Solutions of Pell's Equation From the Fundamental Solution

Theorem (All Solutions if Pell's Equation From Fundamental Solutions)
Let  
\[x_1, \; y_1\]
  be the fundamental - smallest - solutions of  
\[x^2-ny^2=1\]
. For each  
\[k \ge 1\]
,  
\[x=x_k, \; y=y_k\]
  is also a solution, found from  
\[x_k+\sqrt{n} y_k=(x_1+ \sqrt{n} y_1 )^k\]
.
Conversely, the solutions given in this way are the only positive solutions.
Proof
\[\begin{equation} \begin{aligned} (x_1^2-n y_1^2)^k &= (x_k+\sqrt{n} y_k)(x_k-\sqrt{n} y_k) \\ &= (x_1+\sqrt{n} y_1)^k(x_1-\sqrt{n} y_1)^k \\ &= (x_1^2-n y_1^2)^k \\ &= 1^k=1 \end{aligned} \end{equation}\]

So all  
\[x_k, \; y_k\]
  are solutions.
Suppose there are other solutions, say  
\[x=1, \; y=b\]
.  
\[x=x_1, \; y=y_1\]
  is the smallest solution, and  
\[x_1+\sqrt{n} y \gt 1\]
  so there exists  
\[k\]
  such that  
\[(x_1+\sqrt{n} y_1)^k \lt a + b \sqrt{n} \lt (x_1+\sqrt{n} y_1)^{k+1}\]
.
Multiply through by  
\[(x_k- y_k \sqrt{n})= (x_1- y_1 \sqrt{n})^k= \frac{1}{(x_1+ y_1 \sqrt{n})^k} \gt 0\]
.
\[(x_1+\sqrt{n} y_1)^k (x_1- y_1 \sqrt{n})^k \lt (a + b \sqrt{n})(x_k- y_k \sqrt{n}) \lt (x_1+\sqrt{n} y_1)^{k+1}(x_1- y_1 \sqrt{n})^k\]
.
The expression simplifies to
\[1 \lt (a + b \sqrt{n})(x_k- y_k \sqrt{n})=ax_k-by_kn+(bx_k-ay_k) \sqrt{n} \lt x_1+y_1 \sqrt{n}\]
  (1)
Hence  
\[x=ax_k-by_kn. \; y= bx_k-ay_k\]
  are also solutions but (1) implies that this solution is less than the fundamental - smallest - solution, which a contradiction, so the theorem is proved.
Example:  
\[x_1=3, \; y_1=\]
  is a solution of  
\[x^2-8y^2=1\]
  and so are
\[(3+ \sqrt{8})^2=17+6 \sqrt{8} \rightarrow x_2=17, \; y_2=6\]

\[(3+ \sqrt{8})^3=17+6 \sqrt{8} \rightarrow x_3=99, \; y_2=35\]

and so on.

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