Audiophile wrote: Determine all real polynomials $p$ such that $p(x+p(x))=x^2p(x)$ for all $x$. If $P(x)$ is constant, we get $\boxed{\text{S1 : }P(x)=0\quad\forall x}$, which indeed fits If $P(x)$ has degree $1$, we easily get no solution. If degree of $P(x)$ is $n\ge 2$, degree of LHS is $n^2$ while degree of $RHS$ is $n+2$ So $n=2$ and $P(x)=ax^2+bx+c$ and equation is : $a(x+P(x))^2+b(x+P(x))+c=x^2P(x)$ Which is $P(x)+2axP(x)+aP(x)^2+bP(x)=x^2P(x)$ And so $x^2-2ax-b-1=aP(x)=a^2x^2+abx+ac$ and so $(a,b,c)=(1,-2,1)$ And $\boxed{\text{S2 : }P(x)=(x-1)^2\quad\forall x}$, which indeed fits Hemmm, thanks to @biomathematics below, I add the $(a,b,c)=(-1,-2,-1)$ which fits also And $\boxed{\text{S3 : }P(x)=-(x+1)^2\quad\forall x}$, which indeed fits

The only answers are $p(x) = 0$ and $p(x)=(x-1)^2$. First let's assume that $deg(p) > 0$. Then we have that by our equation we must have that $deg(p)^2=2+deg(p) \implies (deg(p)+1)(deg(p)-2)=0$, this easily implies that $deg(p)=2$. Then we must have that $p(x)=ax^2+bx+c$, plugging in what we got we must have that: $$a^3x^4+2ad^2x^3+ad^2x^2+ac^2+2a^2cx^2+2adx+(d-1)ax^2+d(d-1)x+cd=ax^4+(d-1)x^3+cx^2$$where $d=b+1$ This easily implies that $a^3=a$, $ac^2+cd=0$ and $2adc+d(d-1)=0$ Now we just do a case bash on $a^3=a$, to get that $p(x)=(x-1)^2$. Now assume that $deg(p)=0$, then we must have that $c=x^2c$, where $c=p(x)$, giving us that $c=0$.

If $\deg P\ge3$, then the degree of the $\text{LHS}$ is $n^2$ while the degree of the $\text{RHS}$ is $n+2$, no solutions. The only solutions where $\deg P\le2$ are $\boxed{P(x)=0}$ and $\boxed{P(x)=x^2-2x+1}$ and $\boxed{P(x)=-x^2-2x-1}$ after comparing coefficients, which both easily fit.

I am a little confused, because I also get $P(x) = -(x+1)^2$ as a solution, unlike what the above solutions claim. $$P(x+P(x)) = -(x+P(x)+1)^2 = -(x-(x+1)^2+1)^2 = -(-x^2-x)^2 = -x^2(x+1)^2 = x^2P(x)$$

biomathematics wrote: I am a little confused, because I also get $P(x) = -(x+1)^2$ as a solution, unlike what the above solutions claim. Ahhh ! you are perfectly right ! Thanks a lot for checking I edited my post (adding thanks for you)

set $\deg(P(x))=m$ where $m \in \mathbb{Z}^{+}_{0}$ notice if $m=0$ we hae trivially a solution that is $P(x) \equiv 0$ so we consider $m \geqslant 1$ comparing degrees we have $m^2=m+2 \implies m=2$ we set $P(x)=ax^2+bx+c$ plug $x=0$ and notice we have $P(P(0))=0 \implies P(c)=0$ so we have either $c=0$ or $ac+b+1=0$ Claim:- $c \neq 0$ FTSOC consider we have $c=0$ then we have $P(x)=ax^2+bx$ for $a \neq 0$ we notice that plugging back into the given original equation we don't have any such equation hence we have a contradiction and $c \neq 0$. Now we have $a(x+P(x))^2+b(P(x))+x=ax^4+bx^3+cx^2 \implies ax^2+aP(x)^2+2axP(x)+bP(x)+c=ax^{4}+bx^3+cx^2$ comparing degress we have $a^2=1$ $c+b=-1$ and $2a^2(b+1)=b$ which gives only two polynomials works that are $(x-1)^2 , -(x+1)^2$ so all P(x) satisfying the polynomi equation are: $\boxed{-(x+1)^2,0,(x-1)^2}$ $\blacksquare$