More values of $a$ and $D $ on conditions set by me and a way to obtain more values. 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 $\alpha \quad and \quad\beta$.
 A: 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...
