Septic Sums

Let $O$ be the center of the unit circle. By cosine rule , we have $AB^2 = OA^2 + OB^2 - 2\cdot OA \cdot OB \cdot \cos \frac {2\pi}7 = 2\left(1-\cos \frac {2\pi}7 \right) \implies AB^4 = 4\left(1-\cos \frac {2\pi}7 \right)^2$ . Similarly, $AC^4 = 4\left(1-\cos \frac {4\pi}7 \right)^2$ and $AD^4 = 4\left(1-\cos \frac {6\pi}7 \right)^2$ . Then the sum of the six terms is:

$\begin{aligned} S & = 8 \left(\left(1-\cos \frac {2\pi}7 \right)^2 + \left(1-\cos \frac {4\pi}7 \right)^2 + \left(1-\cos \frac {6\pi}7 \right)^2 \right) \\ & = 8 \left(3 - 2 \left(\blue{\cos \frac {2\pi}7 + \cos \frac {4\pi}7 + \cos \frac {6\pi}7}\right) + \cos^2 \frac {2\pi}7 + \cos^2 \frac {4\pi}7 + \cos^2 \frac {6\pi}7 \right) \\ & = 8 \left(3 - 2 \left(\blue{-\frac 12}\right) + \frac 12 \left(3 + \cos \frac {4\pi}7 + \cos \frac {8\pi}7 + \cos \frac {12\pi}7 \right) \right) \\ & = 8 \left(4+ \frac 12 \left(3 - \left(\blue{\cos \frac {3\pi}7 + \cos \frac {\pi}7 + \cos \frac {5\pi}7} \right) \right) \right) \\ & = 32 + 4 \left(3 - \blue{\frac 12} \right) = 32 + 10 = \boxed{42} \end{aligned}$

Instead of $A,B,C,D,E,F,G$ , label the vertices $Z_j$ for $0 \le j \le 6$ . If $z = e^{\frac{2\pi i}{7}}$ , for $0 \le j \le 6$ , then $Z_j$ represents $z^j$ on the Argand diagram, and so $\begin{aligned} \sum_{j=1}^6 Z_jZ_0^4 & = \; \sum_{j=1}^6 \big|z^j - 1\big|^4 \; = \; \sum_{j=1}^6\big(2 - z^j - z^{-j}\big)^2 \\ & = \; \sum_{j=1}^6\big(4 + z^{2j} + z^{-2j} - 4z^{j} - 4z^{-j} + 2\big) \; = \; 36 + \sum_{j=1}^6 \big(z^{2j} + z^{-2j} - 4z^{j} - 4z^{-j}\big) \\ & = \; 42 + \sum_{j=0}^6 \big(z^{2j} + z^{-2j} - 4z^{j} - 4z^{-j}\big) \; =\; \boxed{42} \end{aligned}$

Fine problem. N=6,7,8,9,10,11... S(N)=36,42,48,54,60,66...