Remainders Are Fun 6

$\begin{aligned} x_i & = \frac {i-1}{(\sqrt i+1)(\sqrt [4] i+1)(\sqrt [8] i+1)(\sqrt [16] i+1)} & \small \color{#3D99F6} \text{Let }a = \sqrt [16] i . \\ & = \frac {a^{16}-1}{(a^8+1)(a^4+1)(a^2+1)(a+1)} \\ & = \frac {(a^8+1)(a^8-1)}{(a^8+1)(a^4+1)(a^2+1)(a+1)} \\ & = \frac {(a^8+1)(a^4+1)(a^4-1)}{(a^8+1)(a^4+1)(a^2+1)(a+1)} \\ & = \frac {(a^8+1)(a^4+1)(a^2+1)(a^2-1)}{(a^8+1)(a^4+1)(a^2+1)(a+1)} \\ & = \frac {(a^8+1)(a^4+1)(a^2+1)(a+1)(a-1)}{(a^8+1)(a^4+1)(a^2+1)(a+1)} \\ & = a - 1 \\ & = \sqrt [16] i - 1 \end{aligned}$

Therefore, we have:

$\begin{aligned} \sum_{i=1}^{2017} (x_i + 1)^{48} & = \sum_{i=1}^{2017} (\sqrt [16] i - 1 + 1)^{48} = \sum_{i=1}^{2017} i^3 = \left( \frac {2017 \times 2018}2 \right)^2 = (2017 \times 1009)^2 \\ & \equiv (17 \times 9)^2 \equiv 23409 \equiv \boxed{409} \text{ (mod 1000)} \end{aligned}$