of $C_r$?