, 2019

We suppose that:
$$\alpha^2 + \left\lfloor \frac{4\alpha^2}{\beta} \right\rfloor = (\alpha + \kappa)^2, \kappa \in \mathbb{N}^* \Leftrightarrow \left\lfloor \frac{4\alpha^2}{\beta} \right\rfloor = (2\alpha + \kappa)\kappa, \kappa \in \mathbb{N}^*.$$
By putting $2\alpha = x$, we have: $\left\lfloor \frac{x^2}{\beta} \right\rfloor = (x+\kappa)\kappa, \kappa \in \mathbb{N}^*$. Equivalently, we have supposed that the equation
$$\left\lfloor \frac{x^2}{\beta} \right\rfloor = (x + \kappa)\kappa, \kappa \in \mathbb{N}^* \quad (1)$$
has a solution $(x, \kappa)$ in the positive integers with $x$ even.
We suppose that equation (1) has a solution $(x, \kappa)$, where $\kappa$ is the least possible integer and we will try to find a contradiction.
We have:
$$\begin{gathered} \frac{x^2}{\beta} > \left\lfloor \frac{x^2}{\beta} \right\rfloor - 1 = (x+\kappa)\kappa - 1 = x\kappa + \kappa^2 - 1 \ge x\kappa \Rightarrow \frac{x}{\beta} > \kappa \Rightarrow \\ x > \beta\kappa \end{gathered} \quad (2)$$
Also, we have:
$$\frac{x^2 - \kappa^2}{\beta} < \frac{x^2}{\beta} \le \left\lfloor \frac{x^2}{\beta} \right\rfloor = (x + \kappa)\kappa \Rightarrow \beta\kappa \ge x - \kappa. \quad (3)$$
From relations (2) (3) it follows that: $x > \beta\kappa \ge x - \kappa$, and hence there exists $\upsilon$ such that:
$$x = \beta\kappa + \upsilon, \quad 0 < \upsilon < \kappa \quad (4)$$
From (3) and (4) we have:
$$\left\lfloor \frac{x^2}{\beta} \right\rfloor = \left\lfloor \frac{(\beta\kappa + \upsilon)^2}{\beta} \right\rfloor = \beta\kappa^2 + 2\kappa\upsilon + \left\lfloor \frac{\upsilon^2}{\beta} \right\rfloor \quad (5)$$
$$(x + \kappa)\kappa = (\beta\kappa + \upsilon + \kappa)\kappa = \beta\kappa^2 + 2\kappa\upsilon + \kappa(\kappa - \upsilon) \quad (6)$$
From relations (5) and (6) we get:
$$\left\lfloor \frac{\upsilon^2}{\beta} \right\rfloor = \kappa(\kappa - \upsilon) \iff \left\lfloor \frac{x'^2}{\beta} \right\rfloor = (x' + \kappa')\kappa',$$
From which we conclude that equation (1) has a solution $(x', \kappa')$ in the positive integers with $\kappa' = \kappa - \upsilon < \kappa$, which is absurd.