Question

D={x€Z| 14 supra x+15 € Z}

Answer 1

$\frac{14}{x + 1} \: \in \: \mathbb{Z} = > x + 1 \: \in \: D_{14}$

$D_{14} = \left \{ \pm1,\pm2,\pm7,\pm14 \right \}$

$1)x + 1 = - 1 = > x = - 1 - 1 = - 2$

$\frac{14}{ - 2 + 1} = \frac{14}{ - 1} = - \frac{14}{1} = - 14 \: \in \: \mathbb{Z}$

$2)x + 1 = 1 = > x = 1 - 1 = 0$

$\frac{14}{0 + 1} = \frac{14}{1} = 14 \: \in \: \mathbb{Z}$

$3)x + 1 = - 2 = > x = - 2 - 1 = - 3$

$\frac{14}{ - 3 + 1} = \frac{14}{ - 2} = - \frac{14}{2} = - 7 \: \in \: \mathbb{Z}$

$4)x + 1 = 2 = > x = 2 - 1 = 1$

$\frac{14}{1 + 1} = \frac{14}{2} = 7 \: \in \: \mathbb{Z}$

$5)x + 1 = - 7 = > x = - 7 - 1 = - 8$

$\frac{14}{ - 8 + 1} = \frac{14}{ - 7} = - \frac{14}{7} = - 2 \: \in \: \mathbb{Z}$

$6)x + 1 = 7 = > x = 7 - 1 = 6$

$\frac{14}{6 +1 } = \frac{14}{7} = 2 \: \in \: \mathbb{Z}$

$7)x + 1 = - 14 = > x = - 14 - 1 = - 15$

$\frac{14}{ - 15 + 1} = \frac{14}{ - 14} = - \frac{14}{14} = - 1 \: \in \: \mathbb{Z}$

$8)x + 1 = 14 = > x = 14 - 1 = 13$

$\frac{14}{13 + 1} = \frac{14}{14} = 1 \: \in \: \mathbb{Z}$