Polynomials with P(x²) = P(x)²

Constants. If $P$ is constant, $c=c^2$ gives $P=0$ or $P=1$.

Non-constant case. Let $P(x)=ax^n+cx^k+(\text{terms of degree}<k)$ with $a\ne0$, $n\ge1$, and $k<n$ the degree of the second non-zero term, if any.

*Leading coefficient.* Comparing coefficients of $x^{2n}$: $a=a^2$, so $a=1$.

*Second term.* Suppose $P$ has a second non-zero term, $cx^k$ with $c\ne0$, $0\le k<n$. In $P(x)^2=x^{2n}+2cx^{n+k}+\dots$, the second-highest term is $2cx^{n+k}$. In $P(x^2)=x^{2n}+cx^{2k}+\dots$, the second-highest degree is $2k<n+k$. So the coefficient of $x^{n+k}$ is $2c\ne0$ on the left and $0$ on the right: contradiction.

Hence $P(x)=x^n$, and indeed $P(x^2)=x^{2n}=(x^n)^2$ ✓.

Answer. $P=0$, $P=1$, or $P(x)=x^n$ for some integer $n\ge1$.

Lesson. In polynomial identities, compare leading terms first, then the next term; degrees often give an immediate contradiction.