Total no. of ordered pairs $(x,y)$ in $x^2-y^2=2013$

Question

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

Answer 1

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???)