Question

Aratati ca ecuatia
$6 x^{2} + y^{2} =2016$ nu are solutii in multimea numerelor naturale.

Answer 1

$Presupunem~prin~absurd~ca~ecuatia~are~solutii~naturale. \\ \\ Deoarece~6x^2~si~2016~sunt~multipli~de~6,~rezulta~ca~si~y^2~este \\ \\multiplu~de~6.~ Deci~y=6y_1~(y_1 \in N). ~Ecuatia~devine: \\ \\ 6x^2+36y_1^2=2016 \Leftrightarrow x^2+6y_1^2=336 .~Acum~6y_1^2~si~336~sunt~multipli \\ \\ de~6 \Rightarrow x=6x_1.~Ecuatia~devine:~36x_1^2+6y_1^2=336 \Leftrightarrow 6x_1^2+y_1^2=56. \\ \\ Rezulta~ y_1^2=56-6x_1^2=54+2-6x_1^2=3(18-2x_1^2)+2=M_3+2.$

$Insa~restul~impartirii~unui~patrat~perfect~la~3~poate~fi~0~sau~1,~ \\ \\ iar~cum~y_1^2=M_3+2 \Rightarrow am~ obtinut~ o~ contradictie \Rightarrow ecuatia~nu \\ \\ are~solutii~naturale.$