How to find the number of solutions to the given equation? How do I find the number of integer solutions to the following equation:
$$\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{y}} = \frac{1}{2016}$$
I think $x, y$ should be perfect squares but I am not sure... Can anybody help me with this proof and go ahead with the question?
 A: If $\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{y}} = \frac{1}{2016}$, then $\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}}$ should be rational, since their product is $\frac 1x - \frac 1y$ is also rational. Under the assumption $\frac1{\sqrt x}$ and $\frac1{\sqrt y}$ should be both rational.
A: It is a good assumption that $x$ and $y$ are perfect squares.  
But is it a valid one?
If $\frac 1{\sqrt x} - \frac 1{\sqrt y} = \frac 1{2016}$ then 
$\frac 1{\sqrt x} = \frac 1{2016} + \frac 1{\sqrt y}$ and $
$\frac 1x = \frac 1{2016^2} + 2\frac 1{2016}\frac 1{\sqrt y} + \frac 1y$.
So $\sqrt y$ is rational. so  $y$ is an integer. So $y$ is a perfect square.
$x$ is a perfect square by the same reasoning.
So now substitute $x = a^2$ and $y = b^2$ and solve:
$\frac 1a - \frac 1b = \frac 1{2016}$ or 
$\frac {b-a}{ab} = \frac 1{2016}$
$2016(b-a) = ab$
Case 1:  If we assume $a$ and $b$ are relatively prime then $ab$ will not have any factors in common with $b-a$.  So $b-a = 1$ and wet have
$2016 = a(a+1)$ which is not possible.  Use the quadratic equation.
Case 2: Let $\gcd(a,b) = d$ and $a = cd$ and $b = ed$.  We have
$2016(e-c) = dec$.  Now $e$ and $c$ are relatively prime so $e-c = 1$ and we have
$2016 = dc(c+1)$ .
$2016 = 32*9*7$  we need to find two factors that are only one apart.  To be one apart means one is even and the other odd. There are only so many odd factors.  so we can have:
$1$ and $2$; $3$ and $2$; $3$ and $4$; $7$ and $6$; $7$ and $8$; $9$ and $8$.
I will  do one solution:
If we take $6$ and $7$ then we have $c=6$ and $e=7$ and $d = \frac {2016}{6*7} = 48$.  So $a = 6*48$ and $b = 7*48$.  And $x = (6*48)^2= 82944$ and $y = (7*48)^2 = 112896$.
And $\frac 1{\sqrt{82944}} - \frac 1{\sqrt{112896}} =$
$\frac 1{6*48} - \frac 1{7*48} =$
$\frac 1{48}(\frac 16 - \frac 17) =$
$\frac 1{48}*\frac 1{42} = \frac 1{48*42} = \frac 1{2016}$.
The others will be done the same way.  There will be $6$ pairs of solutions.
