$f(x)$ 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?
 A: Hint $\ $ Much easier: the functional equation allows you to generate further values of $f(x),\,$ e.g. $\,\color{#c00}{f(1)} = 3\,\Rightarrow f(4) = f(\color{#c00}{f(1)}\!+\!1) = \ldots\,$ Then we can easily compute $\,f(x)\,$ by interpolation.
Remark $ $ It turns out that the interpolated polynomial does not satisfy the functional equation, therefore no solution exists (probably there is a typo in the statement). $ $ 
Alternative proof: $ $ for any polynomial $\,f(x)\,$ it is true that $\,f(x)\mid f(x\!+\!f(x))\,$ because $\,{\rm mod}\ f(x)\!:\ f(x\!+\!f(x))\equiv f(x)\equiv 0.\,$ When $\,f\,$ is quadratic the cofactor (quotient) $\,g\,$ has a simple closed form, which can be computed directly or, more simply, by the Lemma below. 
Therefore, by the Lemma, if  $\,g(x)  = af(x\!+\!1/a) = x^2\!+\!4x\!-\!7\,$ then the lead coeff $\,a^2=1,\,$ thus if $\,a=1\,$ then $\,g(x) = f(x\!+\!1)\,$ so $\,f(1)=g(0)=-7\neq 3,\,$ and if $\,a=-1\,$ this yields $\,g(x)= -f(x\!-\!1),\,$ so $\,f(1)=-g(2) = -5\neq 3.\,$ Though unneeded, going further we get $\,f(x) = g(x\!-\!1)= x^2\!+\!2x\!-\!10\,$ if $\,a=1,\,$  $\,f(x) = -g(x\!+\!1) = -x^2\!-\!6x\!+\!2\,$ if $\,a=-1.\,$
Lemma $\,\ f(x\!+\!f(x))\, =\, a\,f(x)f(x\!+\!1/a)\ $ if $\ f(x) = a(x\!-\!r)(x\!-\!r')$
$\!\begin{align}{\bf Proof}\quad  f(x\!+\!f(x))\, &=\, a\,(x\!+\!f(x)\!-\!r)\ (x\!+\!f(x)\!-\!r')\\[.1em]  
&=\, a\,a(x\!-\!r)(1/a\!+\!x\!-\!r')\  a(x\!-\!r')(1/a\!+\!x\!-\!r)\\[.1em] 
&=\, af(x)\,f(1/a\!+\!x)\end{align}$
A: 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.
A: $ac + b + 8=0$, $ac - a - c + 11 = 0$. If $a = 1$, then $10 = 0$. So $a = -1$, hence $c = 6$, $b = -2$, $f(x) = -x^2 -2x + 6$
A: My answer to the question : Quadratic Functional equations. 
Should work in the exact same manner, the extra conditions may or may not contradict the solution obtained, but in the case of a contradiction, then such a function does not exist.
