General solution for nonlinear Diophantine equation

Question

As the title says I'm looking for a general solution to a Diophantine equation of the form:

$y^2 = x^2 + kx - m$

Where $x$ and $y$ are both positive integers. I know that $k$ and $m$ will always be a multiple of 2 if that helps.

Here are a few examples that I solved through brute force.

1. $y^2 = x^2 + 30 x - 28$
   $x = 2$
   $y = 6$
   
   
2. $y^2 = x^2 + 474 x - 554$
   $x = 45$
   $y = 151$
   
   
3. $y^2 = x^2 + 1802 x - 3018$
   $x = 37$
   $y = 255$

I'm a bit of a dummy so a thorough explanation would be much appreciated. Thanks in advance!

Answer 1

Hint: Write $k=2b$. Then $y^2 = (x+b)^2 -b^2 - m$ and so $m+b^2=(x+b)^2 -y^2$ is a difference of two squares. Now use this well-known fact, which is easy to prove:

$N \in \mathbb Z$ is a difference of two squares iff $N$ is odd or a multiple of $4$.

To be explicit, if $N=uv$ with $u$ and $v$ of the same parity, then $$ N=uv=\left(\dfrac{u+v}2\right)^2 - \left(\dfrac{u-v}2\right)^2 $$

Therefore, the general solution of $y^2 = x^2 + kx - m$ follows from factoring $N=m+(\frac k2)^2$ as two factors of the same parity and using the formula above to find $x+\frac k2$ and $y$.

Let's see how this works for your first example: $m+b^2=28+15^2=253=11\cdot 23=17^2-6^2$. Therefore, $x+15=17, y=6$. This is the only solution.

Answer 2

Above equation shown below, has parametric soution:

$y^2=x^2+kx-m$

$(x)= (2w)(p^2-6p+23)$

$(y)= (2w)(3p^2-9p-42)$

$(k)= (30w)(p^2+6p+5)$

$(m)= (28w^2)(p^2+6p+5)(p^2+14)$

Where $w=1/(p^2+14)$

For p=3 we get:

$84^2=28^2 +960(28)-20608$