Can't cut squares from here (in the thickest Mainer accent imaginable)

Consider a generic square. It can be subdivided into smaller sub-squares, not necessarily of equal size so that the smaller squares do not overlap. For example, you can slice the big square into four smaller squares, each a quarter of the area of the big square. Or you could slice it into seven squares, if you take one of those four squares and slice it into four yet smaller squares.

What whole numbers of squares can you not slice the big square into?

Let $S$ be the set of whole numbers $n \in \mathbb{N},$ where the original big square can be partitioned into $n$ not necessarily equally sized smaller squares. The problem setup provides a hint that shows that for any $n \in S,$ by taking any single smaller square and dividing it equally into $4$ even smaller squares, we end up with $n + 3$ subsquares, so $n \in S \Rightarrow n+3 \in S.$ So in particular, since $1 \in S$ (just the original square with no further cuts), we have $(3\mathbb{N} + 1) \subset S.$

However, we needn't focus solely on subdividing into $4$ equally sized sub-squares. For instance for any $k \in \mathbb{N}$ we can subdivide into $k^2$ equally sized sub-squares by equally dividing the length and width of the original square into $k$ segments and forming a $k$ by $k$ grid of subsquares. In this case, for any $n \in S,$ we can take any sub-square and replace it with $k^2$ smaller sub-squares, getting $n + k^2 - 1$ total sub=squares. In particular, for $k = 3,$ we get $(8 \mathbb{N} + 1) \subset S,$ which implies that for instance $9 \in S$ and $17 \in S.$

Since $9 \equiv 0 \mod 3,$ by further subdividing squares into equal quarter squares, we get that any multiple of $3$ greater than or equal to $9$ is included in $S.$ That is, $(3 \mathbb{N} \setminus \{0,3,6\} ) \subset S.$ Similarly, since $17 \equiv 2 \mod 3,$ we get that $((3 \mathbb{N} + 2) \setminus \{2, 5, 8, 11, 14\}) \subset S.$ So since all remainders with respect to $3$ are covered for large enough values, we see that \begin{align*}S &= (3 \mathbb{N} + 1) \cup (3 \mathbb{N} \setminus \{0, 3, 6\}) \cup ((3 \mathbb{N} + 2) \setminus \{2, 5, 8, 11, 14\})\\ &= \mathbb{N} \setminus \{2, 3, 5, 6, 8, 11, 14\}.\end{align*}

Thus the complement of $S$ is given by $S^C = \{2, 3, 5, 6, 8, 11, 14\}.$