IstekOlympiadTeam wrote: Find all polynomials $P(x)$ with real coefficents such that \[P(P(x))=(x^2+x+1)\cdot P(x)\]where $x \in \mathbb{R}$ The only constant solution is $\boxed{\text{S1 : }P(x)=0\text{ }\forall x}$ If $P(x)$ is non constant with degree $n$, equation gives $n^2=n+2$ and so $n=2$ If $z$ is a (real or complex) root of $P(x)$, equation gives $P(0)=0$ and so $P(x)=ax^2+bx$ And equation becomes $aP(x)+b=x^2+x+1$ and so $a(ax^2+bx)+b=x^2+x+1$ and $a=b=1$ And so $\boxed{\text{S2 : }P(x)=x^2+x\text{ }\forall x}$

My solution: For non constant We have $degP(P(x))=(degP(x))^2=2+degP(x)$ which implies $degP(x)=2$($P(x)=a_2x^2+a_1x+a_0$. From $P(x)|P(P(x))$ we have $a_0=0$$\implies $ $P(x)=x(a_2x+a_1)$ $\implies $ $P(P(x))=P(x)(a_2P(x)+a_1)=P(x)(x^2+x+1)$ $\implies $ $a_2^2x^2+a_2a_1+a_1=x^2+x+1$ $\implies $ $ P(x)=x^2+x$ for non constant and $P(x)=0$ for constant.

here is my solution. assume the polynomial is constant. so, c = (x^2 + x + 1)*c or, c = 0. if P(x) is not a constant polynomial, let the degree of P(x) = t. degree of L.H.S is t^2 but degree of R.H.S is t + 2. t^2 = t + 2 by solving this quadratic equation, t = 2. then you can write P(x) = a*x^2 + b*x + c and after calculation you will find a = 1 , b = 1, c = 0 then the solution is P(x) = 0 or P(x) = x^2 + x

am I correct

The only constant solution is $\boxed{P(x)=0}$. Suppose $P$ has degree $n$. Then $n^2=n+2\implies n^2-n-2=0\implies n=2$. Now suppose $a$ is a root of $P$. Then by setting $a$ into the equation, we have $P(0)=0\implies P(x)=ax^2+bx$. So the equation becomes \[a(ax^2+bx)^2+b(ax^2+bx)=(x^2+x+1)\cdot(ax^2+bx)\implies (ax^2+bx)(a(ax^2+bx)+b)=(ax^2+bx)\cdot(x^2+x+1)\implies a(ax^2+bx)+b=x^2+x+1\] So $a^2x^2+abx+b=x^2+x+1\implies a=b=1$. Thus, the only non-constant polynomial is $\boxed{P(x)=x^2+x}$.

If the degree of $P(x)$ is not $2$, then the sides degrees won't match up, so $P(x)$ is a quadratic (but $P(x)=0$ is the only constant solution), if $P(x)$, is a quadratic, we can express it as $ax^2+bx$, by hard bash, $P(x)=x^2+x$, is the only thing that works. So we have $P(x)=0$ or $P(x)=x^2+x$

Let t be the degree of P(x). We know that: $$t^2=t+2$$$$t=0 (1)$$$$t=2 (2)$$ (1) $$c=(x^2+x+1)c \Rightarrow c=0$$ (2) Let $$P=ax^2+bx+c$$ LHS constant is ac^2+bc+c ,while RHS constant is c $$ac^2+bc+c=c \Rightarrow c(ac+b)=0$$ If c=0, then $$P(x)=ax^2+bx$$ $$a(ax^2+bx)^2+b(ax^2+bx)=(x^2+x+1)(ax^2+bx) \Rightarrow x^2(a^2-1)+x(ab-1)+(b-1)=0$$$$a=1\hspace{2mm} b=1 \Rightarrow P(x)=ax^2+bx$$ If b=-ac, then LHS is $$a^3x^4 \hspace{1mm}modulo \hspace{1mm} c $$While RHS is $$ax^4+ax^3+ax^2 \hspace{1mm} modulo \hspace{1mm} c$$$$a^3x^4 \equiv ax^4+ax^3+ax^2\mod c$$$$a^3x^2 \equiv ax^2+ax+a\mod c$$$$a(a^2x^2-ax-x-1)\mod c$$Since this equation is true for every x, then a must be a multiple of c. $$a=mc$$$$P(x)=cmx^2-c^2mx+c$$LHS always gives c modulo c^2 , while it is not true for RHS.