Total no. of ordered pairs $(x,y)$ in $x^2-y^2=2013$ Total no. of ordered pairs $(x,y)$ which satisfy $x^2-y^2=2013$
My try:: $(x-y).(x+y) = 3 \times 11 \times 61$
If we Calculate for positive integers Then $(x-y).(x+y)=1.2013 = 3 .671=11.183=61.33$
my question is there is any better method for solving the given question.
thanks

You can solve this pretty quickly, since you essentially need to solve a bunch of linear systems. One of them is e.g.
$$x-y = 3\times 11$$
$$x+y = 61$$
Just compute the inverse matrix of
$$\left[\begin{array}{cc}
1 & -1\\
1 & 1
\end{array}\right]$$
and multiply that with the vectors corresponding to the different combinations of factors.
(ps. How do I write matrices???)
