mathematics – Mermin’s Lights Out

Bonus question first:

We can treat lights being on or off as a parity constraint, and toggling all lights in a row or column as increasing said lights, inverting their parity. As addition commutes, so do our moves, as in most lights-out puzzles.
Since each move is self-inverse, the possible states after a move are those reached after at most one use of each of the six moves. However, toggling the first row is equivalent to toggling the other two rows and every column, so only five independent binary choices are needed to fix a path from one pattern to an equivalent pattern.
This yields 32 possible changes, all of which are distinct, so each equivalence class has size 32, and there are 16 such classes in total.
Each class has a representative: Toggle rows and columns so that the center and edges are unlit, and each possible pattern of lit and unlit corners corresponds to a unique equivalence class.

Now we can answer the main question:

The empty pattern has no lit corners, the patterns with one lit corner belong to that corner’s class, the patterns with one lit edge have the class with both adjacent corners, and the pattern with a lit center has the class with all corners lit.
The given pattern has class (NE,SW), so it cannot be equivalent to any pattern with fewer than 2 lights. Its class representative, however, has 2 lights, so this bound is sharp.