Determine three different solutions of
$$ x^2− 47y^2 = 1 $$
We first try to find the least positive solution by trying the smallest values of $y$.
Trying $y=1,2,3,4,5,6$ does not give a natural number solution for $x$. The first solution is $y=7, x=48$.
From the least positive solution $r=47+7\sqrt{47}$, we can generate two further solutions $r^2$ and $r^3$.
$$ r^2 = 4607 + 672 \sqrt{47}$$
$$ r^3 = 442224 + 64505 \sqrt{47} $$
So three solutions are $(x,y)=(48,7)$, $(x,y)=(4607,672)$ and $(x,y)=(442224, 64505 )$.