Question

(a) If $2x^2-3xy-2y^2=7$, then prove that there will be only two integral pairs $(x,y)$ satisfying the above relation.
(b) If $2x^2{y}^2+y^2-6x^2-12=0$, then prove that there will be only four integral pairs $(x,y)$ satisfying the above relation.

Answer 1

we are given that $(2x+y)(x-2y)=7$
Since x and y are to be integers, hence L.H.S. isthe product of two integers and R.H.S. is also the product of two integers $7$ and $1$ or $1$ and $7$ or $-7$ and $-1$ or $-1$ and $-7$.
Hence we can choose
$2x+y=7$ and $x-2y=1$.......(1)
$2x+y=1$ and $x-2y=7$.......(2)
$2x+y=-7$ and $x-2y=-1$......(3)
$2x+y=-1$ and $x-2y=-7$......(4)
Solve them as usual and only (1) and (3) give integral solutions as 3,1 for (1) and $-3,-1$ for (3).
Both (2) and (4) when solved do not give integral values of x and y.
(b) The given equation can be written as
$2x^2(y^2-3)=12-y^2$
$\therefore$ $2x^2=\frac{12-y^2}{y^2-3}$ $\therefore$ $2x^2+1=\frac{9}{y^2-3}$.......(1)
Now L.H.S. for integral values of x is clearly an odd positive integer and hence $\frac{9}{y^2-3}$ must also
be an odd positive integer for integral values of y:
By inspection this is possible only
when $y^2=4$ $\therefore$ $y=2,-2$
$\therefore$ $2x^2+1=9$ or $x^2=4$ $\therefore$ $x=2,-2$
Hence $(2,2),(2,-2),(-2,2)$ and $(-2,-2)$
are the only four possible integral solutions.