DiophantinMO

The equation can be written as $(2x - 1997)^2 + (2y + 1997)^2 \; = \; 2 \times 1997^2$ Since $1997$ is a prime number that is congruent to $1$ modulo $4$ , we see that we can factorize this identity into a product of irreducibles in the Gaussian integers $\mathbb{Z}[i]$ $\big|(2x - 1997) + (2y + 1997)i\big|^2 \; = \; (1 + i)(1 - i)(34 + 29i)^2(34 - 29i)^2$ and hence we deduce that $(2x-1997) + (2y - 1997)i$ must be equal to one of $\begin{aligned} u(1+i)(34 + 29i)^2 & = \; u(-1657 + 2287i) \\ u(1+i)(34 + 29i)(34 -29i) & = \; u(1997 + 1997i) \\ u(1+i)(34 - 29i)^2 & = \; u(2287 - 1657i) \end{aligned}$ where $u \in \{1,-1,i,-i\}$ is a unit. Since $x,y > 0$ we deduce that $(2x - 1997) + (2y + 1997)i \; = \; \pm1657 + 2287i$ so that possible solutions in positive integers are $(x,y) = (1827,145)$ and $(170,145)$ . This makes the answer $1827 + 170 + 2\times145 = \boxed{2287}$ .