Finding integers $s$ and $t$ such that $s+t=\frac{\alpha}{b}+b$ and $st=\frac{\beta}{b}+b$ for integers $\alpha$, $\beta$, $b$ In my research I need to find integers $s$ and $t$ with the following properties (respect to $\alpha, \beta, b$) :
\begin{align}
s+t=\frac{\alpha}{b}+b\\
st=\frac{\beta}{b}+b,
\end{align}
where $\alpha, \beta, b\in\mathbb{Z}$ and $b\neq0$. Do there exist any integers $s$ and $t$ with the above properties?
Anyone can help me? Thanks in advance.
 A: $s,t$ are the roots of
$$ X^2-(\tfrac\alpha b+b)X+\tfrac\beta b+b$$
and the formula to find these is well-known:
$$s,t=\frac{\tfrac\alpha b+b\pm\sqrt{(\tfrac\alpha b+b)^2-4(\tfrac\beta b+b)}}{2} 
=\frac{\alpha+b^2\pm\sqrt{(\alpha +b^2)^2-4(\beta b+b^3)}}{2b} 
$$
and for these to be integers, the radicand at least needs to be a perfect square. (and even then, dividing by $2b$ might spoil it).
A: $$s+t=\frac{\alpha}{b}+b \qquad st=\frac{\beta}{b}+b\\
\implies {α + b^2 = b (s + t) \qquad b^2 + β = b s t}\\
\implies α = -b^2 + b s + b t \qquad  β = b s t - b^2\qquad  b\ne0
$$
We can see infinite solutions in terms of $\quad b, s, t\quad$ but it appears $\quad \alpha,\beta\subset\mathbb{Z}.$
Example of $(b,s,t,\alpha, \beta)$
$$(1,1,1,3,0)\quad 
(2,1,1,8,-2)\quad 
(3,1,1,15,-6)\quad 
(4,1,1,24,-12)\\
(1,1,2,4,1)\quad 
(2,1,2,10,0)\quad 
(3,1,2,18,-3)\quad 
(4,1,2,28,-8)\\
(1,2,2,5,3)\quad 
(2,2,2,12,4)\quad 
(3,2,2,21,3)\quad 
(4,2,2,32,0)\\ $$
For $\quad s,t \quad\text{in terms of}\quad b, \alpha, \beta\quad$ as in Hagen von Eitzen's answer, we an immediately see that few of the lower number combinations yield integers. Some $\quad (b, \alpha, \beta )\quad$ values are
$$(1,3,2,3,1)\quad (1,3,3,2,2)\quad (1,4,3,4,1)\quad (1,4,5,3,2)\quad (1,5,4,5,1)\\$$
$$(2,4,2,3,1)\quad (2,4,4,2,2)\quad (2,6,4,4,1)\quad (2,6,8,3,2)\quad (2,8,6,5,1)\\ $$
$$(3,3,3,2,2)\quad (3,6,3,4,1)\quad (3,6,9,3,2)\quad (3,9,6,5,1)\quad (3,9,15,4,2)\\$$
