Cristian Calude, proba pe echipe, R.III, P.II

[Laurian Filip]

Se considera ecuatia: \( x^2+(2+\alpha-\beta)\cdot x + 1 +\alpha=0 \), unde \( \alpha,\beta \in mathbb{Z} \). Stiind ca solutiile sunt \( x_1,x_2\in \mathbb{Z} \) si \( \alpha \cdot \beta=231 \), sa se rezolve ecuatia.

[Mateescu Constantin]

Din relatiile lui Viete avem:
\( \left\{\begin x_1+x_2=\beta-\alpha-2 \\
x_1x_2=\alpha+1\right\ \ \ \Longrightarrow \left\{\begin{array} \beta=x_1+x_2+\alpha+2 \\
\alpha=x_1x_2-1\end{array} \)

Dar \( \alpha\cdot \beta=231\ \Longrightarrow (x_1+1)(x_2+1)(x_1x_2-1)=3\cdot 7\cdot 11 \)

Cum \( x_1,\ x_2\in\mathbb{Z} \) obtinem solutiile \( x_1=2,\ x_2=6 \).