Here I define -: $\alpha=a+ \sqrt D$ and $\beta=a-\sqrt D$
Then find out values of $\alpha$ and $\beta$ satisfying- $$\alpha>1 \quad and \quad -1< \beta <1 $$
and both the variables are integers .I began finding some initial values-
\begin{array}{c|lcr} & \text{a} & \text{D} & \text{} \\ \hline & 1 & 2 \\ & 1 & 3 \\ & 2 & 2 \\ & 2 & 3 \\ & 2 & 5 \\ \end{array} Can anyone make a very big list of such numbers and tell me how to generate the values of $\alpha \quad and \quad\beta$.
Let's first consider that $a$ and $D$ are 2 real positive number. You have $3$ conditions on $a$: \begin{equation} \begin{array}{rcl} a & > & 1 - \sqrt{D} \\ a & > & -1 + \sqrt{D} \\ a & < & 1 + \sqrt{D} \end{array} \end{equation} Since $1 + \sqrt{D}$ is always greater than $1-\sqrt{D}$ and $-1+\sqrt{D}$, then you have the following: $$\max(1-\sqrt{D}, -1 + \sqrt{D}) < a < 1 + \sqrt{D}$$
Now, we have to evaluate $\max(1-\sqrt{D}, -1 + \sqrt{D})$. It's easy to notice that: \begin{equation} \begin{array}{cc} D=1 & \max(1-\sqrt{1}, -1 + \sqrt{1}) = 0 \\ D\geq 2 & \max(1-\sqrt{D}, -1 + \sqrt{D}) = ... \end{array} \end{equation} Since $\sqrt{D} > 1$ when $D \geq 2$, then we can pose that $\sqrt{D} = 1 + \delta$, with $\delta > 0$. Then: $$\max(1-\sqrt{D}, -1 + \sqrt{D}) = \max(1-1-\delta, -1 + 1+\delta) = \max(-\delta, \delta) = \delta = -1 + \sqrt{D}$$ Then, you can say that, for every $D \geq 1$ we have that $\alpha > 1$ and $-1 < \beta <1$ when $$-1+\sqrt{D} < a < 1 + \sqrt{D}$$ Passing to integer when $D \geq 2$, then we have: $$\lceil -1+\sqrt{D} \rceil \leq a \leq \lfloor 1 + \sqrt{D} \rfloor$$ or $$ -1+\lceil \sqrt{D} \rceil \leq a \leq 1 + \lfloor \sqrt{D} \rfloor$$ Then:
if $D=1$, $0 < a < 2$, $a=1$
if $D=2$, $1 \leq a \leq 2$, $a=1, a=2$
if $D=3$, $1 \leq a \leq 2$, $a=1, a=2$
if $D=4$, $1 \leq a \leq 3$, $a=1, a=2, a=3$
and so on...