IMC 2017 · Day 2 · P7

Lemma. If $f(x) = a_m x^m + \dots + a_1 x + a_0$ is a polynomial with $a_m \ne 0$, and all roots of $f$ are real, then $$a_{m-1}^2 - 2 a_m a_{m-2} \ge 0.$$ Proof. Let the roots of $f$ be $w_1, \dots, w_n$. By the Viète-formulas, $$\sum_{i=1}^{m} w_i = -\frac{a_{m-1}}{a_m}, \qquad \sum_{i<j} w_i w_j = \frac{a_{m-2}}{a_m},$$ $$0 \le \sum_{i=1}^{m} w_i^2 = \left( \sum_{i=1}^{m} w_i \right)^2 - 2 \sum_{i<j} w_i w_j = \left( \frac{a_{m-1}}{a_m} \right)^2 - 2 \frac{a_{m-2}}{a_m} = \frac{a_{m-1}^2 - 2 a_m a_{m-2}}{a_m^2}.$$

In view of the Lemma we focus on the asymptotic behavior of the three terms in $q_n(x)$ with the highest degrees. Let $p(x) = a x^k + b x^{k-1} + c x^{k-2} + \dots$ and $q_n(x) = A_n x^{n+k} + B_n x^{n+k-1} + C_n x^{n+k-2} + \dots$; then $$\begin{align*} q_n(x) &= (x+1)^n p(x) + x^n p(x+1) = \\ &= \left( x^n + n x^{n-1} + \frac{n(n-1)}{2} x^{n-2} + \dots \right) (a x^k + b x^{k-1} + c x^{k-2} + \dots) \\ &\quad + x^n \biggl( a \left( x^k + k x^{k-1} + \frac{k(k-1)}{2} x^{k-2} + \dots \right) \\ &\qquad + b \bigl( x^{k-1} + (k-1) x^{k-2} + \dots \bigr) + c \bigl( x^{k-2} \dots \bigr) + \dots \biggr) \\ &= 2a \cdot x^{n+k} + \bigl( (n+k) a + 2b \bigr) x^{n+k-1} \\ &\quad + \left( \frac{n(n-1) + k(k-1)}{2} a + (n+k-1) b + 2c \right) x^{n+k-2} + \dots, \end{align*}$$ so $$A_n = 2a, \quad B_n = (n+k) a + 2b =$$

C_n = \frac{n(n-1) + k(k-1)}{2} a + (n+k-1) b + 2c. If $n \to \infty$ then $$B_n^2 - 2 A_n C_n = \bigl( na + O(1) \bigr)^2 - 2 \cdot 2a \left( \frac{n^2 a}{2} + O(n) \right) = -a n^2 + O(n) \to -\infty,$$ so $B_n^2 - 2 A_n C_n$ is eventually negative, indicating that $q_n$ cannot have only real roots.