The quadratic diophantine $ k^2 - 1 = 5(m^2 - 1)$ Here's the problem.
Find the solutions of the following equation:
$$ k^2 - 1 = 5(m^2 - 1).$$

Here's my idea:
The original equation can be written as:
$$ k^2 = 5m^2 - 4 \Longleftrightarrow k^2 - 5m^2 = -4$$
I know this is Quadratic Diophantine Equation and i've done some searching on the internet and I couldn't find a particular way for solving equations of this type. Also I know that this is a varition of Pell's equation, because instead of 1 we have -4.
I actually found the fundamental solution to this equation (by guessing) and it's (1,1). Then using this algorithm (that works for Pell's equation) I tried to generate another solution and I get:
$$ X_k+_1 = aX_k + nbY_k$$
$$ X_2 = aX_1 + nbY_1$$
$$ X_2 = 1 \times 1 + 5 \times 1 \times 1 = 6$$
$$ Y_k+_1 = bX_k + aY_k$$
$$ Y_2 = bX_1 + aY_1$$
$$ Y_2 = 1 \times 1 + 1 \times 1 = 2$$
We can easily check that (6,2) isn't a solution.
So how can I transform a Quadratic Diophantine Equation into a Pell's equatuon and how can I generate more solution for Pell's equation if the constant isn't 1 (in this case it's -4)?

$$ A = 
 \left(  \begin{array}{cc}
  9  &  20  \\
   4   &  9  
\end{array} 
  \right)  ,
 $$ and 
$$ A^{-1} = 
 \left(  \begin{array}{cc}
  9  &  -20  \\
   -4   &  9  
\end{array} 
  \right).
 $$ 
$$  
 \left(  \begin{array}{cc}
  9  &  20  \\
   4   &  9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  1  \\
   1  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  29  \\
   13  
\end{array} 
  \right), 
 $$
$$  
 \left(  \begin{array}{cc}
  9  &  20  \\
   4   &  9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  29  \\
   13  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  521  \\
   233  
\end{array} 
  \right), 
 $$
$$  
 \left(  \begin{array}{cc}
  9  &  20  \\
   4   &  9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  521  \\
   233  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  9349  \\
   4181  
\end{array} 
  \right), 
 $$
Switching to $-A^{-1},$ we get
$$  
 \left(  \begin{array}{cc}
  -9  &  20  \\
   4   &  -9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  1  \\
   1  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  11  \\
   -5  
\end{array} 
  \right), 
 $$
$$  
 \left(  \begin{array}{cc}
  -9  &  20  \\
   4   &  -9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  11  \\
   -5  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  -199  \\
   89  
\end{array} 
  \right), 
 $$
$$  
 \left(  \begin{array}{cc}
  -9  &  20  \\
   4   &  -9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  -199  \\
   89  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  3571  \\
   -1597  
\end{array} 
  \right), 
 $$
If you want to allow common factors,
$$  
 \left(  \begin{array}{cc}
  9  &  20  \\
   4   &  9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  4  \\
   2  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  76  \\
   34  
\end{array} 
  \right), 
 $$
$$  
 \left(  \begin{array}{cc}
  9  &  20  \\
   4   &  9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  76  \\
  34  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  1364  \\
   610  
\end{array} 
  \right), 
 $$
$$  
 \left(  \begin{array}{cc}
  9  &  20  \\
   4   &  9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  1364  \\
   610  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
 24476  \\
 10946  
\end{array} 
  \right). 
 $$
Switching to $-A^{-1},$ we get
$$  
 \left(  \begin{array}{cc}
  -9  &  20  \\
   4   &  -9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  4  \\
   2  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  4  \\
   -2  
\end{array} 
  \right), 
 $$
$$  
 \left(  \begin{array}{cc}
  -9  &  20  \\
   4   &  -9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  4  \\
   -2  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  -76  \\
   34  
\end{array} 
  \right), 
 $$
$$  
 \left(  \begin{array}{cc}
  -9  &  20  \\
   4   &  -9  
\end{array} 
  \right)  
 \left(  \begin{array}{c}
  -76  \\
   34  
\end{array} 
  \right) =
 \left(  \begin{array}{c}
  1364  \\
   -610  
\end{array} 
  \right), 
 $$
so you see nothing new happens this time. 
