Originally Posted by
angel.white I'll take a stab at it.
Premise: x, y are integers
$\displaystyle x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}$
$\displaystyle 2x = -7y \pm \sqrt{49y^2 + 12}$
setup case 1: $\displaystyle -7y + \sqrt{49y^2 + 12} = 2x$
$\displaystyle -7y + \sqrt{49y^2 + 12} = 2x$
$\displaystyle \sqrt{49y^2 + 12} = 2x+7y$
$\displaystyle 49y^2 + 12 = 4x^2+28xy+49y^2$
$\displaystyle 12 = 4x^2+28xy$
$\displaystyle 3 = x^2+7xy$
setup case 2: $\displaystyle -7y - \sqrt{49y^2 + 12} = 2x$
$\displaystyle \sqrt{49y^2 + 12} = -2x-7y$
$\displaystyle 49y^2 + 12 = 4x^2 +28xy+49y^2$
$\displaystyle 12 = 4x^2 +28xy$
$\displaystyle 3 = x^2 +7xy$
Continued from $\displaystyle 3 = x^2 +7xy$
So it can be seen that regardless of which setup case we choose, positive or negative, we end up evaluating the same equation of $\displaystyle 3 = x^2 +7xy$
Note that if x = 0 then 3 = 0, which is false, so x cannot be zero.
$\displaystyle \frac 3x =x + 7y$
Now, because x and y are both integers, x + 7y must be an integer. Which means that 3/x must be an integer.
So $\displaystyle x \in \{ -3, -1, 1, 3\}$
proof case 1: x = -3
$\displaystyle \frac 3x =x + 7y$
$\displaystyle \frac 3{-3} =-3 + 7y$
$\displaystyle \frac 27 = y$
But this contradicts that y is an integer, so $\displaystyle x \not{=} -3$
proof case 2: x = -1
$\displaystyle \frac 3x =x + 7y$
$\displaystyle \frac 3{-1} =-1 + 7y$
$\displaystyle -\frac 27 = y$
But this contradicts that y is an integer, so $\displaystyle x \not{=} -1$
proof case 3: x = 1
$\displaystyle \frac 3x =x + 7y$
$\displaystyle \frac 3{1} =1 + 7y$
$\displaystyle \frac 27 = y$
But this contradicts that y is an integer, so $\displaystyle x \not{=} 1$
proof case 4: x = 3
$\displaystyle \frac 3x =x + 7y$
$\displaystyle \frac 3{3} =3 + 7y$
$\displaystyle -\frac 27 = y$
But this contradicts that y is an integer, so $\displaystyle x \not{=} 3$
Therefore there are no integer values of x for which y can be an integer also.
So the solution to the problem "find all integer solutions (x, y)" is the empty set.