Let $a,b,c\in\mathbb{Z}$, $1Let $a,b,c\in\mathbb{Z}$, $1<a<10$, $c$ is a prime number and $f(x)=ax^2+bx+c$. If $f(f(1))=f(f(2))=f(f(3))$, find $f'(f(1))+f(f'(2))+f'(f(3))$
My attempt:
\begin{align*}
f'(x)&=2ax+b\\
(f(f(x)))'&=f'(f(x))f'(x)\\
f'(f(x))&=\frac{(f(f(x)))'}{f'(x)}\\
\end{align*}
 A: (Edited to add details)
Due to the symmetry of vertical parabola, for distinct $x_1, x_2,$ $f(x_1)=f(x_2) \Rightarrow x_1+x_2=-b/a$ and $f'(x_1)+f'(x_2)=0$
For a quadratic, w cannot have $f(x_1)=f(x_2)=f(x_3)$ for distinct three $x_i$ since this would imply, $f(x)−f(x_1)$ has three distinct roots. There arise three conditions of arguments being pairwise equal.
$f(f(1))=f(f(2))=f(f(3)) \Rightarrow$ following cases :

*

*$f(1)=f(2) \Rightarrow 1+2=-b/a$.

Now $f(1) \neq f(3)$. But $f(f(1))=f(f(3))$.  $\Rightarrow f(1)+f(3)=-b/a=3$
Here $f(1)+f(2) \neq -b/a$ since $f(1),f(2)$ are identical. Similarly,

*

*$f(2)=f(3) \Rightarrow 2+3=-b/a$$f(1) \neq f(3) \Rightarrow f(1)+f(3)=-b/a=5$

*$f(1)=f(3) \Rightarrow 1+3=-b/a$$f(1) \neq f(2) \Rightarrow f(1)+f(2)=-b/a=4$
But $f(1)+f(3)=10a+4b+2c$ is even. So only the third case holds: $f(1)=f(3)$ and $f'(1)+f'(3)=0$.
Thus $b=-4a$. $f(x)=ax^2-4ax+c \Rightarrow f'(x)=2a(x-2)$
Also $f'(f(x)) = (ay^2-4ay+c)'=2a(y-2)y'=2af'(x)(f(x)-2)$
Evaluating, $f'(f(1))+f(f'(2)+f'(f(3))=c$
where $a,c$ have been computed as below.

Remark :
Inspired from the other answer, even the numerical value of $c$ can be calculated. Subbing $f(x)=ax^2-4ax+c$ into $f(f(1))=f(f(2))$, one obtains $$c=\dfrac{7a+4}{2}$$
Quickly checking for $a\in \{2,4,6,8\}$, $c=23$ a prime, only for $a=6$.
A: Okay here's the outline of how you solve this.
first, you write down the system of equations
$$ 
0=f(f(2))-f(f(1))=(3a+b)*(b+5a^2+3ab+2ac) \\
0=f(f(3))-f(f(2))=(5a+b)*(b+13a^2+5ab+2ac)
$$
and this leaves you with four possibilities:
(a) $(3a+b=0)$ and $(5a+b=0)$ which is impossible since $a \ne 0$
(b) $(3a+b=0)$ and $(b+13a^2+5ab+2ac=0)$
in which case we obtain $b=-3a$, then our second condition turns into $2ac-2a^2-3a=0$, which further yields $c=a+\frac{3}{2}$, and we see no acceptable solutions here since both $a$ and $c$ are asked to be integers.
(c) $(b+13a^2+5ab+2ac=0)$ and $(b+5a^2+3ab+2ac=0)$ which, if we subtract one of the equations from the other, gives us $b=-4a$ and, shortly after, $p=\frac{7a+4}{2}$, and we can just check the values of $a$ from $2$ to $9$ to find the only solution at $a=6, b=-24, c=23$.
(d) $(5a+b=0)$ and $(b+5a^2+3ab+2ac=0)$, in which case we get $b=-5a$, the second condition simplifies to $10a-2c+5$, which gives us $c=5a+\frac{5}{2}$, and we also see that $c$ will never be integer if $a$ is integer, and $a$ is always integer.
This leaves us with the only acceptable solution found in (c). As I already mentioned above, putting $(a,b,c)=(6,-24,23)$ into $f'(f(1))+f(f'(2))+f'(f(3))$ gives you $95$.
