Nr. irational (OMN-Spania-2012)

[mihai miculita]

Aratati ca: $\sqrt{3n^2+2n+2}\notin\mathbb{Q};\,(\forall)\, n\in\mathbb{N}.$

[anamariaradu]

Daca n=par atunci $3n^2+2n +2= M_4+M_4+2=M_4+2$ $\implies 3n^2+2n+2$ nu poate fi patrat perfect.$\implies \sqrt{3n^2+2n +2}\notin\mathbb{Q}$.
Daca n=impar atunci $3n^2+2n +2= M_4+3+M_4+2+2=M_4+3$ $\implies 3n^2+2n+2$ nu poate fi patrat perfect.$\implies \sqrt{3n^2+2n +2}\notin\mathbb{Q}$.