Generalized Problem

In general:

Let $a$ be a prime number.

Find largest positive integer $x$ such that $x^2 + a^2 x = j^2$ , where $j$ is a positive integer.

$x^2 + a^2 x = j^2 \implies (2x + a^2)^2 - a^4 = (2j)^2 \implies (2x + a^2)^2 - (2j)^2 = a^4 \implies (2x - 2j + a^2) (2x + 2j + a^2) = a^4$

$2x + 2j = a^4 - a^2$

$2x - 2j = 1 - a^2$

$\implies x = (\dfrac{a^2 - 1}{2})^2 = k^2 \in \mathbb{N}$ and $j = \dfrac{(a - 1)(a^3 + a^2 + 1)}{4} = m * l \in \mathbb{N}$

$2x + 2j = a^3 - a^2$

$2x - 2j = a - a^2$

$\implies x = a (\dfrac{a - 1}{2})^2 = 2 h n^2 \in \mathbb{N}$ and $j = a (\dfrac{a^2 - 1}{2})^2 = 2 q s^2 \in \mathbb{N}$

$2x + 2j = 0$

$2x - 2j = 0$

$\implies x = j = 0$ clearly $x = 0 \neq x_{max}.$

For $a > 1$

$(a^2 - 1)^2 - a(a - 1)^2 = (a - 1)^2 (a + 1)^2 - a (a - 1)^2 = (a - 1)^2 (a^2 + a + 1) > 0 \implies (a^2 - 1)^2 > a(a - 1)^2 \implies$ $(\dfrac{a^2 - 1}{2})^2 > a (\dfrac{a - 1}{2})^2$

$\therefore x = x_{max} = (\dfrac{a^2 - 1}{2})^2$

Using $a = 7919 \implies x_{max} = \boxed{983153583878400}$ .