International Mathematics Competition for University Students

July 31 – August 6 2017, Blagoevgrad, Bulgaria

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Problem 7

7. Let $p(x)$ be a nonconstant polynomial with real coefficients. For every positive integer~$n$, let $$q_n(x) = (x+1)^np(x)+x^n p(x+1) .$$

Prove that there are only finitely many numbers $n$ such that all roots of $q_n(x)$ are real.

Proposed by: Alexandr Bolbot, Novosibirsk State University

Solution.

Lemma. If $f(x)=a_mx^m+\ldots+a_1x+a_0$ is a polynomial with $a_m\ne0$, and all roots of $f$ are real, then $$ a_{m-1}^2-2a_ma_{m-2} \ge0. $$

Proof. Let the roots of $f$ be $w_1,\ldots,w_n$. By the ViƩte-formulas, $$ \sum_{i=1}^m w_i = -\frac{a_{m-1}}{a_m}, \qquad \sum_{i<j} w_iw_j = \frac{a_{m-2}}{a_m}, $$ $$ 0 \le \sum_{i=1}^m w_i^2 = \left(\sum_{i=1}^m w_i\right)^2 -2\sum_{i<j} w_iw_j = \left(\frac{a_{m-1}}{a_m}\right)^2-2\frac{a_{m-2}}{a_m} = \frac{a_{m-1}^2-2a_ma_{m-2}}{a_m^2}. $$

In view of the Lemma we focus on the asymptotic behavior of the three terms in $q_n(x)$ with the highest degrees. Let $p(x)=ax^k+bx^{k-1}+cx^{k-2}+\dots$ and $q_n(x)=A_nx^{n+k}+B_nx^{n+k-1}+C_nx^{n+k-2}+\dots$; then \begin{align*} q_n(x) &= (x+1)^np(x)+x^n p(x+1) = \\ &= \bigg(x^n+nx^{n-1}+\frac{n(n-1)}2x^{n-2}+\ldots\bigg) (ax^k+bx^{k-1}+cx^{k-2}+\ldots) \\ &+x^n \bigg(a\bigg(x^k+kx^{k-1}+\frac{k(k-1)}2x^{k-2}+\ldots\bigg) \\ &\qquad\qquad+b\Big(x^{k-1}+(k-1)x^{k-2}+\dots\Big) +c\Big(x^{k-2}\ldots\Big)+\ldots\bigg) \\ &= 2a \cdot x^{n+k} + \big((n+k)a+2b\big) x^{n+k-1} \\ &\qquad\qquad + \bigg(\frac{n(n-1)+k(k-1)}2 a + (n+k-1)b + 2c\bigg) x^{n+k-2} + \ldots, \end{align*} so $$ A_n=2a, \quad B_n=(n+k)a+2b = \quad C_n=\frac{n(n-1)+k(k-1)}2 a + (n+k-1)b + 2c. $$ If $n\to\infty$ then $$ B_n^2-2A_nC_n = \big(na+O(1)\big)^2-2\cdot2a\bigg(\frac{n^2a}2+O(n)\bigg) = -an^2 + O(n) \to-\infty, $$ so $B_n^2-2A_nC_n$ is eventually negative, indicating that $q_n$ cannot have only real roots.


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