any hints to this problem?

Or this way: Give deg(P(x))=d Then we have: d=3d means: d=0 Now pick P(x)=c (because deg=0) So we have: c=c^3 => c=0 or c=1 or c=-1 I don't know this is right or not though

PhoenixPhat wrote: Or this way: Give deg(P(x))=d Then we have: d=3d means: d=0 Now pick P(x)=c (because deg=0) So we have: c=c^3 => c=0 or c=1 or c=-1 I don't know this is right or not though The degree of both sides are 3d so why must d=3d

jasperE3 wrote: Find all polynomials $p(x)$ satisfying $p(x^3+1)=p(x+1)^3$ for all $x$. Let $p(x+1) = q(x)$. Then we have $q(x^3) = q(x)^3$. Clearly $q(x)=0$ works. Otherwise, let the leading term of $q$ be $ax^n$ where $n \ge 0$ and $a \neq 0$. We then have $a^3x^{3n} = ax^{3n}$, so $a = 1$ or $a=-1$. We can assume that $a=1$, because $-q(x)$ will also work if $q(x)$ works. Let the next term in $q(x)$ be $bx^m$, where $m<n$. Then the second highest degree term in $q(x^3)$ is $bx^{3m}$, whereas the second highest degree term in $q(x)^3$ is $3bx^{2n+m}$. However, these must be the same, meaning $m=n$, a contradiction. Hence there are no other terms in $q(x)$ other than $x^n$. Therefore $q(x) = \pm x^n$ or $q(x) = 0$. This gives $p(x) = (x-1)^n$, $p(x) = -(x-1)^n$, or $p(x) = 0$ where $n$ is a nonnegative integer. All work.

CircleGeometryGang wrote: PhoenixPhat wrote: Or this way: Give deg(P(x))=d Then we have: d=3d means: d=0 Now pick P(x)=c (because deg=0) So we have: c=c^3 => c=0 or c=1 or c=-1 I don't know this is right or not though The degree of both sides are 3d so why must d=3d Ups , I read wrong title:)