We will prove that all solutions to Pell's equation are given by powers of the smallest positive solution. Let's assume it to be

$ x_{0} + y_{0} \cdot \sqrt{d} $

[//]:#(We claim that every solution to the equation is given by $(x_{0} + y_{0} \cdot \sqrt{d})^{n} $ for some integer $ n $. The idea is that any other solution)

[//]:#( $( x + \sqrt{d} \cdot y)$ must be of this form, meaning that all solutions are generated by repeated multiplication of the smallest solution. Here we use Brahmagupta's identity of quadratic decomposition.)

[//]:#( $(a^{2} - n \cdot b^{2}) \cdot(c^{2} - n \cdot d^{2}) =(a \cdot c + n \cdot b \cdot d)^{2} - n \cdot(a \cdot d + b \cdot c)^{2}$)

[//]:#()

[//]:#(Here, $a = x_{1}$, $b = y_{1}$, $c = x_{2}$, $d = y_{2}$)

[//]:#( So if(x_{1}, y_{1}) and(x_{2}, y_{2}) are solutions, then $(x_{1} \cdot x_{2} + n \cdot y_{1} \cdot y_{2}, x_{1} \cdot y_{2} + y_{1} \cdot x_{2})$ is also a solution.)

[//]:#(First, we prove that product of two solutions is also a solution.)

We use method of descent to prove it. suppose there is a solution $u + v \cdot \sqrt{d}$ such that $u^{2} - d \cdot v^{2} = 1 $

Therefore, $ ( x_{0} + \sqrt{d} \cdot y_{0} )^{n} < u + v \cdot \sqrt{d} < ( x_{0} + \sqrt{d} \cdot y_{0} )^{n+1} $