Solving second-degree diophantine equation

Question

I'm interested in the following Diophantine equation:

$y^2 - y - x^2 + x = 2xy$

I've managed to find some solutions like $(6, 15)$ and $(35, 85)$, but I need some general method of solving this type of equations. As I understood, it's a hyperbola, which makes it harder to study.

Any help?

Answer 1

Solution is:

$x=m(1-2n)$

$y=mn$

Where:

$m=(3n-1)/(n^2+2n-1)$

For, $n=(5/12)$ we have:

$(x,y)=(6,15)$

Answer 2

The equation can be solved for either variable in terms of the other by $y$ is easier than x.

$$y = \frac{\pm\sqrt{8 x^2 + 1} + 2 x + 1}{2}\rightarrow$$ $$ x\in\{0,1,6,35,204,1189,5008,40391...\}\implies y\in\{1,3,15,85,493,2871,16731,97513...\}$$

I believe these are all the integer pairs for $x\le100000$.

Answer 3

$y^2 - y - x^2 + x = 2xy \implies (2 y - 2 x - 1)^2 - 2 (2 x)^2 = 1$

This Pell equation.

gp-code:

pell_d2c1()=
{
 D= 2; C= 1;
 Q= bnfinit('x^2-D, 1);
 fu= Q.fu[1]; \\print("Fundamental Unit: "fu);
 N= bnfisintnorm(Q, C); \\print("Fundamental Solutions (Norm): "N"\n");
 for(i=1, #N, ni= N[i];
  for(j=0, 48,
   s= lift(ni*fu^j);
   X= abs(polcoeff(s, 0)); Y= abs(polcoeff(s, 1));  
   if(X^2-D*Y^2==C,
    x= Y/2; y= (X+2*x+1)/2;
    if(x==floor(x) & y==floor(y),
     print("("x", "y")")
    )
   )
  )
 )
};

Some first solutions:

(0, 1)
(1, 3)
(6, 15)
(35, 85)
(204, 493)
(1189, 2871)
(6930, 16731)
(40391, 97513)
(235416, 568345)
(1372105, 3312555)
(7997214, 19306983)
(46611179, 112529341)
(271669860, 655869061)
(1583407981, 3822685023)
(9228778026, 22280241075)
(53789260175, 129858761425)
(313506783024, 756872327473)
(1827251437969, 4411375203411)
(10650001844790, 25711378892991)
(62072759630771, 149856898154533)
(361786555939836, 873430010034205)
(2108646576008245, 5090723162050695)
(12290092900109634, 29670908962269963)
(71631910824649559, 172934730611569081)
(417501372047787720, 1007937474707144521)