Quadratic Diophantine equations in two variables

Question

I am interested in integer solutions of the following bivariate quadratic equation

$$x^2-y^2=(x+y)(x-y)=c,$$ with $x>y$. I know that the usual way is for one to assume that if there exist two positive integers $c_1$ and $c_2$ such that $c=c_1c_2$, then by setting $$x+y=c_1\\ x-y=c_2$$ or vice versa, one obtains $x=\frac{c_1+c_2}{2}$, from which the value of $y$ can also be obtained. For very large $c$, this method of factorizing is not efficient. My question is whether there are some quick ways to solve equations of the above form.

Answer 1

Every integral solution corresponds to a factorization $c=u\cdot v$ with $u\equiv v\pmod{2}$. Finding all integral solutions is therefore (nearly) equivalent to factoring $c$ completely, which is hard for large $c$. However, finding some integral solutions is easy:

First note that if $c\equiv2\pmod{4}$ then there are no integral solutions. If $c$ is odd then $c=c\cdot1$ and so $$x:=\tfrac{c+1}{2},\qquad y:=\tfrac{c-1}{2},$$ is an integral solution. If $c\equiv0\pmod{4}$ then $c=(2d)\cdot2$ for some integer $d$, and so $$x:=d+1,\qquad y=d-1,$$ is an integral solution.

Answer 2

Hint: If your main equation is $x^2 - y^2 = c$ , express $y$ in terms of $x$ (you'll get $y = \pm\sqrt{x^2-c}$ ). If you can express $c$ as $2xm$ for integers $x$ and $m$, you get: $$y = \pm\sqrt{(x-m)^2-m^2}$$

Now , from a constant $x$ (for eg., $x = 1$ or whatever), you can one way or the other deduce the solution by varying $m$.

This is what I believe as a 15 year old. Please correct me if I went wrong anywhere, or even if the whole answer is useless.

Here's what I did on Geogebra. This kind of thought may also work, I think.

NB : I used that circle to see what value of $x$ would be feasible to begin with as per the above method.