Greatest integer.

Consider the following expression.

$\begin{aligned} X & = \sqrt n + \frac 1{\sqrt n + \sqrt{n+2}} \\ & = \sqrt n + \frac {\sqrt{n+2}-\sqrt n}2 \\ & = \frac {\sqrt{n+2} + \sqrt n}2 \\ & = \frac {\sqrt{(\sqrt{n+2} + \sqrt n)^2}}2 \\ & = \frac {\sqrt{2n+2+2\sqrt{\color{#3D99F6}n^2+2n}}}2 \\ & {\color{#D61F06}<} \frac {\sqrt{2n+2+2\sqrt{\color{#D61F06}n^2+2n+1}}}2 \\ & = \sqrt{n+1} \end{aligned}$

So we note that $\lfloor X \rfloor - \lfloor \sqrt n \rfloor < \lfloor \sqrt{n+1} \rfloor - \lfloor \sqrt n \rfloor < \sqrt {n+1} - \sqrt n$ . Now let us check the maximum value of $\sqrt {n+1} - \sqrt n$ . Let us consider $n$ is continuous instead being a discreet integer. Then $\dfrac d{dn} \left(\sqrt{n+1}-\sqrt n\right) = \dfrac 1{2\sqrt{n+1}} - \dfrac 1{2\sqrt n} < 0$ for all $n > 0$ , which is a decreasing function. Therefore the maximum value of $\sqrt {n+1} - \sqrt n = 1$ , when $n=0$ . Hence we have $\left \lfloor \sqrt n + \dfrac 1{\sqrt n + \sqrt{n+2}}\right \rfloor - \left \lfloor \sqrt n \right \rfloor < 1 = \boxed{0}$ .

$\frac { 1 }{ \sqrt { 2018! } + \sqrt { 2018! + 2 } }$ is too close to 0 that can be ignored,so $\ \left\lfloor \sqrt { 2018! } + \frac { 1 }{ \sqrt { 2018! } + \sqrt { 2018! + 2 } } \right\rfloor - \left\lfloor \sqrt { 2018! } \right\rfloor=\left\lfloor \sqrt { 2018! } \right\rfloor-\left\lfloor \sqrt { 2018! } \right\rfloor=0$ .

The term $\lfloor \sqrt{n}+\frac{1}{\sqrt{n}+\sqrt{n+2}}\rfloor$ can be written as

$\lfloor \sqrt{n}+\frac{1}{\sqrt{n}+\sqrt{n+2}}\rfloor= \lfloor \sqrt{n}+\frac{-\sqrt{n}+\sqrt{n+2}}{(\sqrt{n+2})^2-(\sqrt{n})^2}\rfloor =\lfloor \sqrt{n}+\frac{-\sqrt{n}+\sqrt{n+2}}{2}\rfloor=\lfloor \frac{\sqrt{n}+\sqrt{n+2}}{2} \rfloor$

Then write the main expression as $\lfloor \frac{\sqrt{n+2}+\sqrt{n}}{2} \rfloor -\lfloor \frac{\sqrt{n}+\sqrt{n}}{2} \rfloor = \lfloor a \rfloor- \lfloor b\rfloor= \lfloor m+\epsilon_1\rfloor -\lfloor n+\epsilon_2\rfloor$ , where $m,n$ are positive integers and $0 \leq \epsilon_1 , \epsilon_2 <1$ .

As the first fact, one can show, is that $a-b = \lfloor \frac{\sqrt{n+2}-\sqrt{n}}{2} \rfloor = \frac{\sqrt{3}-1}{2} < 0.5$ .

Also, $\ 0 \leq \epsilon_1 , \epsilon_2 <0.5$ , because both $a$ and $b$ are results of a real number divided by two, which makes the fractional part less than $0.5$ . Consequently, $-0.5<\epsilon_1-\epsilon_1<0.5 \implies 0.5<1-\epsilon_1-\epsilon_1<1.5$ .

Finally, in case $m=n+1$ then $a-b=1+\epsilon_1-\epsilon_2>0.5$ , which contradicts the first fact. Therefore, $m=n$ .