Solve for x,y,z in integer. x+y+z=1x+y+z=1 , x^3+y^3+z^2=1x3+y3+z2=1?

2 Answers
Sep 24, 2016

(x,y,z) = (1,-1,1) or (-1,1,1)(x,y,z)=(1,1,1)or(1,1,1)

Sep 24, 2016

{y=-3,x=-2,z=6}{y=3,x=2,z=6}
{y=-2,x=-3,z=6}{y=2,x=3,z=6}
{y=-2,x=0,z=3}{y=2,x=0,z=3}
{y=0,x=-2,z=3}{y=0,x=2,z=3}
{y=0,x=1,z=0}{y=0,x=1,z=0}
{y=1,x=0,z=0}{y=1,x=0,z=0}

Explanation:

x+y=1-zx+y=1z
x^3+y^3=1-z^2x3+y3=1z2

Dividing term to term the second equation by the first we have

(x^3+y^3)/(x+y) = ((1-z)(1+z))/(1-z)x3+y3x+y=(1z)(1+z)1z or

x^2-xy+y^2=1+zx2xy+y2=1+z

Adding this equation with the first we have

x^2-x y+y^2+x+y=2x2xy+y2+x+y=2. Solving for xx we obtain

x = 1/2 (-1 + y pm sqrt[3] sqrt[3 - 2 y - y^2])x=12(1+y±332yy2)

Here

3 - 2 y - y^2 ge 032yy20 so

-3 le y le 13y1 but y in NN so y in {-3,-2,-1,0,1}

Checking we have

{y=-3,x=-2,z=6}
{y=-2,x=-3,z=6}
{y=-2,x=0,z=3}
{y=0,x=-2,z=3}
{y=0,x=1,z=0}
{y=1,x=0,z=0}

for y = -1 the solutions, are not integer solutions.