Determine the least positive solution for the following values of $N$ in Pell’s equation
$$ x^2 − Ny^2 = 1$$
by substituting appropriate integer values for $y$:
(a) $N = 3$
(b) $N = 7$
(c) $N = 8$
(d) $N = 11$
(e) $N = 10$
(f) $N = 12$
The approach is trial and error, starting with the smallest candidate solutions.
(a) $x^2 - 3y^2 = 1$
We try $y=1$, which immediately gives us $x=2$. So the solution is $(x,y)=(2, 1)$.
(b) $x^2 - 7y^2 = 1$
$y=1, 2$ doesn't give us a perfect square for $1+7y^2$. $y=3$ works and gives a solution $(x,y)=(8, 3)$.
(c) $x^2 - 8y^2 = 1$
$y=1$ immediately gives a solution $(x,y)=(3, 1)$.
(d) $x^2 - 11y^2 =1$
$y=1, 2$ doesn't give us a perfect square for $1+11y^2$. $y=3$ works and gives us $(x,y)=(10, 3)$.
(e) $x^2 - 10y^2 = 1$
$y = 1, 2, 3, 4, 5, $ do not give us a perfect square for $1+10y^2$. $y=6$ gives us $(x,y)=(19, 6)$.
(f) $x^2 - 12y^2=1$
$y=1$ does not give us a perfect square for $1+12y^2$. $y=2$ gives $(x,y) = (7,2)$.