$F(x , y)=\cos ^{y}\left(\frac{\pi}{x}\right)+\cos ^{y}\left(\frac{3 \pi}{x}\right)+\cos ^{y}\left(\frac{5 \pi}{x}\right)$

Question

I am stuck with this problem , may you help me , it is from Spanish math olympiad ....

Given $F(x , y)=\cos ^{y}\left(\frac{\pi}{x}\right)+\cos ^{y}\left(\frac{3 \pi}{x}\right)+\cos ^{y}\left(\frac{5 \pi}{x}\right)$

Calculate

$M=F(7 , 2)+F(7 , 3)+F(7 , 5)-F(7 , 6)$

a) $37 / 32$

b) $7 / 4$

c) $19 / 16$

d) $53 / 32$

e)$41 / 32$

Answer 1

A way to kill the problem, not to solve:
Let $w=e^{\frac{\pi}{7}i}$, $\cos\left(\frac{k\pi}{7}\right)=\frac{w^k+w^{-k}}{2}$, $w^{14}=1$, $$F(7,2)=\frac{w^{20} + w^{16} + w^{12} + 6 w^{10} + w^8 + w^4 + 1}{4 w^{10}}\\= \frac{1}{4}\left(w^{10} + w^6 + w^2 + 6 w^0 + w^{12} + w^8 + w^4\right)$$ $$F(7,3)=\frac{w^{30} + w^{24} + 3 w^{20} + 4 w^{18} + 3 w^{16} + 3 w^{14} + 4 w^{12} + 3 w^{10} + w^6 + 1}{8 w^{15}}\\ =\frac{1}{8}\left(4 w^{13} + 4 w^{11} + 4 w^9 + 4 w^5 + 4 w^3 + 4 w\right)$$ $$F(7,5)=\frac{1}{32 w^{25}}\left( w^{50} + 6 w^{40} + 5 w^{34} + 11 w^{30} + 15 w^{28} + 10 w^{26} + 10 w^{24} + 15 w^{22} + 11 w^{20} + 5 w^{16} + 6 w^{10} + 1\right)\\ =\frac{1}{32}\left(16 w^{13} + 16 w^{11} + 16 w^9 + 16 w^5 + 16 w^3 + 16 w\right)$$ $$F(7,6)=\frac{1}{64 w^{30}}\left(w^{60} + 6 w^{50} + w^{48} + 6 w^{42} + 15 w^{40} + 16 w^{36} + 6 w^{34} + 15 w^{32} + 60 w^{30} + 15 w^{28} + 6 w^{26} + 16 w^{24} + 15 w^{20} + 6 w^{18} + w^{12} + 6 w^{10} + 1\right)\\ =\frac{1}{64}\left(22 w^{12} + 22 w^{10} + 22 w^8 + 22 w^6 + 22 w^4 + 22 w^2 + 60 \right)$$ $$\hbox{So }F(7,3)+F(7,5)=w^{13} + w^{11} + w^9 + w^5 + w^3 + w =w\frac{w^{14}-1}{w^2-1}-w^7=1,\\ F(7,2)-F(7,6)=\frac{5}{4}-\frac{60-22}{64}+\left(\frac{1}{4}-\frac{22}{64}\right)\frac{w^{14}-1}{w^2-1}=\frac{21}{32}$$ and the whole sum $=\dfrac{53}{32}$, answer d)

Answer 2

First let's define

$$p_n:=\left(2\cos\frac\pi7\right)^n+\left(2\cos\frac{5\pi}7\right)^n+\left(2\cos\frac{3\pi}7\right)^n\quad(n\in\mathbb{Z}) $$

$$p_0=3,\quad p_1=1,\quad p_2=5$$

$$\forall n\in\mathbb{Z}c\quad p_{n+3}-p_{n+2}-2p_{n+1}+p_n=0$$

Then your sum is :

$$M=\dfrac{p_2}{4}+\dfrac{p_3}{8}+\dfrac{p_5}{32}-\dfrac{p_6}{64} $$

So

$$ M= \dfrac{5}{4}+\dfrac{4}{8}+\dfrac{16}{32}-\dfrac{38}{64}=\dfrac{106}{64}=\dfrac{53}{32} $$

Nota bene:

The expression of $p_n$ can be deduced to the fact that :

$$x^3-\frac{x^2}2-\frac x2+\frac18=0$$

has for roots $$\cos\frac\pi7,\quad-\cos\frac{2\pi}7,\quad\cos\frac{3\pi}7$$

and using the relation between coefficient of your polynom and the roots.