The nested radicals are back for 2016!

Let us first try to generalize a few things,

$\sqrt{ a+\sqrt{ a- \sqrt{a+ \sqrt{a-\ldots } }}}=m \\ \sqrt{ a-\sqrt{ a+ \sqrt{a- \sqrt{a+\ldots } }}}=n \\ \Rightarrow \begin{cases} \sqrt{a+n}=m \\ \sqrt{a-m}=n \end{cases}$

Squaring both the equations,

$a+n=m^2 \\ a-m=n^2 \\ m^2-n^2=m+n \\ m-n=1$

Now placing $m=\sqrt{a+n}$ in and squaring,

$a+n=n^2+2n+1 \\ n^2+n+1-a=0 \\ n=\frac{-1+\sqrt{1-4(1-a)}}{2}$

Similarly,

$m= \frac{1+\sqrt{1-4(1-a)}}{2}$

So, in our original question, $a=x^2+x+1$ and therefore,

$\begin{aligned} n & = \frac{-1+\sqrt{1-4(1-x^2-x-1)}}{2} \\ &= \frac{-1+\sqrt{4x^2+4x+1}}{2} \\ &= \frac{2x}{2}=x \end{aligned}$

Similarly,

$m=x+1$

So,

$=\frac{\displaystyle\sum_{x=1}^{2016} \sqrt{\left[x^{2} + x + 1 \right]+\sqrt{\left[x^{2} + x + 1\right]-\sqrt{\left[x^{2} + x + 1 \right]+\sqrt{\left[x^{2} + x + 1 \right]-\cdots}}}}}{\displaystyle \sum_{x=1}^{2016} \sqrt{\left[x^{2} + x + 1 \right]-\sqrt{\left[x^{2} + x + 1\right]+\sqrt{\left[x^{2} + x + 1 \right]-\sqrt{\left[x^{2} + x + 1 \right]+\cdots}}}}} \\ = \frac{\displaystyle \sum^{2016}_{x=1}(x+1)}{\displaystyle \sum^{2016}_{x=1}x } = \frac{\displaystyle \sum^{2017}_{x=2}x}{\displaystyle \sum^{2016}_{x=1}x } = \frac{2019}{2017} \\ \Rightarrow 2017-2019=\boxed{-2}$