For every integer $n\ge1$, if the regular $n^2$-gon is constructible, then $n$ has no odd prime divisors.
I know is has something to do with the fact the output Euler's Totient function on $n^2$ needs to be a power of $2$ for $n^2$ to be constructible. I works up through $81=9^2$ as far as I can tell, but I can figure out how to nail down a proof.
By the Gauss-Wantzel theorem, a regular $k$-gon is constructable if and only if $k$ is a product of a power of $2$ and zero-or-more distinct Fermat primes. If $n$ has any Fermat prime divisors then $k=n^2$ has two identical Fermat prime divisors and hence the $n^2$-gon is not constructable. If $n$ has any odd prime divisor that is not a Fermat prime then $n^2$ has that same odd non-Fermat prime divisor hence the $n^2$-gon is not constructable. Taken backwards, this means if the $n^2$-gon is constructable then $n$ has no odd prime divisors, Fermat or non-Fermat -- hence $n$ is just a power of $2$ and thus has no odd prime divisors.