Tilings of a 5×5 square with smaller squares. There is a $5 × 5$ array of lights, such that at each step, we may toggle all the lights in any
$2 × 2, 3 × 3, 4 × 4$ or $5 × 5$ sub-square. Initially all the lights are switched off. After a certain
number of toggles, exactly one light is switched on. Find all the possible positions of the
light
Here's what I have done. We can safely ignore the $4 × 4$ squares as they are just a bunch of $2 × 2$ squares put together. Also the sum of the number of $3$ sided squares and $5$ sided squares must be odd as they are the only squares that can change the parity of the total number of turned on lights. As each $3$ sided as well as each $5$ sided square must contain the middle square, it must be turned on after placing these squares. How do I proceed from here?
 A: You have made some good observations! As you have observed, we need to place an odd number of $3\times 3$ or $5\times 5$ squares, each of which toggles the central cell. We can also forget about the $4\times 4$ square, since it arises from $2\times 2$ square.
Now, observe that placing a $2\times 2$ does not change the parity of the cells in any row or column. So if we are to have a single cell at the end, we must have only one row and only one column with an odd number of cells after placing the $3\times 3$ and $5\times 5$ squares.
But all of these squares flip the parity of the central row and the central column, so since we've placed an odd number of them, those rows and columns must be odd, and hence have to contain our final square (if it exists).
So we've shown that the only possibility is the center cell. Can this actually be done? Yes it can! Flip a $5\times 5$, a $3\times 3$ nestled in two opposite corners, and a $2\times2$ at each of the other two corners. Here is a picture of everything but the $5\times 5$, showing how the central cell is covered twice:

