Show that if $N=n^2− 1$ where $n$ is a natural number then the least positive solution of $x^2− Ny^2 = 1$ is given by $x = n, y= 1$.
Determine the least positive solutions of $x^2 − Ny^2 = 1$ where
(a) $N = 15$
(b) $N = 24$
(c) $N = 35$
(d) $N = 48$
We proceed as follows
$$ x^2 - Ny^2 = x^2 - (n^2-1)y^2 = 1 $$
If $y=1$ then a solution is a candidate for the least positive solution. Let's try it
$$ x^2 - n^2 + 1 = 1 \quad \implies \quad x = n $$
So $(x,y) = (n,1)$ is a solution.
We need to check that $x=1$ isn't part of a least positive solution. We will use Proposition (8.20),
Let $r = a + b\sqrt{N}$ and $s = c + d\sqrt{N}$ both produce positive solutions of Pell’s equation $x^2− Ny^2 = 1$. Then $r < s \iff a < c$.
For the purpose of contradiction let's assume a positive solution $a+b\sqrt{N}$ exists and is smaller than the known solution $n+\sqrt{N}$, that is, $a+b\sqrt{N}<n+\sqrt{N}$. By Proposition (8.20) this means $a<n$. From $a^2 - Ny^2=1$ we have $1+(n^2-1)y^2<n^2$, which gives $y^2<\frac{n^2-1}{n^2-1}=1$. This is a contradiction because $y$ must be greater than or equal to 1. And so there is no smaller solution than $n+\sqrt{N}$.
And so $(x,y)=(n,1)$ is a solution, and is the least positive solution.
(a) Since $15=4^2-1$ we have the least positive solution $(x,y) = (4,1)$.
(b) Since $24=5^2-1$ we have the least positive solution $(x,y) = (5,1)$.
(c) Since $35=6^2-1$ we have the least positive solution $(x,y) = (6,1)$.
(d) Since $48=7^2-1$ we have the least positive solution $(x,y) = (7,1)$.