$x_2 + y_2 \sqrt d = (x_1 + y_1 \sqrt d)^2$ is the smallest positive solution of $x^2 - dy^2 = 1$ 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.
I have posted It about 2 weeks ago. Has anyone got any idea about the solution.
Thank You.
 A: From the web you can find rich resources on Pell's equation. For example, wiki would give you a good overview. 
Anyway here I give an elementary treatment, which is of my own. 
Clearly $x_2 + y_2 \sqrt d = (x_1 + y_1 \sqrt d)^2$ is a positive solution of $x^2-dy^2=1$. We will show it is the smallest. Assume that $x_2' + y_2'\sqrt{d}$ is another positive solution of $x^2-dy^2=1$  and that $y_2'<y_2$, and we will reach a contradiction. 
Let $$x_1'+y_1'\sqrt{d} = \frac{x_2' + y_2'\sqrt{d}}{x_1 + y_1 \sqrt d} = (x_2'+y_2'\sqrt d)(y_1\sqrt d - x_1) = (dy_1y_2'-x_1x_2') + (x_2'y_1-x_1y_2')\sqrt d, $$
so $x_1'=dy_1y_2'-x_1x_2'$ and $y_1'=x_2'y_1-x_1y_2'$. It is easy to verify that $x_1'+y_1'\sqrt d$ is a solution of $x^2-dy^2=-1$. We want to further show that $0<y_1'<y_1$ and thus have a contradiction. 
 $$y_1'>0   \iff x_2'y_1-x_1y_2' > 0  \iff \frac{x_2'}{y_2'} > \frac{x_1}{y_1}, $$ and we have $$\frac{x_2'}{y_2'} > \sqrt d > \frac{x_1}{y_1}.$$ So $y_1' > 0$. 
On the other hand, $$y_1'=x_2'y_1-x_1y_2' <y_1 = x_2y_1 - x_1y_2 \iff \frac{x_2-1}{y_2}=\frac{x_1}{y_1} > \frac{x_2'-1}{y_2'},  $$ which is true by proving that $\frac{\sqrt {dy^2+1} - 1 }{y}$ is an increasing function (note that $y_2'<y_2$). The proof of $\frac{\sqrt {dy^2+1} - 1 }{y}$ is an increasing function is not very complicated. For example, we can let $z=\sqrt d y$ and show $f(z)=\frac{\sqrt{z^2+1} - 1}{z}$ is an increasing function. A little calculus can be useful here, but I suspect that an elementary solution is also available (for example, let $z=\tan \theta$). 
We have shown that  $x_1'+y_1'\sqrt{d}$ is a smaller positive solution for $x^2-dy^2=-1$, contradicting to that $x_1+y_1\sqrt{d}$ is the smallest solution. (well, there is one missing step: we have not shown $x_1'>0$. It is ok by ignoring it, as otherwise we can use $-x_1'$.) Therefore $x_2+y_2\sqrt d$ is indeed the smallest positive solution for $x^2-dy^2=1$.
The same idea can be used to prove the remaining problems. 
