Bijective proof that $8+1=9$, or really $3^2-1=2^3$ Catalan's conjecture states that $8$ and $9$ are the only consecutive powers. This suggests to me that the identity $3^2-1=2^3$ might be purely "accidental". So here's the challenge: Is there any natural bijection between a set of $3^2-1$ things and a set of $2^3$ things?
As examples of what I'm looking for, here are some structures that don't seem to work:


*

*The field $\mathbb F_9$ has $3^2-1$ nonzero elements. They form a group isomorphic to $\mathbb Z/8\mathbb Z$. If this group were isomorphic to $(\mathbb Z/2\mathbb Z)^3$ instead, it would be a good answer.

*Gluons, which carry the force between $SU(3)$-charged particles, come in $3^2-1$ colors. I don't know how to see that there are $2^3$ of them.


Is there a better example, along the same lines, that witnesses $3^2-1=2^3$ or $2^3+1=3^2$?
 A: I want to argue that $\mathbb{F}_9$ can actually be an answer to this question. We just need to see why its unit group has order $8 = 2^3$ without computing the size of the unit group as $9 - 1$. 
First, $\mathbb{F}_9$ has a unit $-1$ of order $2$. So the size of the unit group is at least even. Next let's try to take its square root: observe that $\mathbb{F}_9 \cong \mathbb{F}_3[x]/(x^2 + 1) \cong \mathbb{F}_3[i]$ because $x^2 + 1$ is irreducible over $\mathbb{F}_3$. It follows that $\mathbb{F}_9$ has a unit of order $4 = 2^2$ (one $2$ for the order of $-1$ and one $2$ from trying to take its square root), namely $i$. To exhibit a unit of order $8$ we need to find a square root of $i$. So we compute:
$$(a + bi)^2 = (a^2 - b^2) + 2ab i = i.$$
If we set $a = 1, b  = -1$ then $a^2 - b^2 = 0$ and $2ab = 1$, so we conclude that $1 - i$ is a square root of $i$ and hence is a unit of order $8$ as desired. 
In fancy terms we are exhibiting the unit group as an iterated extension of $3$ copies of $C_2$ (in general, any finite $p$-group of order $p^n$ is an iterated central extension of $n$ copies of $C_p$). This is still bijective insofar as an extension of $H$ by $N$ has the same order $|H| |N|$ as the trivial extension $H \times N$. 
A: I am not sure if this is an acceptable answer (one could argue this proves $3^2 - 1 = 2^2 + 2^2$ instead):
For a 3 x 3 square grid with the center removed, we need 3 bits to describe uniquely any of the remaining points:
The first bit is 1 if the point is on a corner, 0 if on a midpoint.
If on a corner, the second bit chooses between left and right edge, third bit chooses between top and bottom edge.
If on a midpoint, the second bit chooses between touching the top-right corner or touching the bottom-left corner. The third bit chooses between touching the top-left corner or touching the bottom-right corner.
Hence on the one hand we have $3^2 - 1$ points, on the other hand we have $2^3$ ways to describe the points.
