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. $\alpha \ne \beta$ 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 $$\begin{gathered} \alpha \beta (3\alpha +16\beta )={\alpha }^2\omega \Rightarrow \beta (3\alpha +16\beta )=\alpha \omega \Rightarrow \alpha |\beta (3\alpha +16\beta ) \\ \Rightarrow \alpha |(3\alpha +16\beta )\Rightarrow \alpha |16\beta \Rightarrow \alpha |16=2^4\Rightarrow \alpha =2. \end{gathered}$$ 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 $$\begin{gathered} \alpha \beta (3\alpha +16\beta )={\beta }^2\tau \Rightarrow \alpha (3\alpha +16\beta )=\beta \omega \Rightarrow \beta |\alpha (3\alpha +16\beta ) \\ \Rightarrow \beta |(3\alpha +16\beta )\Rightarrow \beta |3\alpha \Rightarrow \beta |3\Rightarrow \beta =3. \end{gathered}$$ 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.