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

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Answer 1 · 2016-09-24T08:14:51

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

Answer 2 · 2016-09-24T11:17:01

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

$x+y=1-z$
$x^3+y^3=1-z^2$

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

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

$x^2-xy+y^2=1+z$

Adding this equation with the first we have

$x^2-x y+y^2+x+y=2$ . Solving for $x$ we obtain

$x = 1/2 (-1 + y pm sqrt[3] sqrt[3 - 2 y - y^2])$

Here

$3 - 2 y - y^2 ge 0$ so

$-3 le y le 1$ 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.

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