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 \(16\).

(i) Suppose that \(P(0)=0\). Then \(P(x)=xQ(x)\) for some
polynomial \(Q\) and \(16x^2Q(x^2)=4x^2Q(2x)^2\), which reduces to
\(4Q(x^2)=Q(2x)^2\). Now setting \(4Q(x)=R(x)\) gives us \(16R(x^2)=
R(2x)^2\). Hence, \(P(x)=\frac 14xR(x)\), with \(R\) satifying the same
relation as \(P\).
 (ii) Suppose that \(P(0)=16\). Putting \(P(x)=xQ(x)+16\)
in the given relation we obtain \(4xQ(x^2)=xQ(2x)^2+16Q(2x)\); hence
\(Q(0)=0\), i.e. \(Q(x)=xQ_1(x)\) for some polynomial \(Q_1\).
Furthermore, \(x^2Q_1(x^2)=x^2Q_1(2x)^2+8Q_1(2x)\), implying that
\(Q_1(0)=0\), so \(Q_1\) too is divisible by \(x\). Thus \(Q(x)=
x^2Q_1(x)\). Now suppose that \(x^n\) is the highest degree of \(x\)
dividing \(Q\), and \(Q(x)=x^nR(x)\), where \(R(0)\neq0\). Then \(R\)
satisfies \(4x^{n+1}R(x^2)=2^{2n}x^{n+1}R(2x)^2+2^{n+4}R(2x)\),
which implies that \(R(0)=0\), a contradiction. It follows that
\(Q\equiv0\) and \(P(x)\equiv16\).
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 \(P(x)/x\).
Let \(P(x)=a_nx^n+a_{n1}x^{n1}+\cdots+a_0\),
\(a_n\neq0\). The coefficient at \(x^{2n1}\) is \(2a_na_{n1}\), from
which we get \(a_{n1}=0\). Now the coefficient at \(x^{2n3}\) equals
\(2a_na_{n3}\); hence \(a_{n3}=0\), and so on. Continuing in this
manner we conclude that \(a_{n2k1}=0\) for \(k=0,1,2,\dots\), i.e.
\(P(x)=a_nx^n+a_{n2}x^{n2}+a_{n4}x^{n4}+\cdots\).
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!).