Question

determinați tripletele(x,y,z)de numere naturale pentru care fractia 9 pe x²±y²+x² este :a) supraunitară ; b) echiunitara

Answer 1

a) 9 >x²+y²+z² ⇒

x=1; y=0 ;  z=0⇒ 9/1
x=0; y=1; z=0⇒ 9/1
x=0 ; y=0; z=1⇒ 9/1

x=1; y=1;z=0⇒ 9/2
x=1; y=0; z=1⇒9/2
x=0;y=1;z=1⇒9/2

x=1 ; y=1 ;z=1⇒ 9/3

x=2 ; y=0 ;z=0⇒ 9/4
x=0 ; y=2 ;z=0⇒ 9/4
x=0 ; y=0 ;z=2⇒ 9/4 

x=1;y=2;z=0⇒9/5
x=1;y=0;z=2⇒9/5
x=0;y=2;z=1⇒9/5
x=0;y=1;z=2⇒9/5
x=2;y=0;z=1⇒9/5
x=2;y=1;z=0⇒9/5

x=2;y=1;z=1⇒9/6
x=1;y=1;z=2⇒9/6
x=1;y=2;z=1⇒9/6

x=2;y=2;z=0⇒9/8
x=2;y=0;z=2⇒9/8
x=0;y=2;z=2⇒9/8

(x;y;z) ∈ { (0,0,1); (0,1,0); (1,0,0) ; (1,1,0);(1,0,1); (0,1,1);(1;1;1);(1,2,0);
(1,0,2);(0,1,2);(0,2,1);(2,1,0);(2,0,1);(2;1;1);(1,2,1);(1,1,2);(2,0,2);(0,2,2);(2,2,0);(2,0,0); (0,2,0);(0,0,2)}

b) x²+y²+z²= 9
x=2;y=2;z=1⇒9/9
x=1;y=2;z=2⇒9/9
x=2;y=1;z=2⇒9/9
x=3;y=0;z=0⇒9/9
x=0;y=3;z=0⇒9/9
x=0;y=0;z=3⇒9/9

(x;y;z) ∈{(2,2,1);(1,2,2);(2,1,2);(3,0,0); (0,3,0);(0,0,3)}