If $P(a)=0$, than $P(a^2)=P(a)P(a-1)=0$, so $P(a^{2^k})=0$ for all $k$. And $P((a+1)^2)=P(a)P(a+1)=0$, so $P((a+1)^{2^k})=0$ for all $k$. If $P(0)=0$, than $P((0+1)^2)=P(1)=0$, and so $P(2^{2^k})=0$ for all $k$,contradiction. So if $P(a)=0$, than $\left|a\right|=\left|a+1\right|=1$, hence $a\overline a=1$, $(a+1)\overline {(a+1)}=1$, so $(x-a)(x-\overline a)=x^2+x+1$. So $P(x)=(x^2+x+1)^n$ for all $n\in{N_0}$.

yunxiu wrote: If $P(a)=0$, than $P(a^2)=P(a)P(a-1)=0$, so $P(a^{2^k})=0$ for all $k$. And $P((a+1)^2)=P(a)P(a+1)=0$, so $P((a+1)^{2^k})=0$ for all $k$. If $P(0)=0$, than $P((0+1)^2)=P(1)=0$, and so $P(2^{2^k})=0$ for all $k$,contradiction. So if $P(a)=0$, than $\left|a\right|=\left|a+1\right|=1$, hence $a\overline a=1$, $(a+1)\overline {(a+1)}=1$, so $(x-a)(x-\overline a)=x^2+x+1$. So $P(x)=(x^2+x+1)^n$ for all $n\in{N_0}$. If $P(x)=(x^2+x+1)^n$ then $(x^2+x+1)^n(((x+1)^2+(x+1)+1)^n)=(x^4+x^2+1)^n$ which is not true maybe

ShahinBJK wrote: Find all polynomials $P(x)$ with real coefficients such that $P(x)P(x + 1) = P(x^2)$ for all real $x$. If $P(x)$ constant , we get the solutions $P(x)=0$ and $P(x)=1$ If $P(x)$ is non constant : $z$ root implies $z^2$ root and so $|z|=0,1$ else infinitely many distinct roots. $z\ne 0$ root implies $(z-1)^2$ root and so $|z-1|=0,1$ and so : either $z=1$, either $|z-1|=1$ and $|z|=1$ and so $z=e^{i\frac{\pi}3}$ or $z=e^{-i\frac{\pi}3}$ : $z=e^{i\frac{\pi}3}$ implies $z_1=(z-1)^2=e^{i\frac{2\pi}3}$ root and so $(z_1-1)^2$ root, impossible since $|z_1-1|>1$ $z=e^{-i\frac{\pi}3}$ implies $z_1=(z-1)^2=e^{-i\frac{2\pi}3}$ root and so $(z_1-1)^2$ root, impossible since $|z_1-1|>1$ So the only possible roots are $0,1$ and $P(x)=x^n(x-1)^m$ Plugging this in original equation, we get $n=m$ Hence the solutions : $P(x)=0$ $\forall x$ $P(x)=1$ $\forall x$ $P(x)=x^n(x-1)^n$ $\forall x$ and for any $n\in\mathbb N$

pco wrote: ShahinBJK wrote: Find all polynomials $P(x)$ with real coefficients such that $P(x)P(x + 1) = P(x^2)$ for all real $x$. If $P(x)$ constant , we get the solutions $P(x)=0$ and $P(x)=1$ If $P(x)$ is non constant : $z$ root implies $z^2$ root and so $|z|=0,1$ else infinitely many distinct roots. $z\ne 0$ root implies $(z-1)^2$ root and so $|z-1|=0,1$ and so : either $z=1$, either $|z-1|=1$ and $|z|=1$ and so $z=e^{i\frac{\pi}3}$ or $z=e^{-i\frac{\pi}3}$ : $z=e^{i\frac{\pi}3}$ implies $z_1=(z-1)^2=e^{i\frac{2\pi}3}$ root and so $(z_1-1)^2$ root, impossible since $|z_1-1|>1$ $z=e^{-i\frac{\pi}3}$ implies $z_1=(z-1)^2=e^{-i\frac{2\pi}3}$ root and so $(z_1-1)^2$ root, impossible since $|z_1-1|>1$ So the only possible roots are $0,1$ and $P(x)=x^n(x-1)^m$ Plugging this in original equation, we get $n=m$ Hence the solutions : $P(x)=0$ $\forall x$ $P(x)=1$ $\forall x$ $P(x)=x^n(x-1)^n$ $\forall x$ and for any $n\in\mathbb N$ Can't we just say that because there cannot be infinitely many solutions, the only solutions of the Polynomial P(x) should be 0 or 1. As for every root r there is root $r^2$. Which means that for all $r$ there will be infinitely many $r^2$ as $r^2>=r$. From this fact we get that r=0 or r=1 which means that $P(x)=[x(x-1)]^n$ Is this solution correct or should we take further cases like the fact that $(r-1)^2$ should also be a root??

Let $A(x)$ be the assertion \[P(x)P(x + 1) = P(x^2), \quad \forall x \in \mathbb R.\] $A(0) \implies P(0)P(1)=P(0)$ $1^{st}$ case $P(0)=0$ $A(-1) \implies P(1)=0$ Then $P(x)=x(x-1)Q(x)$ such that $Q \in \mathbb{R}[X]$ plugging it in we get $x^2(x-1)(x+1)Q(x)Q(x+1)=x^2(x^2-1)Q(x^2)$ Then $Q(x)Q(x+1)=Q(x^2)$ $\forall x \not= 0,1$ $Q$ is continuous so $Q(0)Q(1)=Q(0)$ If $Q(0)=0$ we will get $Q(1)=0$ so we will do the same thing until we stop at some polynomial $T$ such that $T(x)T(x+1)=T(x^2)$ and $T(0) \not =0$ (This case will be treated in $2^{nd}$ case). So by infinite denscente we will get that $0,1 $ are roots of $P$ multiple times so $P(x)=0$ $\forall x \in \mathbb{R}$ or $P(x)=ax^n(x-1)^n$ $\forall x \in \mathbb{R}$ $n \in \mathbb{N} $ $a \in \mathbb{R}$ Plugging them back we get $P(x)P(x+1)=0=P(x^2)$ $P(x)P(x+1)=a^2x^{2n}(x^2-1)^n=ax^{2n}(x^2-1)^n$ hence $a=1$ In the case we get $P=0$ or $P=x^n(x-1)^n$ $n \in \mathbb{N}$ $2^{nd}$ case $P(0) \not=0$ Hence $P(1)=1$ Suppose that $P$ is not constant Let $z$ be a complexe root of $P$ $A(z) \implies P(z^2)=0$ hence $P(z^{2^n})=0$ $\forall n \in \mathbb{N}$ since $z \not=0,1$ hence $P$ has infinitly many roots or $|z|=1$ if $|z|=1$ $A(z-1)$ $\implies z-1$ root of $P$ hence $|z-1|=1$ thus $|z-1|=1$ and $|z|=1$ and so $z=e^{i\frac{\pi}3}$ or $z=e^{-i\frac{\pi}3}$ impossible since $|(z-1)^2-1 |\not=1$ thus $P$ has infinitly many roots so $P=0$ contradiction ($P(0)\not=0$) Thus $P$ is constant so $P=1$ Trying it back we get $P(x)P(x+1)=1=P(x^2)$ Thus $P=0$ or $P=x^n(x-1)^n$ $n \in \mathbb{N}$ or $P=1$ $Q.E.D$