A certain sum

Since $a$ and $b$ are the roots of $x^2 - x + c = 0$ , by Vieta's formula , $a+b = 1$ and $ab=c$ . Using Newton's method , we have:

$\begin{aligned} a+b & = 1 \\ a^2 + b^2 & = (a+b)^2 - 2ab = 1 - 2c \\ a^3 + b^3 & = (a+b)(a^2+b^2) - ab(a+b) = 1 - 3c \\ a^4 + b^4 & = (a+b)(a^3+b^3) - ab(a^2+b^2) =\color{#3D99F6} 1 - 4c + 2c^2 = 7 \\ a^5 + b^5 & = (a+b)(a^4+b^4) - ab(a^3+b^3) = 7 - c(1-3c) \end{aligned}$ .

From $a^4+b^4 = 7$ :

$\begin{aligned} \implies \color{#3D99F6} 2c^2 - 4x + 1 & \color{#3D99F6} = 7 \\ 2c^2 - 4x - 6 & = 0 \\ c^2 - 2x - 3 & = 0 \\ (c+1)(c-3) & = 0 \\ \implies c & = \begin{cases} - 1 \\ 3 \end{cases} \end{aligned}$

$\implies a^5+ b^5 = 7-c(1-3c) = \begin{cases} 11 & \text{for }c = -1 \\ 31 & \text{for }c = 3 \end{cases}$

Therefore, the sum of possible values is $11+31 = \boxed{42}$ .

Expand (a+b)^5 = a^5 + b^5 + 5ab(a^3 + b^3) + 10 a^2b^2(a+b) = 1, but , a+b = 1, and ab=c. also (a+b)^3 = a^3 + b^3+ 3ab(a+b) = 1, so a^3 + b^3 = 1 – 3c. Substitute to a^5 + b^5 + 5c(1 – 3c) + 10 c^2 = 1,
But a^5 + b^5 = (a^4 + b^4)(a+b) – ab(a^3 + b^3) = 7-c+3c^2 So 7-c+3c^2 + 5c - 5 c^2 = 1, or c^2 -2c – 3 = 0. The solution : c=3, or -1. a^5 + b^5 = 31, or 11

Let $S_n=a^n+b^n$ . Then, by Newton's Sums ,

$1\cdot S_1-1=0\Rightarrow S_1=1$

$1\cdot S_2-1\cdot S_1+2c=0\Rightarrow S_2=1-2c$

$1\cdot S_3-1\cdot S_2+c\cdot S_1=0\Rightarrow S_3=1-3c$

$1\cdot S_4-1\cdot S_3+c\cdot S_2=0\Rightarrow S_4=2c^2-4c+1$

$S_4$ also equals to $7$ , so $2c^2-4c+1=7\Rightarrow c= -1,3$ .

Then, through one more iteration of Newton's Sums,

$1\cdot S_5-1\cdot S_4 + c\cdot S_3 = 0\Rightarrow S_5+c\cdot S_3 = 7$

Plugging in $c=-1$ yields $S_5=11$ and $c=3$ yields $S_5=31$ , so the answer is $\boxed{11+31=42}$ .