Integer solution of $ y-x = \sqrt{y+x} $ I have a problem with how to find integer solution of  $\ \ \ y-x = \sqrt{y+x} \ \ \ which \ \ \ \ y>x$ 
$ y^2 -2xy + x^2 -y -x = 0 $
$ y^2 - (2x+1)y + (x^2 - x) = 0 $
$ y =  \frac{(2x+1) \pm \sqrt{8x+1}}{2}  $ and then ? I cannot solve this anymore.
because If it had a solution , I will bring it to apply for  $\ \ (x^b \ mod  \ \ y )= x    $   
It's so cool. Thank you so much for every comments. 

Set $a=\sqrt{y+x}$.  Then $a^2=y+x$ and $a=y-x$ by the condition.  Since $y>x$ in fact $a$ is a positive integer.  Adding and subtracting, respectively, we have $2y=a^2+a$ and $2x=a^2-a$.  Whether $a$ is even or odd, $a^2+a$ and $a^2-a$ are even, so we get integer $x,y$.  Hence the solution set is given by the following (where $a$ is any positive integer):
$$(x,y)=\left(\frac{a^2-a}{2},\frac{a^2+a}{2}\right)$$
