Find all polynomials $p(x)$ such that $p(x^2)={p(x)}^2$

Question

The question is

Find all polynomials $p(x)$ such that $p(x^2)=[p(x)]^2$.

First of all i saw that $p(x)=0$ and $p(x)=1$ are two polynomials satisfying the condition. Next I tried putting some values and observed that Since $p\big((-a)^2\big)=[p(-a)]^2$. Also, $p\big((a)^2\big)=[p(a)]^2$ Hence we can conclude - $p(-x)=p(x)$

Now my doubt is what to do next to find all possible polynomials or whether my approach is right or not .Any help would be appreciated.Thanks in advance.

You get $p(-a)^2=p(a)^2$ so $p(-a)=\pm p(a)$ – kingW3 Sep 16 '19 at 20:16 — kingW3, Sep 16 '19 at 20:16

score 1 · Accepted Answer · answered Sep 16 '19 at 20:17

1

hint

begin by remarking that

$$p(0)=p(0)^2 \implies p(0)=0 \text{ or } p(0)=1$$

$$p(1)=p(1)^2$$

also, $p(x)=x^n$ is a solution.

answered Sep 16 '19 at 20:17

hamam_Abdallah

62,951

Find all polynomials $p(x)$ such that $p(x^2)={p(x)}^2$

1 Answers1