37th Hellenic Mathematical Olympiad 2020

Since the equation is symmetric with respect to $\alpha ,\beta ,\gamma$ we suppose that $\alpha \ge \beta \ge \gamma$. Then we have: $$\alpha \le \alpha +\beta +\gamma \le 3\alpha \iff \alpha \le v\alpha \beta \gamma \le 3\alpha \Rightarrow 1\le v\beta \gamma \le 3.$$

We distinguish the following cases: • $v>3$. Then $v\beta \gamma >3$, absurd. • $v=3$. Then $1\le 3\beta \gamma \le 3\Rightarrow \beta \gamma =1\Rightarrow \beta =\gamma =1$, and hence $$\alpha +2=3\alpha \iff \alpha =1.$$ Therefore we get the solution: $(\alpha ,\beta ,\gamma )=(1,1,1)$. • $v=2$. Then $1\le 2\beta \gamma \le 3\Rightarrow \beta \gamma =1\Rightarrow \beta =\gamma =1$, and $$\alpha +2=2\alpha \iff \alpha =2.$$ Therefore we get the solution $(\alpha ,\beta ,\gamma )=(2,1,1)$ and using symmetry we obtain the solutions $(\alpha ,\beta ,\gamma )=(1,2,1)$ and $(\alpha ,\beta ,\gamma )=(1,1,2)$. • $v=1$. Then $1\le \beta \gamma \le 3\Rightarrow \beta \gamma \in \{1,2,3\}$. If $\beta \gamma =1$, then $\beta =\gamma =1$ and $\alpha +2=1$, impossible. If $\beta \gamma =2$, then $\beta =2,\gamma =1$ and $\alpha +3=2\alpha \iff \alpha =3$. Therefore $(\alpha ,\beta ,\gamma )=(3,2,1)$ and by symmetry $$\begin{array}{rl}(\alpha ,\beta ,\gamma )& =(2,1,3),\\ (\alpha ,\beta ,\gamma )& =(3,2,1),\\ (\alpha ,\beta ,\gamma )& =(3,1,2),\\ (\alpha ,\beta ,\gamma )& =(1,2,3),\\ (\alpha ,\beta ,\gamma )& =(2,3,1).\end{array}$$ If $\beta \gamma =3$, then $\beta =3,\gamma =1$ and $\alpha +4=3\alpha \iff \alpha =2$. (rejected, $\alpha <\beta$). Hence $v\notin \{1,2,3\}$.