2 Quadratics, Both Alike in Dignity

Firstly, observe that the equation can be restated into: $y = \frac{x^2 + 19(53)(2x-1)}{x^2 - 19(53)(2x-1)}$ , for integers $x,y$ . Let $a = 2x-1$ . $y = \frac{x^2 + 19(53)a}{x^2 - 19(53)a} = 1 + \frac{2 \cdot 19(53)a}{x^2 - 19(53)a}$ Notice for $y \in \mathbb{Z}$ , $(x^2 - 19(53)a) | (2 \cdot 19(53)a)$ . From here, there are 2 cases to consider:

Case 1: When the denominator is equal to 1 or -1 Thus, $x^2 - 19(53)a = \pm 1$ . But substituting $a = 2x-1$ in, we get $x^2 - 2 \cdot 1007x + 1007 = \pm 1$ . Using the quadratic formula, we see that these 2 equations have no integer solution for $x$ . Thus, no solution for Case 1.

Case 2: When denominator $\neq 1, -1$

That implies that $19 | x^2$ and $53 | x^2$ . Since $19$ and $53$ are both primes, $19 | x, 53 | x$ .

Firstly, $x$ can be $0$ since any number divides $0$ . Then, $y = -1$ .

Suppose $x\neq 0$ , $19 \cdot 53 = 1007 | x$ . Let $x=1007k$ , for some integer $k$ . Then we see that the expression (of the fraction - we ignore the 1 first) becomes: $\frac{2 \cdot 1007(2 \cdot 1007k - 1)}{(1007k)^2 - 1007(2\cdot 1007k - 1)}$ $= \frac{1007(4 \cdot 1007k - 2))}{1007(1007k^2 - 2\cdot 1007k + 1)}$ $= \frac{4 \cdot 1007k - 2}{1007k^2 - 2\cdot 1007k + 1}$ . Once again, we want the above to be an integer, hence denominator divides numerator evenly. This implies that: $1007k^2 - 2\cdot 1007 + 1 \leq 4 \cdot 1007k - 2$ $1007k^2 - 6\cdot 1007k + 3 \leq 0$ . Using the quadratic formula, we see that the roots, $k$ , of the equations are approximately (larger than) $0$ and (smaller than) $6$ . We also know that the function is an increasing one (meaning that the shape of this quadratic is a U shape - you can use differentiation or test out some small values to verify).

Hence, to find the other solutions for Case 2, you just have to try $k =1,2,3,4,5$ . You will realise that $k=2$ yields, $x=2\cdot 1007 = 2014$ and $y = \frac{4 \cdot 1007(2) - 2}{1007(2)^2 - 2\cdot 1007(2) + 1} + 1=8055$ . Hence sum of all integral $y$ is $-1 + 8055 = \boxed{8054}$ .