Equations in polynomials (Table of contents)

Introduction to Equations in Polynomials

This article deals with determining polynomials in one or more variables (e.g. with real or complex coefficients) which satisfy some given relation(s).

The following example illustrates some basic methods:

Example 1. Determine the polynomials \(P\) for which \(16P(x^2)=P(2x)^2\).

First method: Evaluating at certain points and reducing degree

Plugging \(x=0\) in the given relation yields \(16P(0)=P(0)^2\), i.e. \(P(0)=0\) or \(16\).

We conclude that \(P(x)=16\left(\frac14x\right)^n\) for some \(n\in \mathbb{N}_0\).

Second method: investigating coefficients

We start by proving the following lemma (to be used frequently):

Lemma 2. If \(P(x)^2\) is a polynomial in \(x^2\), then so is either \(P(x)\) or \(P(x)/x\).

Since \(P(x)^2=16P(x^2/4)\) is a polynomial in \(x^2\), we have \(P(x)=Q(x^2)\) or \(P(x)=xQ(x^2)\). In the former case we get \(16Q(x^4)=Q(4x^2)^2\) and therefore \(16Q(x^2)=Q(4x)^2\); in the latter case we similarly get \(4Q(x^2)=Q(4x)^2\). In either case, \(Q(x)=R(x^2)\) or \(Q(x)=xR(x^2)\) for some polynomial \(R\), so \(P(x)=x^iR(x^4)\) for some \(i\in\{0,1,2,3\}\). Proceeding in this way we find that \(P(x)=x^iS(x^{2^k})\) for each \(k\in\mathbb{N}\) and some \(i\in\{0,1,\dots,2^k\}\). Now it is enough to take \(k\) with \(2^k > \deg P\) and to conclude that \(S\) must be constant. Thus \(P(x)=cx^i\) for some \(c\in\mathbb{R}\). A simple verification gives us the general solution \(P(x)=16\left(\frac14x\right)^n\) for \(n\in\mathbb{N}_0\).

Investigating zeroes of the unknown polynomial is also counted under the first method.

A majority of problems of this type can be solved by one of the above two methods (although some cannot, making math more interesting!).