Track / Stage 5 / 369 of 400 #969 of 1964

Problem 969

AIME late

Algebra Difficulty 5.9 Prove it

Kuzneuov A.

On the board, several reduced polynomials of the 37th degree are written, all coefficients of which are non-negative. It is allowed to choose any two written polynomials $f$ and $g$ and replace them with two reduced polynomials of the 37th degree $f_{1}$ and $g_{1}$ such that $f+g=f_{1}+g_{1}$ or $f g=f_{1} g_{1}$ . Prove that after applying any finite number of such operations, it cannot happen that each polynomial on the board has 37 different positive roots.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Note that when replacing polynomials $f$ and $g$ with $f_{1}$ and $g_{1}$ , the sum of the coefficients of the $36-x$ polynomials does not change. For the first type of replacement (where $f+g=f_{1}+g_{1}$ ), this is obvious, and for the second type of replacement, it follows from the equality

$\left(x^{37}+a x^{36}+\ldots\right)\left(x^{37}+b x^{36}+\ldots\right)=x^{74}+(a+b) x^{73}+\ldots$ But initially, the sum of all such coefficients $\mathrm{y}$

of the polynomials written on the board is non-negative, so it will always be such. If at the end all polynomials have 37 roots, then by Vieta's theorem, this sum is equal to the sum of all these roots with the opposite sign.

Therefore, among the roots, there will be non-positive ones.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.