Problem A. 867. (December 2023)

A. 867. Let \(\displaystyle p(x)\) be a monic integer polynomial (polynomial with integer coefficients and leading coefficient 1) of degree \(\displaystyle n\) that has \(\displaystyle n\) real roots, \(\displaystyle \alpha_1, \alpha_2,\ldots, \alpha_n\). Let \(\displaystyle q(x)\) be an arbitrary integer polynomial that is relatively prime to polynomial \(\displaystyle p(x)\) (i.e. it's not possible to find an integer polynomial different from constant 1 and \(\displaystyle -1\) that divides both \(\displaystyle p(x)\) and \(\displaystyle q(x)\)). Prove that \(\displaystyle \sum\limits_{i=1}^{n} \big|q(\alpha_i)\big|\ge n\).

Submitted by Dávid Matolcsi, Berkeley

(7 pont)

Deadline expired on January 10, 2024.

Notice that product

\(\displaystyle \prod_{i=1}^{n} q(x_i)\)

is a symmetric polynomial of variables \(\displaystyle x_i\) with integer coefficients, since swapping any two of its variables will result in the same polynomial. By the fundamental theorem of symmetric polynomials it can be expressed as a polynomial of the elementary symmetric polynomials of variables \(\displaystyle x_i\) with integer coefficients.

The leading coefficient of \(\displaystyle p(x)\) is \(\displaystyle 1\), thus by the Vieta's formulas the the values of the elementary symmetric polynomials of values \(\displaystyle \alpha_i\) will be the coefficients of \(\displaystyle p\) with the appropriate signs. The coefficients of \(\displaystyle p\) are all integers, therefore the values of the elementary symmetric polynomials of \(\displaystyle \alpha_1, \alpha_2, \ldots , \alpha_n\) will take integer values. Thus, combining these two observations the value \(\displaystyle \prod_{i=1}^{n} q(\alpha_i)\) is also integer.

It's well know that minimal polynomial of an algebraic number divides all the rational polynomials where the given algebraic number is a root. Therefore, if \(\displaystyle p\) and \(\displaystyle q\) would have a common root, the minimal polynomial of it would divide both \(\displaystyle p\) and \(\displaystyle q\) in the ring of rational polynomials, and thus by Gauss's lemma they would also have a non-trivial common divisor among the integer polynomials. Therefore, \(\displaystyle p\) and \(\displaystyle q\) cannot have a common a root, consequently \(\displaystyle \prod_{i=1}^{n} q(\alpha_i)\) is not \(\displaystyle 0\), however, it's an integer, so its absolute value is at least \(\displaystyle 1\).

Applying the AM-GM inequality we get that

\(\displaystyle 1\le \sqrt[n]{\prod_{i=1}^{n} |q(\alpha_i)|}\le \frac{1}{n}\sum_{i=1}^{n} |q(\alpha_i)|,\)

consequently

\(\displaystyle \sum_{i=1}^{n} |q(\alpha_i)|\ge n.\)

Statistics:

11 students sent a solution.
7 points:	Duchon Márton, Simon László Bence, Szakács Ábel, Varga Boldizsár.
6 points:	Bodor Mátyás, Czanik Pál, Diaconescu Tashi, Fleischman Illés, Nguyen Kim Dorka, Wiener Anna.
0 point:	1 student.

Problems in Mathematics of KöMaL, December 2023