Double Quadratic Equation

Since $\alpha$ and $\beta$ are roots of $x^2+x+c = 0$ , by Vieta's formula we have $\alpha + \beta = -1$ and $\alpha \beta = c$ . Then,

$\begin{aligned} (\alpha + \beta)^5 & = {\color{#3D99F6}\alpha^5} + 5 \alpha^4 \beta + 10 \alpha^3 \beta^2 + 10 \alpha^2 \beta^3 + 5 \alpha \beta^4 + {\color{#3D99F6}\beta^5} \\ (-1)^5 & = {\color{#3D99F6}2017} + 5 \alpha \beta \left(\alpha^3 + \beta^3\right) + 10 \alpha^2 \beta^2 \left(\alpha + \beta\right) \\ -1 & = 2017 + 5c \left((\alpha + \beta)(\alpha^2 -\alpha\beta + \beta^2) \right) + 10 c^2 (-1) \\ -2018 & = - 5c \left((\alpha + \beta)^2 -3\alpha\beta \right) - 10 c^2 \\ 2018 & = 5c \left(1 -3c \right) + 10 c^2 \\ 2018 & = 5c - 5 c^2 \end{aligned}$

$\implies 5c^2 - 5c + 2018 =0$ , by Vieta's formula, the sum of the possible values of $c$ is $\dfrac 55 = \boxed{1}$ .

I did the problem slightly differently than Chew-Seong Cheong, so I'll post my solution. It's late at night, so I might be a bit sloppy and skip some steps. Throughout, I use the following facts: $\alpha + \beta = -1$ and $\alpha\beta=c$ . $2017 = \alpha^5+\beta^5 = (\alpha + \beta)(\alpha^4-\alpha^3\beta+\alpha^2\beta^2-\alpha\beta^3+\beta^4)=(-1)(\alpha^4-\alpha^2c+c^2-\beta^2c+\beta^4)$ $=-(\alpha^4+\beta^4)+c(\alpha^2+\beta^2)-c^2=-((\alpha+\beta)^4-4\alpha^3\beta-6\alpha^2\beta^2-4\alpha\beta^3)+c(\alpha^2+\beta^2) -c^2$ $=-1+4c\alpha^2+6c^2+4c\beta^2+c(\alpha^2+\beta^2)-c^2=-1+5c(\alpha^2+\beta^2)+5c^2=-1+5c((\alpha+\beta)^2-2\alpha\beta)+5c^2$ $=-1+5c(1-2c)+5c^2$ $\Rightarrow -5c^2+5c -2018 = 0$

The sum of the roots of this quadratic equation are easily calculated as $-\frac{5}{-5} = 1$ .