# 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

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

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!).

2005-2018 IMOmath.com | imomath"at"gmail.com | Math rendered by MathJax