An algebra problem by Ilham Saiful Fauzi

$\begin{aligned} S & = \frac{2017}{2\sqrt{1}+1\sqrt{2}}+\frac{2017}{3\sqrt{2}+2\sqrt{3}}+\frac{2017}{4\sqrt{3}+3\sqrt{4}}+\cdots+\frac{2017}{2017\sqrt{2016}+2016\sqrt{2017}} \\ & = \sum_{n=1}^{2016} \frac {2017}{(n+1)\sqrt n + n \sqrt{n+1}} \\ & = 2017 \sum_{n=1}^{2016} \frac {(n+1)\sqrt n - n \sqrt{n+1}}{\left((n+1)\sqrt n + n \sqrt{n+1}\right)\left((n+1)\sqrt n - n \sqrt{n+1}\right)} \\ & = 2017 \sum_{n=1}^{2016} \frac {(n+1)\sqrt n - n \sqrt{n+1}}{n(n+1)^2 - n^2(n+1)} \\ & = 2017 \sum_{n=1}^{2016} \frac {(n+1)\sqrt n - n \sqrt{n+1}}{n(n^2 + 2n + 1 - n^2-n)} \\ & = 2017 \sum_{n=1}^{2016} \frac {(n+1)\sqrt n - n \sqrt{n+1}}{n(n+1)} \\ & = 2017 \sum_{n=1}^{2016} \left(\frac 1{\sqrt n} - \frac 1{\sqrt{n+1}} \right) \\ & = 2017 \left(\frac 1{\sqrt 1} - \frac 1{\sqrt{2017}} \right) \\ & = 2017 - \sqrt{2017} \end{aligned}$

$\implies a + b + c = 2017-1+2017 = \boxed{4033}$