sororak wrote: Find all polynomials $P$ with real coefficients such that $\forall{x\in\mathbb{R}}$ we have: \[ P(2P(x))=2P(P(x))+2(P(x))^2. \] If $P(x)$ is constant, we get the two solutions $P(x)=0$ and $P(x)=-\frac 12$ If $P(x)$ is not constant, we get $P(2x)=2P(x)+2x^2$ for any $x$ in some non empty open interval, and so $\forall x\in\mathbb R$ Then, if degree is $>2$, comparaison of higher degrees lead to contradiction ($2^na_n=2a_n$), and so $P(x)=ax^2+bx+c$ Plugging this in the equation, we get the solution $P(x)=x^2+ax$ Hence the three solutions : $P(x)=0$ $P(x)=-\frac 12$ $P(x)=x^2+ax$

pco wrote: sororak wrote: Find all polynomials $P$ with real coefficients such that $\forall{x\in\mathbb{R}}$ we have: \[ P(2P(x))=2P(P(x))+2(P(x))^2. \] If $P(x)$ is not constant, we get $P(2x)=2P(x)+2x^2$ for any $x$ in some non empty open interval, and so $\forall x\in\mathbb R$ Then, if degree is $>2$, comparaison of higher degrees lead to contradiction ($2^na_n=2a_n$), and so $P(x)=ax^2+bx+c$ Hello, Would you please explain that a little more?

AgentC wrote: pco wrote: sororak wrote: Find all polynomials $P$ with real coefficients such that $\forall{x\in\mathbb{R}}$ we have: \[ P(2P(x))=2P(P(x))+2(P(x))^2. \] If $P(x)$ is not constant, we get $P(2x)=2P(x)+2x^2$ for any $x$ in some non empty open interval, and so $\forall x\in\mathbb R$ Then, if degree is $>2$, comparaison of higher degrees lead to contradiction ($2^na_n=2a_n$), and so $P(x)=ax^2+bx+c$ Hello, Would you please explain that a little more? set $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}$ , suppose $\deg(P)=n>2$ then clearly we have the leading term of $P(2P(x))=a_{n}^{n+1} 2^{n^2}$ , and in $2P(P(x))$ it will have largest degree as $n>2$ so we have $2^{n^2}=2$ , but that is a contradiction for $n>2$, hence we have $n \leqslant 2$ $P(x)=ax^2+bx+c \implies P(x)=0,\frac{-1}{2} ,x^2+ax$ as the polynomials satisfying the polynomial equation.

lifeismathematics wrote: set $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}$ , suppose $\deg(P)=n>2$ then clearly we have the leading term of $P(2P(x))=a_{n}^{n+1} 2^{n^2}$ , and in $2P(P(x))$ it will have largest degree as $n>2$ so we have $2^{n^2}=2$ , but that is a contradiction for $n>2$, hence we have $n \leqslant 2$ $P(x)=ax^2+bx+c \implies P(x)=0,\frac{-1}{2} ,x^2+ax$ as the polynomials satisfying the polynomial equation. YeahThank you I understood that part, what i don't understand is the " $P(2x)=2P(x)+2x^2$ " part.

AgentC wrote: lifeismathematics wrote: set $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}$ , suppose $\deg(P)=n>2$ then clearly we have the leading term of $P(2P(x))=a_{n}^{n+1} 2^{n^2}$ , and in $2P(P(x))$ it will have largest degree as $n>2$ so we have $2^{n^2}=2$ , but that is a contradiction for $n>2$, hence we have $n \leqslant 2$ $P(x)=ax^2+bx+c \implies P(x)=0,\frac{-1}{2} ,x^2+ax$ as the polynomials satisfying the polynomial equation. YeahThank you I understood that part, what i don't understand is the " $P(2x)=2P(x)+2x^2$ " part. let $P(x) = t$ and $Q(t) = P(2t)- 2P(t)-2(t)^2$ this is a polynomial relation which holds for infinitely many nos $t=P(x)$ this holds for all i.e. Q is 0 polynomial everywhere