Question

Aratati ca a= $\sqrt{n^{3} + 14n + 14 }$ ∈R\Q (∀)n∈N

Answer 1

$\displaystyle\it\\presupunem~prin~absurd~ca~a\in\mathbb{Q}.\\a=\sqrt{n^3+14n+14}\in\mathbb{Q} \Leftrightarrow \exists~m\in\mathbb{N}~a.i.~n^3+14n+14=m^2.\\acum,~incercam~sa~aratam~ca~ecuatia~de~mai~sus~nu~are~solutii~in\\multimea~numerelor~naturale,~considerand~ecuatia~modulo~3.\\n^3+14n+14\equiv n+2n+2 \equiv 3n+2\equiv2 (mod3),~dar~un~patrat~perfect\\nu~poate~fi~niciodata~congruent~cu~2~modulo~3,~si~doar~cu~0~sau~1.\\n^3 \equiv n(mod3),~14n+14\equiv 2n+2(mod3).$

$\displaystyle\it\\prin~urmare~a~este~un~numar~irational.$