grandes-ecoles 2014 QIV.B
Proof of polynomial identity or inequality involving roots
We introduce the Dickson polynomials of the first and second kind, $(D_n)_{n \in \mathbb{N}}$ and $(E_n)_{n \in \mathbb{N}}$, defined in the form of polynomial functions of two variables by $$D_0(x,a) = 2 \quad D_1(x,a) = x \quad E_0(x,a) = 1 \quad E_1(x,a) = x$$ then, for every integer $n \in \mathbb{N}$, $$D_{n+2}(x,a) = x D_{n+1}(x,a) - a D_n(x,a) \quad \text{and} \quad E_{n+2}(x,a) = x E_{n+1}(x,a) - a E_n(x,a)$$
Justify the following relation with Chebyshev polynomials $$\forall (x,a) \in \mathbb{C}^2, \quad D_n\left(2xa, a^2\right) = 2a^n T_n(x) \quad \text{and} \quad E_n\left(2xa, a^2\right) = a^n U_n(x)$$ as well as the following two relations, valid for every natural integer $n$ and every $(x,a) \in \mathbb{C}^* \times \mathbb{C}$ $$D_n\left(x + \frac{a}{x}, a\right) = x^n + \frac{a^n}{x^n} \quad \text{and} \quad \left(x - \frac{a}{x}\right) E_n\left(x + \frac{a}{x}, a\right) = \left(x^{n+1} - \frac{a^{n+1}}{x^{n+1}}\right)$$