Problem: subdividing a square into other squares, doesn’t have to be all the same size. to be specific, no overlapping squares, every region must be part of a square. For what values of $n$ can a square be subdivided into $n$ squares?
Theorem: For any $n \geq 6$, there is a way to subdivide a square into $n$ smaller squares.
Proof: Let $P(n)$ be the statement “there is a way to subdivide a square into $n$ smaller squares.” We will prove by induction that $P(n)$ holds for all $n \geq 6,$ from which the theorem follows.
As our base cases, we prove $P(6)$, $P(7)$, and $P(8)$, that a square can be subdivided into $6,7,$ and $8$ squares. This is shown here: (image from slides)
For the inductive step, assume that for some arbitrary $k \geq 6$ that $P(k)$ is true and that there is away to subdivide a square into $k$ squares. We prove $P(k+3)$, that there is a way to subdivide a square into $k+3$ squares. To see this, start by obtaining (via the inductive hypothesis) a subdivision of a square into $k$ squares. Then, choose, any of the squares and split into four equal squares. This removes one of the $k$ squares add adds four more, so there will be a net total of $k+3$ squares. Thus $P(k+3)$ holds, completing the induction. $\blacksquare$