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

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 \[3699^2- 74 \times 430^2=1\]

.