Question

20 de puncte:))) nu prea mai am:))

Answer 1

sper că te-am ajutat, cumva

Answer 2

$\displaystyle\it\\x^2+y^2+z^2=16(x+y+z) \Leftrightarrow x^2+y^2+z^2=16x+16y+16z \Leftrightarrow\\\Big(x^2-16x\Big)+\Big(y^2-16y\Big)+\Big(z^2-16z\Big)=0~\bigg|+64\cdot3 \implies\\\Big(x^2-16x+64\Big)+\Big(y^2-16y+64\Big)+\Big(z^2-16z+64\Big)=192 \Leftrightarrow\\\Big(x-8\Big)^2+\Big(y-8\Big)^2+\Big(z-8\Big)^2=192 \implies\\\Bigg(\Big(x-8\Big)^2+\Big(y-8\Big)^2+\Big(z-8\Big)^2\Bigg)\equiv 0(mod~4)\Leftrightarrow\\tinem~cont~de~faptul~ca~un~patrat~perfect~este~congruent~cu~0~sau~1~\\modulo~4.$

$\displaystyle\it\\ca~sa~se~respecte~conditia~de~mai~sus,~evident,~termenii~sumei~\\sunt~congruenti~cu~0~modulo~4.\\\implies x-8\equiv 0(mod~4) \implies x-8=4x_0,~x_0\in\mathbb{Z}\\\implies y-8\equiv0(mod~4) \implies y-8=4y_0,~y_0\in\mathbb{Z}.\\\implies z-8\equiv0(mod~4) \implies z-8=4z_0,~z_0\in\mathbb{Z}.\\inlocuim~in~relatia~de~mai~sus,~si~ajungem~la~:~\\16x_0^2+16y_0^2+16z_0^2=192~\Big|:16 \implies x_0+y_0+z_0=12,\\analog~:~$

$\displaystyle\it\\x_0=2x_1,~x_1\in\mathbb{Z}.\\y_0=2y_1,~y_1\in\mathbb{Z}.\\z_0=2z_1,~z_1\in\mathbb{Z}.\\\implies x_1^2+z_1^2+y_1^2=3 \implies ecuatia~fiind~simetrica~ordonam~astfel~:~\\x_1\leq y_1\leq z_1 \implies x\leq y\leq z.\\\boxed{\it x_1=0=y_0=z_1=0 \implies x=y=z=0}~.\\\boxed{\it x_1=0,~y_1=z_1=1 \implies x=0,~y=z=16,~si~permutarile}~.\\\boxed{\it x_1=y_1=z_1=1 \implies x_1=y_1=z_1=16}~.$