New Year's Countdown Day 18: Calc and NT for 2018

For simplicity, let $a = 2018.$ Since

$\begin{aligned} \displaystyle \int \left ( \sqrt{\dfrac{a + x}{x}} + \sqrt{\dfrac{x}{a + x}} \right ) \, dx &= \displaystyle \int \dfrac{(a + x) + x}{\sqrt{x(a + x)}} \, dx \\ &= \displaystyle \int \dfrac{a + 2x}{\sqrt{x(a + x)}} \, dx \\ &= \displaystyle \int \dfrac{1}{\sqrt{u}} \, du \quad (\text{Let } u = x(a + x), du = a + 2x) \\ &= 2 \sqrt{u} + C \\ &= 2 \sqrt{x(a + x)} + C \\ &= 2 \sqrt{x(2018 + x)} + C, \end{aligned}$

we have

$\begin{aligned} \displaystyle \int_0^p \left ( \sqrt{\dfrac{2018 + x}{x}} + \sqrt{\dfrac{x}{2018 + x}} \right ) \, dx &= \lim_{n \rightarrow 0^+} \displaystyle \int_n^p \left ( \sqrt{\dfrac{2018 + x}{x}} + \sqrt{\dfrac{x}{2018 + x}} \right ) \, dx \\ &= \lim_{n \rightarrow 0^+} \left [2 \sqrt{x(2018 + x)} \right]_n^p \\ &= \lim_{n \rightarrow 0^+} \left (2 \sqrt{p(2018 + p)} - 2 \sqrt{n(2018 + n)} \right) \\ &= 2 \sqrt{p(2018 + p)}. \end{aligned}$

Since $p$ is an integer, the expression $2 \sqrt{p(2018 + p)}$ is an integer if $\sqrt{p(2018 + p)}$ is. Let $b$ be a positive integer. We have

$\begin{aligned} \sqrt{p(2018 + p)} &= k \\ p(2018 + p) &= k^2 \\ p^2 + 2018p - k^2 &= 0. \end{aligned}$

In order for this quadratic to have integer solutions, its discriminant must equal a square number. Letting $c$ be another positive integer, we have

$\begin{aligned} 2018^2 - 4(1)(-k^2) &= c^2 \\ 2018^2 + (2k)^2 &= c^2 \\ c^2 - (2k)^2 &= 2018^2 \\ (c + 2k)(c - 2k) &= 2018^2. \end{aligned}$

Both $c + 2k$ and $c - 2k$ are positive integers, and their sum is $(c + 2k) + (c - 2k) = 2c,$ which is a multiple of 2. Because $2018^2 = 2^2 \cdot 1009^2,$ we must partition the 2s between the two factors as such:

$\begin{aligned} c + 2k &= 2 \cdot 1009^2 \\ c - 2k &= 2. \end{aligned}$

Solving this system for $c$ yields $c = 1009^2 + 1.$ Plugging this value into the quadratic formula yields $p = \dfrac{-2018 \pm (1009^2 + 1)}{2}.$ Since we want positive $p,$ we take the plus sign to get $p = \dfrac{-2018 + 1009^2 + 1}{2} = 508032,$ which has digit sum $5 + 8 + 3 + 2 = \boxed{15}.$