<s>[INST] $f(x)$ is a quadratic polynomial with $f(0)\neq 0$ and $f(f(x)+x)=f(x)(x^2+4x-7)$ 
$f(x)$ is a quadratic polynomial with $f(0) \neq 0$ and $$f(f(x)+x)=f(x)(x^2+4x-7)$$
It is given that the remainder when $f(x)$ is divided by $(x-1)$ is $3$.
Find the remainder when $f(x)$ is divided by$(x-3)$.


My Attempt:
Let $f(x)=ax^2+bx+c$ and $a,c \neq 0 $
I got $a+b+c=3$
And by the functional equation
$a[ax^2+(b+1)x+c]^2+b[ax^2+(b+1)x+c]+c= [ax^2+bx+c][x^2+4x-7]$
Then by putting $x=0$ , $ac^2 +bc+c=-7c$
Since $c \neq 0 $ , we have $ac+b+8=0$
Then comparing the coeffiecient of $x^4$ , we get $a^3=a$
Since $a \neq 0$ , $a=-1 $ or $a =1 $
Then how to proceed with two values of $a$ ?
or Is there a polynomial satisfying these conditions?
 [/INST] In general, if $F$ is a field, whose algebraic closure is $\bar{F}$, and $q(X)\in F[X]$ is a polynomial of degree $2$ with leading coefficient $k\neq 0$, then all solutions to $$f\big(\alpha\,f(X)+X\big)=f(X)\,q(X)\,,\tag{*}$$ where $\alpha\neq 0$ is a fixed element of $F$ and $f(X)\in \bar{F}[X]$, are given by $$f(X)=0\,,$$ $$f(X)=+\frac{1}{\alpha\sqrt{k}}\,q\left(X-\frac{1}{\sqrt{k}}\right)\,,$$ and $$f(X)=-\frac{1}{\alpha\sqrt{k}}\,q\left(X+\frac{1}{\sqrt{k}}\right)\,,$$ where $\sqrt{k}$ is one of the square roots of $k$.  Thus, if $X^2-k$ is irreducible in $F[X]$, then the only solution to (*) in $F[X]$ is $f(X)=0$.

Hint:  It is easy to see that nonconstant solutions to (*) must be of degree $2$.  Suppose that $$q(X)=k\,(X-u)(X-v)\,,$$ where $u,v\in\bar{F}$.  Then, it follows immediately that $$f(X)=\pm \frac{\sqrt{k}}{\alpha}\,(X-p)(X-q)$$ for some $p,q\in\bar{F}$.  Compute $$\frac{f\big(\alpha\, f(X)+X\big)}{k\,f(X)}$$ in terms of $X,p,q,\sqrt{k},\alpha$.  This should be the same as $$\frac{q(X)}{k}=(X-u)(X-v)\,.$$

In particular, if $F=\mathbb{C}$, $q(X)=X^2+4\,X-7$ (so that $k=1$), and $\alpha=1$, then the nonzero solutions to the functional equation (*) are $$f(X)=+q(X-1)=+X^2+2\,X-10$$ and $$f(X)=-q(X+1)=-X^2-6\,X+2\,.$$  Note that the extra condition that $f(1)=3$ is incompatible with all such polynomials, whence there are no solutions to the OP's conditions.
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