40th Hellenic Mathematical Olympiad

Since $\beta$ divides $\alpha -1$ and $2\alpha +1$ divides $5\beta -3$, with $\alpha >1$, $$x=\frac{\alpha -1}{\beta }\text{and}y=\frac{5\beta -3}{2\alpha +1}$$ are positive integers such that: $$0<xy=\frac{\alpha -1}{\beta }\cdot \frac{5\beta -3}{2\alpha +1}<\frac{\alpha }{\beta }\cdot \frac{5\beta }{2\alpha }=\frac{5}{2}\Rightarrow xy\in \{1,2\}.$$ If $xy=1$, then $x=y=1$, and so we have the system: $$\{\begin{array}{l}\alpha -1=\beta \\ 2\alpha +1=5\beta -3\end{array}\iff \{\begin{array}{l}\alpha =\beta +1\\ 2(\beta +1)+1=5\beta -3\end{array}\iff \alpha =3,\beta =2.$$ If $xy=2$, then $x=1,y=2$ or $x=2,y=1$, and so we have the systems: $$\{\begin{array}{l}\alpha -1=\beta \\ 4\alpha +2=5\beta -3\end{array}\iff \{\begin{array}{l}\alpha =\beta +1\\ 4(\beta +1)+2=5\beta -3\end{array}\iff \alpha =10,\beta =9.$$

$$\{\begin{array}{l}\alpha -1=2\beta \\ 2\alpha +1=5\beta -3\end{array}\Rightarrow \{\begin{array}{l}\alpha =2\beta +1\\ 2(2\beta +1)+1=5\beta -3\end{array}\Rightarrow \alpha =13,\beta =6.$$

Since $\beta |\alpha -1$, we have: $$\alpha -1=\beta \kappa (1),$$ For some positive integer $\kappa$. Moreover, $2\alpha +1|5\beta -3$, and hence there exists $\lambda \in {\mathbb{N}}^{+}$ such that $$5\beta -3=(2\alpha +1)\lambda ,$$ Which, because of (1), becomes: $$5\beta -3=\lambda (2\beta \kappa +3)\Rightarrow \beta (2\lambda \kappa -5)=-3\lambda -3(2)$$ Since $-3\lambda -3<0$, we get $2\lambda \kappa -5<0$, and hence $$2\lambda \kappa \le 4\iff \lambda \kappa \le 2.$$ Therefore, we have: $$\lambda =\kappa =1\text{ or }\lambda =2,\kappa =1,\text{ or }\lambda =1,\kappa =2.$$ By substitution in (2), in the first case we find $\beta =2$ and $a=3$. In the second case we find $\beta =9$ and $a=10$. Finally, in the third case we find $\beta =6$, $a=13$.