SELECTION EXAMINATION

First of all, it is easy to see that: $${\alpha }_1\sqrt{2}+{\alpha }_2\sqrt{3}={\alpha }_3\sqrt{2}+{\alpha }_4\sqrt{3},\text{ with }{\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4\in {\mathbb{Q}}^{\ast }\iff {\alpha }_1={\alpha }_3\text{ and }{\alpha }_2={\alpha }_4.$$ In fact, we can write the first relation in the form $$({\alpha }_1-{\alpha }_3)\sqrt{2}=({\alpha }_4-{\alpha }_2)\sqrt{3},$$ and if ${\alpha }_1-{\alpha }_3\ne 0$, then $\frac{{\alpha }_4-{\alpha }_2}{{\alpha }_1-{\alpha }_3}=\frac{\sqrt{3}}{\sqrt{2}}$, absurd. Hence ${\alpha }_1={\alpha }_3$ and ${\alpha }_2={\alpha }_4$. The converse is clear. Let now $\frac{\alpha \sqrt{2}+\beta \sqrt{3}}{\beta \sqrt{2}+\gamma \sqrt{3}}=\kappa \in \mathbb{Q}$. Then $\alpha =\kappa \beta$ and $\beta =\kappa \gamma$, that is $$\frac{\alpha }{\beta }=\frac{\beta }{\gamma }\iff {\beta }^2=\alpha \gamma .$$ Hence we have:

$$\begin{array}{rl}{\alpha }^2+{\beta }^2+{\gamma }^2& ={\alpha }^2+\alpha \gamma +{\gamma }^2={\alpha }^2+2\alpha \gamma +{\gamma }^2-\alpha \gamma \\ & =(\alpha +\gamma {)}^2-{\beta }^2=(\alpha +\beta +\gamma )(\alpha -\beta +\gamma ),\end{array}$$ and so $B=\alpha -\beta +\gamma$.