Greece — Selection Examination, 2019

We distinguish the cases:

1. $\alpha = \beta$
Let: $A = 3\alpha^2\alpha + 16\alpha\alpha^2 = 19\alpha^3 = \kappa^2, \kappa \in \mathbb{Z}, \alpha$ prime.

Then $19|\kappa^2 \Rightarrow 19|\kappa \Rightarrow 19^2|\kappa^2 \Rightarrow \kappa^2 = 19^2\omega, \omega \in \mathbb{Z}$, and hence:
$$19\alpha^3 = 19^2\omega \Rightarrow \alpha^3 = 19\omega \Rightarrow 19|\alpha^3 \Rightarrow 19|\alpha \Rightarrow \alpha = 19.$$
Hence, the pair $(\alpha, \beta) = (19,19)$ maybe a solution. Since for $\alpha = \beta = 19$ we have $A = 19 \cdot 19^3 = 19^4$, the pair $(\alpha, \beta) = (19,19)$ is a solution.

2. $\alpha \neq \beta$
Let $A = 3\alpha^2\beta + 16\alpha\beta^2 = \alpha\beta(3\alpha + 16\beta) = \kappa^2, \kappa \in \mathbb{Z}, \alpha, \beta$ primes.
Then: $\alpha|\kappa^2 \Rightarrow \alpha|\kappa \Rightarrow \alpha^2|\kappa^2 \Rightarrow \kappa^2 = \alpha^2\omega, \omega \in \mathbb{Z}$, and hence
$$\alpha\beta(3\alpha + 16\beta) = \alpha^2\omega \Rightarrow \beta(3\alpha + 16\beta) = \alpha\omega \Rightarrow \alpha|\beta(3\alpha + 16\beta) \\ \Rightarrow \alpha|(3\alpha + 16\beta) \Rightarrow \alpha|16\beta \Rightarrow \alpha|16 = 2^4 \Rightarrow \alpha = 2.$$
From above we have $\beta|\kappa^2 \Rightarrow \beta|\kappa \Rightarrow \beta^2|\kappa^2 \Rightarrow \kappa^2 = \beta^2\tau, \tau \in \mathbb{Z}$, and hence
$$\alpha\beta(3\alpha + 16\beta) = \beta^2\tau \Rightarrow \alpha(3\alpha + 16\beta) = \beta\omega \Rightarrow \beta|\alpha(3\alpha + 16\beta) \\ \Rightarrow \beta|(3\alpha + 16\beta) \Rightarrow \beta|3\alpha \Rightarrow \beta|3 \Rightarrow \beta = 3.$$
Hence the pair $(\alpha, \beta) = (2,3)$ is a probable solution. Since
$$A = \alpha\beta(3\alpha + 16\beta) = 6 \cdot (6 + 48) = 6 \cdot 54 = 18^2,$$
the pair $(\alpha, \beta) = (2,3)$ is a solution.