How to solve this Pell's equation $x^{2} - 991y^{2} = 1 $ How to solve the following Pell's equation?
$$x^{2} - 991y^{2} = 1 $$
where $(x, y)$ are naturals.
The answer is $$x = 379,516,400,906,811,930,638,014,896,080$$ 
$$y = 12,055,735,790,331,359,447,442,538,767$$
I can't think of any way apart from brute force. Please help.
Also, what is the general way of solving any Pell's equation?
I read the wiki article on it but couldn't get any general method to solve it.
 A: Following the  link, taking  (379,12,  which was calculated by brute force) as the fundamental solution,  we can get the generic solution.
I'm trying to replace the brute force with human intelligence(!) using the followings:
first ,  second and continued fraction of $\sqrt{991}$
A: I think it can be solved without using complex operators,and the solution is here: When is $991n^2 +1$ a perfect square?
A: All you really have to do is brute force the smallest pair (x, y) and the rest is cake. Once you find it, throw it into a 2×2 matrix with both diagonal entries that are the x coordinate and the Anti-diagonal entries being y and y*991 (doesnt matter). The determinant of this guy is 1. Diagonalize the matrix, take to a power s, where s is n integer, and then multiply it back out. The determinant is still one. Thats how you get all other solutions.
A: Solving method from Pell’s equation without irrational numbers by Norman Wildberger
in pari/gp:
njw(d)=
{
 L= [1,0;1,1]; R= [1,1;0,1];
 A= [1,0;0,-d]; Q= A; N= 1;
 while(1,
  a= Q[1,1]; b= Q[1,2]; c= Q[2,2];
  t= a+2*b+c;
  if(t<0, Q= R~*Q*R; N= N*R, Q= L~*Q*L; N= N*L);
  if(Q==A, break())
 );
 return(N[,1]~)
};

? njw(991)
%5 = [379516400906811930638014896080, 12055735790331359447442538767]

Also this method online in Geogebra
A: $$x^2 - 991y^2 = 1$$ -------(1)
Assuming $y \not= 0$ divide both sides of the equation by $y^2$
i.e. $$(x/y)^2 - 991 = (1/y)^2$$
i.e. $(x/y)^2 - (1/y)^2 = 991$
i.e. $(x/y - 1/y)(x/y + 1/y) = (1)(991)$
i.e. $(x/y - 1/y)(x/y + 1/y) = (496 - 495)(496 + 495)$
i.e. $x/y = 496$ and $1/y = 495$
i.e. $x = (496/495)$ and $y = 1/495$
The equation represents a hyperbola of the form $$x^2/a^2 - y^2/b^2 = 1$$
i.e. the equation is $$x^2/(1^2) - y^2/(\sqrt {1/991})^2 = 1$$
foci of this hyperbola are the co ordinates $(\pm c, 0)$
Here $c = \sqrt{a^2 + b^2}$
The asymptotes are $y = \pm (b/a)x$
i.e. The asymptotes are $y = \pm (1/\sqrt{991})x$
There exists a $CONJUGATE$  $HYPERBOLA$ of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$, for the $HYPERBOLA$ $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Here the equation of the $conjugate$  $hyperbola$ is $x^2 - 991(y^2) = -1$
$$x^2 - 991y^2 = -1$$ -------(2)
Now solve equations (1) and (2) for $x$ and $y$
i.e. $(2)(x^2) - (2)(991)(y^2) = 0$
i.e. $(x^2) - (991)(y^2) = 0$
i.e. $(x^2) = (991)(y^2)$
i.e. $\frac{x^2}{y^2} = 991$
i.e. $\left(\frac{x^2}{y^2} - (\sqrt{991})^2\right) = 0$
i.e. $\left(\frac{x}{y} - \sqrt{991}\right)\left(\frac{x}{y} + \sqrt{991}\right) = 0$
i.e. $(\frac{x}{y} = \pm\sqrt{991})$
i.e. $\frac{y}{x} = \pm\frac{1}{\sqrt{991}}$ , the equations of the asymptotes to the hyperbola
