Let $\xi$ be the primitive root of unity.
If $n=5$, then the minimal polynomial of $xi$ over rationals would have degree $5-1=2^2$ which is a fermat prime so $5$-gon is constructible.
If $n=8$, then the min polynomial will have degree $8-1$ which is not a power of two so it is not constructible.
Is this all there is to whether $n$-gons can be constructible? Just minus one to the $n$ and see if it is a power of $2$?
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If $n=8$, the minimal polynomial has degree $\phi(8)=4$, so a regular octagon is constructible.
In general, a regular $n$-gon is constructible if and only if $n=2^kp_1\cdots p_m$ where the $p_j$ are distinct Fermat primes.
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No, it has to be a certain kind of "power of 2." For a regular polygon with a prime number $n\;$ of sides to be constructible using Euclid-approved methods, $n\;$ must be of the form $2^{2^m}+1\;$. The only ones known to date are
$\begin{array}{r|rrrrr} m&0&1&2&3&4\\ n&3&5&17&257&65537 \end{array}$
Practically speaking, though, the 257-gon and 65537-gon are only theoretically constructible. (Both are indistinguishable from circles when drawn to fit a sheet of paper.)
The $n$-gon is constructible if and only if $n$ has the form
$$2^k\cdot p_1\cdot...\cdot p_m$$ , where $k$ is a non-negative integer and $p_1,...,p_m$ are distinct fermat primes.
You can also say : The $n$-gon is constructible if and only if $\phi(n)$ is a power of $2$. $\phi(n)$ is the euler-phi-function
The only known fermat primes are $3,5,17,257$ and $65537$. These are probably the only fermat-primes, but it is open whether there are more. The next possible fermat-prime is $F_{33}=2^{2^{33}}+1$ , having $$2,585,827,973$$ digits. It would be far larger than the currently largest known prime.