An interesting problem in Number Theory

$\begin{aligned} a^2 - b^2 & = 2018 - 2a \\ a^2 +2a + 1 - b^2 & = 2018 + 1 \\ (a+1)^2 - b^2 & = 2019 \\ (a+b+1)(a-b+1) & = 2019 = \begin{cases} 2019 \times 1 \\ 673 \times 3 \end{cases} \end{aligned}$

For $a, b > 0$ , $a+b+1 > a-b+1$ , therefore

$\begin{cases} a+b+1 = 2019 & \implies a+b = 2018 \\ a - b+1 = 1 & \implies a - b = 0 & \implies \red{a = b} = 1009, & \small \red{\text{unacceptable}} \end{cases}$

$\begin{cases} a+b+1 = 673 & \implies a+b = \boxed{672} \\ a - b+1 = 3 & \implies a - b = 2 & \implies \blue{a = 337 \ne b = 335}, & \small \blue{\text{acceptable}} \end{cases}$

a^2 - b^2 = 2018- 2a

a^2 - b^2 + 2a = 2018 =

a^2 - b^2 + 1 + 2a = 2019 =

(a^2 + 2a + 1) - (b^2) =

(a + 1)*2 - (b^2) = (a+1+b)(a+1-b) = 2019

Prime factorization of 2019 = 673 * 3

As a, and b are positive integers a + b + 1 > a -b + 1

Therefore a+1+ b = 673 , follows that a + b = 672