grandes-ecoles 2025 Q6

grandes-ecoles · France · mines-ponts-maths2__mp Roots of polynomials Proof of polynomial identity or inequality involving roots
For every integer $j \in \llbracket 1, n \rrbracket$, we denote by $f_j$ the polynomial $$f_j = a_n \prod_{k=j+1}^{n}\left(1 - \alpha_k X\right) \prod_{k=1}^{j-1}\left(X - \alpha_k\right)$$ with, according to standard conventions, $\prod_{k=n+1}^{n}(1-\alpha_k X) = \prod_{k=1}^{0}(X - \alpha_k) = 1$.
Show that if there exist two integers $i, k$ such that $1 \leq i < k \leq n$ and $\alpha_i \alpha_k = 1$, then $\alpha_i$ is a root of each polynomial $f_j$, where $j \in \llbracket 1, n \rrbracket$, and that the family $(f_1, \ldots, f_n)$ is linearly dependent.
For every integer $j \in \llbracket 1, n \rrbracket$, we denote by $f_j$ the polynomial
$$f_j = a_n \prod_{k=j+1}^{n}\left(1 - \alpha_k X\right) \prod_{k=1}^{j-1}\left(X - \alpha_k\right)$$
with, according to standard conventions, $\prod_{k=n+1}^{n}(1-\alpha_k X) = \prod_{k=1}^{0}(X - \alpha_k) = 1$.

Show that if there exist two integers $i, k$ such that $1 \leq i < k \leq n$ and $\alpha_i \alpha_k = 1$, then $\alpha_i$ is a root of each polynomial $f_j$, where $j \in \llbracket 1, n \rrbracket$, and that the family $(f_1, \ldots, f_n)$ is linearly dependent.