Summing Complex Fractional Exponents

If we consider the Gauss sum $G(a,c) \; = \; \sum_{r=0}^{c-1} \mathrm{exp}\left( \frac{2 \pi i a r^2}{c} \right)$ then we can use standard identities concerning Gauss sums to deduce that $A \; = \; G(1008,2017) \; =\; \epsilon_{2017} \sqrt{2017} \left( \frac{1008}{2017}\right) \; = \; \sqrt{2017}\left(\frac{1008}{2017}\right)$ where $\epsilon_m = 1$ if $m \equiv 1 \pmod{4}$ , while $\epsilon_m \equiv i$ if $m \equiv 3 \pmod{4}$ , and $\left(\frac{p}{q}\right)$ is the Jacobi symbol, while $\begin{array}{rcl} B^\star & = & \displaystyle \sum_{r=0}^{2015} \mathrm{exp}\left( \frac{i 2017 \pi r^2}{2016}\right) \; = \; \sum_{r=0}^{2015} \mathrm{exp} \left(\frac{2\pi i 2017 r^2}{4032}\right) \\ & = & \displaystyle \frac12\left[ \sum_{r=0}^{2015} \mathrm{exp}\left(\frac{2\pi i 2017 r^2}{4032}\right) + \sum_{r=0}^{2015} \mathrm{exp}\left(\frac{2\pi i 2017(2016+r)^2}{4032}\right)\right] \\ & = & \displaystyle \frac12\sum_{r=0}^{4031} \mathrm{exp}\left(\frac{2\pi i 2017 r^2}{4032}\right) \; = \; \tfrac12G(2017,4032) \\ & = & \displaystyle \tfrac12(1+i)\epsilon_{2017}^{-1}\sqrt{4032}\left(\frac{4032}{2017}\right) \; = \; (1+i)\sqrt{1008}\left(\frac{1008}{2017}\right) \end{array}$ and hence $\frac{A}{B^\star} \; = \; \frac{1}{1+i}\sqrt{\frac{2017}{1008}}$ so that $\mathfrak{Re} \left(\frac{A}{B}\right) \; = \; \tfrac12\sqrt{\frac{2017}{1008}} \; = \; \sqrt{\frac{2017}{4032}}$ making the answer $2017 + 4032 = \boxed{6049}$ .