Let $a,b,c\in\mathbb{Z}$, $1<a<10$, $c$ is a prime number and $f(x)=ax^2+bx+c$. If $f(f(1))=f(f(2))=f(f(3))$, find $f'(f(1))+f(f'(2))+f'(f(3))$

(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.

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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 $a=6$.