AoPS - Problem

Source: Indonesian Mathematical Olympiad 2014 Day 1 Problem 4

Tags: algebra, polynomial, calculus, integration, algebra unsolved

chaotic_iak

03.09.2014 18:13

Determine all polynomials with integral coefficients $P(x)$ such that if $a,b,c$ are the sides of a right-angled triangle, then $P(a), P(b), P(c)$ are also the sides of a right-angled triangle. (Sides of a triangle are necessarily positive. Note that it's not necessary for the order of sides to be preserved; if $c$ is the hypotenuse of the first triangle, it's not necessary that $P(c)$ is the hypotenuse of the second triangle, and similar with the others.)

First suppose that $P(x)$ is not identically null. Suppose that $P(x)=a_nx^n+...+a_0x^0$. Taking the numbers $x,x,\sqrt2x$ as the sides of a right-angled triangle we get that either $2P(x)^2 = P(\sqrt2x)^2$ or $P(x)^2 + P(\sqrt2x)^2 = P(x)^2 \Leftrightarrow P(\sqrt2x)=0$. Because $P(x)$ is of degree $n$ it can have at most $n$ roots, thus we get that $2P(x)^2 = P(\sqrt2x)^2$ for infinitely many $x$. So $2P(x)^2 - P(\sqrt2x)^2$ must be identically null. Thus $\sqrt2P(x) = \pm P(\sqrt2x)$ or consequently $\sqrt2 [x^i] P(x) = \sqrt2a_i = \pm [x^i] P(\sqrt2x) = \pm (\sqrt2)^ia_i$ which implies that $a_i=0$ for all $i\not=1$. So we get that $P(x) = nx$ for any integer $n$.