Solutions of Diophantine Equations x^2-ny^2=pm 1

Theorem
Suppose the Diophantine equation  
\[x^2-ny^2=-1\]
  has a solution  
\[x=x_1, \; y=y_1\]
, Then  
\[x=2x_1^2+1, \; y=2x_1y_1\]
  satisfies the Diophantine equation  
\[x^2-ny_2=1\]
.
Proof
\[(2x_1^2+1)^2-n(2x_1y_1)^2=4x_1^4+4x_1^2+1-4nx_1^2y_1^2=4x_1^2(x_1^2-ny_1^2+1)+1=1\]
  since  
\[x_1^2-ny_1^1+1=0\]
  in the statement of the theorem.
Suppose for example that  
\[\sqrt{n}=\sqrt{74}\]
  has continued fraction  
\[[ 8 \lt 1,1,1,1,16 \gt ]\]
.
The first solution will be found from the convergent  
\[[ 8,1,1,1,1]\]
. This is equal to  
\[\frac{43}{5}\]
  and the solution is  
\[x=43, \; y=5\]
. Another solution according to the theorem is  
\[x=2 \times 43^2+1=3699, \; y=2 \times 43 \times 5=430\]
  and  
\[3699^2- 74 \times 430^2=1\]
.

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