The fifth cyclotomic polynomial $\Phi_5(z)$ factors as $$ \Phi_5(z)=(z^2+\varphi z+1)(z^2+(1-\varphi)z+1) $$ where $\varphi$ and $1-\varphi$ are the solutions to $x^2-x-1$. Of course, $\varphi$ is the ratio of side length to diagonal in a regular pentagon. This can be derived from the Gauss's Cyclotomic formula (https://en.wikipedia.org/wiki/Cyclotomic_polynomial#Gauss.27s_formula): $$ 4 \Phi_5(z)=(2z^2+z+2)^2-5z^2 $$ The seventh cyclotomic polynomial $\Phi_7(z)$ factors as $$ \Phi_7(z)=(z^2+\rho_1 z+1)(z^2+\rho_2 z+1)(z^2+\rho_3 z+1) $$ where $\rho_1,\rho_2$ and $\rho_3$ are the solutions to $x^3-x^2-2x+1$. Here, the largest $\rho$ is the ratio of side length to (shorter) diagonal in a regular heptagon. (see http://www.jstor.org/pss/2691048) Yet, Gauss's formula doesn't admit an obvious factorization: $$ 4\Phi_7(z)=(2z^3+z^2-z-2)^2+7z^2(z+1)^2 $$
Can anyone offer any insight?
Edit: It also hold for $\Phi_{11}(z)$ summing over the roots of $x^5-x^4-4x^3+3x^2+3x+1$ (found in http://arxiv.org/pdf/1210.1018v1.pdf). Also this paper may provide a clue http://www.math.umn.edu/~garrett/m/v/kummer_eis.pdf