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

Sep 24, 2016

$\left(x , y , z\right) = \left(1 , - 1 , 1\right) \mathmr{and} \left(- 1 , 1 , 1\right)$

Sep 24, 2016

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

#### Explanation:

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

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

$\frac{{x}^{3} + {y}^{3}}{x + y} = \frac{\left(1 - z\right) \left(1 + z\right)}{1 - z}$ or

${x}^{2} - x y + {y}^{2} = 1 + z$

Adding this equation with the first we have

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

$x = \frac{1}{2} \left(- 1 + y \pm \sqrt{3} \sqrt{3 - 2 y - {y}^{2}}\right)$

Here

$3 - 2 y - {y}^{2} \ge 0$ so

$- 3 \le y \le 1$ but $y \in \mathbb{N}$ so $y \in \left\{- 3 , - 2 , - 1 , 0 , 1\right\}$

Checking we have

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

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