Contest math integer doublet equation Can anyone help me with this?
Find all ordered pairs $(x, y)$ of positive integers $x$, $y$ such that $$x^2 + 4y^2 = (2xy − 7)^2$$
 A: We have 
$$\left(2y(x-1)-7\right)\left(2y(x+1)-7\right)=x^2.$$ Obviously $x \ne 1$ and $x$ is odd, i.e. $x \ge 3$. Since $y \ge 3$ implies $2y(x-1)-7 \ge x+5x-13 > x$, you only need to check $y=1$ and $y=2$.
A: $$x^2 + 4y^2 - 4xy +4xy = (2xy-7)^2$$
$$(x+2y)^2 = (2xy-7)^2 + 4xy$$
$$         = 4x^2y^2 - 28xy +4xy +49$$
         $$= 4x^2y^2 - 24xy +49$$
         $$= (4x^2y^2 - 24xy +36) +13$$
$$(x+2y)^2 - (2xy -6)^2 +13$$
$$(x+2y +2xy -6) (x+2y - 2xy +6) = 13$$
 The product of these two terms can only be 13 and 1
either
$(x+2y +2xy -6) = 13$ and $(x+2y - 2xy +6) = 1$
or 
$(x+2y+ 2xy - 6) = 1$ and $(x+2y - 2xy + 6) = 13$
Case i)
$x+2y+2xy - 6 = 13$
$x+2y - 2xy + 6) = 1$
Adding these two
We get
$2x+4y = 14$ 
Subtracting 
We get
$4xy -12 = 12$ This gives $xy = 6$
Case ii)
$x+2y+2xy - 6 = 1$
$x+2y - 2xy + 6 = 13$
Adding these two
We get
$2x+4y = 14$ 
Subtracting 
We get
$4xy -12 = -12$ This gives $xy = 0$
Case ii) is not possible because x and y postive integers
Case i) will lead to possible combinations of $x= 6$ or $y =1$, or $x=3$ and $y = 2$ or $x=2$ and $y = 3$ or $x= 1$ and $y = 6$ as xy are positive integers
Substituting back you will see that the ordered pair(x,y) is (3,2) is the only solution
Thanks
Satish
