Diophantine solution to a fraction

Question

How can we find solutions to the following equation:

$$ y=\dfrac{x^2-1085}{14718-2x}$$

where $x,\ y$ are integers.

Answer 1

Manipulate the equation as follows:

$y = \dfrac{x^2-1085}{14718-2x}$

$14718y-2xy = x^2-1085$

$y^2+14718y + 7359^2 = x^2+2xy+y^2+7359^2-1085$

$(y+7359)^2 = (x+y)^2+(7359^2-1085)$

$(y+7359)^2 - (x+y)^2 = 54153796$

$(x+2y+7359)(7359-x) = 2^2 \cdot 1993 \cdot 6793$

Since $x+2y+7359$ and $7359-x$ have the same pairity, either both are even or both are odd. We quickly see that they can't both be odd. So, we have just $8$ cases to check:

$(x+2y+7359,7359-x) = (2,2\cdot 1993 \cdot 6793)$

$(x+2y+7359,7359-x) = (2\cdot 1993,2 \cdot 6793)$

$(x+2y+7359,7359-x) = (2\cdot 6793,2 \cdot 1993)$

$(x+2y+7359,7359-x) = (2\cdot 1993 \cdot 6793,2)$

$(x+2y+7359,7359-x) = (-2,-2\cdot 1993 \cdot 6793)$

$(x+2y+7359,7359-x) = (-2\cdot 1993,-2 \cdot 6793)$

$(x+2y+7359,7359-x) = (-2\cdot 6793,-2 \cdot 1993)$

$(x+2y+7359,7359-x) = (-2\cdot 1993 \cdot 6793,-2)$

Solve each of these $8$ systems to get the solutions.

Answer 2

If this is a standard diophantine equation, the solutions have to be positive integers. Therefore $0 < 2x < 14718$ (if the numerator and denominator are positive) or $x^2 < 1085$ (if they are negative), so the possible values for $x$ are immediately limited.

You can simply try all $x$ in this range, or you can try to be clever and write $x^2-y(14718-2x)-1085 = 0$ and look at the discriminant of this quadratic.