n is an even number

If n is even, start with a 4x4 defective square with a hole in the upper right corner. Then it is possible to extend the 4x4 to a 7x7 by building a border that's 3 units wide in a similar way to the odd case:

+-----+-----+-----+
|     |     |     |
|     +--+  |  +--+--+
| 3x2 |  |  |  |     |
|     |  +--+--+--+  |
|     |     |     |  |
+-----+--+--+--+  +--+
|        |     |  |  |
|        +--+  +--+  |
| previous  |  |     |
|   4x4     +--+-----+               
|  square   |        |
|           |  2x3   |
|           |        |
+-----------+--------+

To extend an 8x8 defective square (since by formula 1 a 6x6 defective square can't be done) just append two 3x2 squares onto the ends of the borders. It is easy to see that this pattern can be used to extend any odd defective square by 3 units.

With these two constructions, it's now possible to prove that formula 1 is a sufficient condition (i.e. if $n^2 -1 \mbox{ mod }3 = 0$ then it is possible to tile the defective square).

$$n^2 -1 \mbox{ mod }3 = 0$$

$$(n-1)(n+1) \mbox{ mod }3 = 0$$

which tells us that either n-1 is divisible by three (and consequently that $n \mbox{ mod }3 = 2$) or n+1 is divisible by three (therefore $n \mbox{ mod }3 = 1$). Looking at the first case, that $n \mbox{ mod }3 = 2$, it's already been shown that 5x5 ( $5=1 \cdot 3 + 2$) can be done, and that by extending it using the odd case, that 8x8 ( $8 = 2 \cdot 3 + 2$) can be done. This can be extended by the even case to show that 11x11 ( $11 = 3 \cdot 3 + 2$) can be done, and so on to show that for any $n \geq 5$, if $n \mbox{ mod }3 = 2$ then it can be tiles.

Proving it's true for $n \mbox{ mod }3 = 1$ is similar: a 4x4 ( $4 = 1 \cdot 3 + 1$) can be extended to a 7x7 ( $7 = 2 \cdot 3 + 1$) using the even rule, and 7x7 can be extended to 10x10 ( $10 = 3 \cdot 3 + 1$) using the odd rule and so on.

The two together prove that if and only if $n \mbox{ mod }3 \neq 0$ then it's possible to tile a defective square of size n.