Prove Theorem (8.21).
Let's remind ourselves of Theorem (8.21).
Let $r = a + b\sqrt{N}$ produce the least positive solution of Pell’s equation $x^2 − Ny^2 = 1$.
Then this equation has all its positive integer solutions produced by $r^n = (a + b\sqrt{N})^n$ where $n$ is a natural number.
We are given that $r=a+b\sqrt{N}$ is the least positive solution of $x^2 - Ny^2=1$. We need to prove two things:
- Step 1: That $r^n$ produces solutions of Pell's equation.
- Step 2: All solutions are produced by $r^n$. That is, there are no solutions not produced by $r^n$.
Step 1
We have already proved Proposition (8.19) in Exercise (8.3).11 that $s^n$ produces solutions of Pell's equation if $s$ does, where $n$ is a natural number. Note that $s$ in Proposition (8.19) does not need to be the least positive solution.
Step 2
For the purpose of contradiction let's assume $q=c + d\sqrt{N}$ produces a positive solution to $x^2-Ny^2=1$, but that $r^n \ne c+d\sqrt{N}$ for any natural $n$, where $r$ is the least positive solution.
We note that $q$ is not less than or equal to $r^1$ because $r$ is the least positive solution.
This means there exists a natural number $x$ such that
$$ r^x < q < r^{x+1} $$
Dividing by the positive number $r^x$ gives
$$ 1 < \frac{q}{r^x} < r $$
By Proposition (8.17), since $r$ is a solution, so is $r^{-1}$. Further more, by Proposition (8.19), if $r^{-1}$ is a solution, so is $(r^{-1})^x = r^{-x}$. By Proposition (8.18) the product of two solutions is also a solution, that is $q \times r^{-x} = \frac{q}{r^x}$ is a solution.
This means $ 1 < \frac{q}{r^x} < r $ is contradiction because $r$ is the least positive solution, and the positive solution $\frac{q}{r^x}$ can't be less than $r$.
And so the assumption that there exists a solution that is not a natural number power of the least positive solution is false. That is, all positive solutions are natural number powers of the least positive solution.