$P$ be a convex polygon in the plane with a prime number $p$ of sides, all angles equal, and all sides of rational length. Show that $P$ is regular. Let $P$ be a convex polygon in the plane with a prime number $p$ of sides, all angles equal, and all sides of rational length. Show that $P$ is regular (i.e. all sides also have equal length).
It's hard for me to connect the dots between rational side lengths, prime number of sides, and equal angles. Some hints would be greatly appreciated.
 A: My idea involves complex numbers, polynomials and some basic field theory. Let $l_1,\ldots,l_p\in\mathbb Q$ the lengths of the sides and write
$$\zeta=e^{\frac{2i\pi}{p}}.$$
If we assign to each vertex of the polygon a complex argument $z_1,\ldots,z_p$, we could choose $z_1=0$ and $z_1$ on the (positive) real line, so that
$$z_1=0,\:z_2=l_1,\:z_3=z_2+l_2\zeta,\:\ldots z_p=z_{p-1}+l_{p-1}\zeta^{p-2}$$
and the last relation is
$$z_1=z_p+l_p\zeta^{p-1}=0.$$
Compiling all these together, we obtain
$$l_1+l_2\zeta+\ldots+l_p\zeta^{p-1}=0.$$
So the polynomial
$$P(X)=l_1+l_2X+\ldots+l_pX^{p-1}\in\mathbb Q[X]$$
annulates $\zeta$.

By Eisentsein's criterion, you can prove that the polynomial
$$Q(X)=1+X+\ldots+X^{p-1}\in\mathbb Q[X]$$
is irreducible (this is the part where the primality of $p$ is essential). Indeed, remark that $Q(X)=\dfrac{X^p-1}{X-1}$ and that $Q(X)$ is irreducible if and only if $Q(X+1)$ is irreducible. But
$$Q(X+1)=\dfrac{(X+1)^p-1}{X}=X^{p-1}+\sum_{k=1}^{p-1}\binom{p}{k}X^{p-k-1}$$
so $p$ divides each coefficient of $Q(X)$ other than the dominating one, but $p^2$ does not divide its constant coefficient, since it is $p$.

Since $Q$ annulates $\zeta$ it is the minimal polynomial of $\zeta$, hence $Q(X)\mid P(X)$, but they are of same degree, so
$P(X)=\lambda Q(X)$ for some $\lambda\in\mathbb Q$. We can thus conclude that
$$\lambda=l_1=l_2=\ldots=l_p,$$
making the polygon regular.
