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How to solve this equation? Find all polynomials $P$ such that $P(x)^2+P(\frac1x)^2= P(x^2)P(\frac1{x^2})$

Please step by step

FMath
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2 Answers2

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Step 1: Let $p(t) = at^n + \cdots + b$. What happens for large $x$?

Step 2: Your turn.

Aritra Das
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gnasher729
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Here is an algebraic solution to the problem. I suppose that the coefficients are complex numbers here. Suppose, therefore, that the complex polynomial $P\in\mathbf C[x]$ satisfies the equation $$ P(x)^2+P(\tfrac1x)^2=P(x^2)P(\tfrac1{x^2}) $$ in the fraction field $\mathbf C(x)$.

Let us first prove that $P(1)=0$ and $P(-1)=0$. Indeed, evaluating the equation that $P$ satisfies at $1$ gives $$ P(1)^2+P(1)^2=P(1)P(1). $$ Hence $P(1)=0$. Next, evaluate at $-1$: $$ P(-1)^2+P(-1)^2=P(1)P(1). $$ Since $P(1)=0$, one gets $P(-1)=0$.

Assuming that $P$ is not the zero polynomial, let $(x^2-1)^n$ be the highest power of $x^2-1$ that divides the polynomial $P$ in $\mathbf C[x]$. By the preceding observations, $n\geq1$. Write $$ P=(x^2-1)^nQ, $$ with $Q\in\mathbf C[x]$. Substituting in the equation that $P$ satisfies, one obtains $$ (x^2-1)^{2n}Q(x)^2+(\tfrac1{x^2}-1)^{2n}Q(\tfrac1x)^2=(x^4-1)^nQ(x^2)(\tfrac1{x^4}-1)^nQ(\tfrac1{x^2}). $$ Rewriting a little bit gives: $$ (x^2-1)^{2n}Q(x)^2+\frac{(x^2-1)^{2n}}{x^{4n}}Q(\tfrac1x)^2=(x^4-1)^nQ(x^2)(-1)^n\frac{(x^4-1)^n}{x^{4n}}Q(\tfrac1{x^2}). $$ Dividing by $(x^2-1)^{2n}$ gives $$ Q(x)^2+\frac{1}{x^{4n}}Q(\tfrac1x)^2=(x^2+1)^nQ(x^2)(-1)^n\frac{(x^2+1)^n}{x^{4n}}Q(\tfrac1{x^2}). $$ Let us show that $Q(1)=0$ and $Q(-1)=0$. This will contradict the fact that one has either $Q(1)\neq0$ or $Q(-1)\neq0$ by maximality of $n$, and prove that $P$ is the zero polynomial.

Evaluate the above expression at $1$: $$ Q(1)^2+Q(1)^2=2^nQ(1)(-1)^n2^nQ(1). $$ Since $n\geq1$, one obtains $Q(1)=0$. Evaluating the expression above at $-1$ gives: $$ Q(-1)^2+Q(-1)^2=2^nQ(1)(-1)^n2^nQ(1). $$ Since $Q(1)=0$, one gets $Q(-1)=0$.

As observed above, it follows that the only polynomial $P$ with complex coefficients that satisfies the equation is the $0$ polynomial.

Johannes Huisman
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    Over any field of characteristic not equal to $2$, the only polynomial solution to $\big(P(x)\big)^2+\big(P(1/x)\big)^2=P\left(x^2\right),P\left(1/x^2\right)$ is $P(x)=0$. Over a field of characteristic $2$, though, the solutions are $P(x)=0$ and $P(x)=x^k+1$ for $k=1,2,\ldots$. – Batominovski May 19 '16 at 15:36