Let $P(x)$ be a 3rd-degree polynomial with integer coefficients, one of whose roots is $\cos(\pi/7)$. Compute $\frac{P(1)}{P(-1)}$
I saw this question in a contest math problem, and I know that it has something to do with Chebeskev(? spelling) Polynomials, but I have no idea what those are.
Best Answer
I will use a method without Chebyshev polynomials.
Let $a = e^{\pi i/7}$. Let $b = \cos \pi/7 = (a + 1/a)/2$.
We have $0 = a^7 + 1 = (a + 1)(a^6 - a^5 + a^4 - a^3 + a^2 - a + 1)$. Since $a, a + 1 \ne 0$, we can divide by $(a + 1)a^3$ to get $$a^3 + \frac{1}{a^3} - a^2 - \frac{1}{a^2} + a + \frac{1}{a} - 1 = 0,$$ which can be rewritten as $$8b^3 - 4b^2 - 4b + 1 = 0.$$
The polynomial $P(b)$ on the left can easily be checked to have no rational roots, by the rational root theorem. Since it is of degree three, it cannot be factored any further using rational coefficients. Therefore any polynomial $P$ satisfying the conditions of the problem must be a constant multiple of this polynomial, hence the ratio $P(1)/P(-1) = -1/7$ is independent of the polynomial chosen.