Question

If $2x^2{y}^2+y^2-6x^2-12=0,$ then number of integral pairs $(x,y)$ satisfying is/are

• A. $1$
• B. $2$
• C. $4$
• D. $6$

Answer 1

The correct option is C $4$

$2x^2{y}^2+y^2-6x^2-12=0$
$\Rightarrow 2x^2(y^2-3)=12-y^2$
$\Rightarrow 2x^2=\frac{12-y^2}{y^2-3}$
$\Rightarrow 2x^2+1=\frac{9}{y^2-3}$

Now, LHS is an odd positive number for integral values of $x.$

$\therefore \frac{9}{y^2-3}$ must be an odd positive integer for integral values of $y$

$\therefore y^2=4$ is the only possibility.
$\Rightarrow y=-2,2$
$\therefore 2x^2+1=9$
$\Rightarrow x^2=4$
$\Rightarrow x=2,-2$

$\therefore$ Possible pairs are $(2,2),(2,-2),(-2,2),(-2,-2)$