SELECTION EXAMINATION 2019

We suppose that: $${\alpha }^2+\lfloor \frac{4{\alpha }^2}{\beta }\rfloor =(\alpha +\kappa {)}^2,\kappa \in {\mathbb{N}}^{\ast }\iff \lfloor \frac{4{\alpha }^2}{\beta }\rfloor =(2\alpha +\kappa )\kappa ,\kappa \in {\mathbb{N}}^{\ast }.$$ By putting $2\alpha =x$, we have: $\lfloor \frac{x^2}{\beta }\rfloor =(x+\kappa )\kappa ,\kappa \in {\mathbb{N}}^{\ast }$. Equivalently, we have supposed that the equation $$\lfloor \frac{x^2}{\beta }\rfloor =(x+\kappa )\kappa ,\kappa \in {\mathbb{N}}^{\ast }(1)$$ 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: $$\begin{array}{c}\frac{x^2}{\beta }>\lfloor \frac{x^2}{\beta }\rfloor -1=(x+\kappa )\kappa -1=x\kappa +{\kappa }^2-1\ge x\kappa \Rightarrow \frac{x}{\beta }>\kappa \Rightarrow \\ x>\beta \kappa \end{array}(2)$$ Also, we have: $$\frac{x^2-{\kappa }^2}{\beta }<\frac{x^2}{\beta }\le \lfloor \frac{x^2}{\beta }\rfloor =(x+\kappa )\kappa \Rightarrow \beta \kappa \ge x-\kappa .(3)$$ From relations (2) (3) it follows that: $x>\beta \kappa \ge x-\kappa$, and hence there exists $\upsilon$ such that: $$x=\beta \kappa +\upsilon ,0<\upsilon <\kappa (4)$$ From (3) and (4) we have: $$\lfloor \frac{x^2}{\beta }\rfloor =\lfloor \frac{(\beta \kappa +\upsilon {)}^2}{\beta }\rfloor =\beta {\kappa }^2+2\kappa \upsilon +\lfloor \frac{{\upsilon }^2}{\beta }\rfloor (5)$$ $$(x+\kappa )\kappa =(\beta \kappa +\upsilon +\kappa )\kappa =\beta {\kappa }^2+2\kappa \upsilon +\kappa (\kappa -\upsilon )(6)$$ From relations (5) and (6) we get: $$\lfloor \frac{{\upsilon }^2}{\beta }\rfloor =\kappa (\kappa -\upsilon )⟺\lfloor \frac{x^{\prime 2}}{\beta }\rfloor =(x^{\prime }+{\kappa }^{\prime }){\kappa }^{\prime },$$ From which we conclude that equation (1) has a solution $(x^{\prime },{\kappa }^{\prime })$ in the positive integers with ${\kappa }^{\prime }=\kappa -\upsilon <\kappa$, which is absurd.