Assuming that $x^2 - dy^2 = -1$ is solvable, let $x_l, y_1$ be the smallest positive solution. Prove that $x_2, y_2$ defined by $x_2 + y_2 \sqrt d = (x_1 + y_1 \sqrt d)^2$ is the smallest positive solution of $x^2 - dy^2 = 1$. Also prove that all solutions of $x^2 - dy^2 = -1$ are given by $x_n, y_n$ where $x_n + y_n \sqrt d = (x_1 + y_1 \sqrt d)^n$, with $n = 1,3,5,7,··$, and that all solutions of $x^2 - dy^2 = 1$ are given by $x_n, y_n$ with $n = 2,4,6,8, ...$.
I am unable to the problem. Help Needed.
Hint - this leaves you with a bit of work to do:
For the first, consider $(x_1-y_1\sqrt d)(x_2+y_2\sqrt d)$ using the information you have been given.
If you have $p^2-dq^2=\pm 1$ consider $(x_1+y_1\sqrt d)(p+q\sqrt d)=(x_1p+dy_1q)+(x_1q+y_1p)\sqrt d$ together with $(x_1-y_1\sqrt d)(p-q\sqrt d)$. Multiply these together in two different ways to get another solution. You can do a similar thing with $(x_1-y_1\sqrt d)$. Using this you can get from any positive solution to a smaller solution except when you are at the smallest.
As well as the Pell Equation reference you are working with the multiplicative group of units in a quadratic field.