a Diophantine Equation (Fermat)

The ring $\mathbb{Z}[i \sqrt{2}]$ is a Euclidean domain, and we want to find $x,y \in \mathbb{Z}$ such that $x^3 = (y + i\sqrt{2})(y - i\sqrt{2})$ . The multiplicative norm function for $\mathbb{Z}[i \sqrt{2}]$ is $N(a + ib\sqrt{2}) = a^2 + 2b^2$ .

Suppose that $z$ is an element of $\mathbb{Z}[i \sqrt{2}]$ which is a common factor of $y + i\sqrt{2}$ and $y - i\sqrt{2}$ . Then $z$ must divide $2i\sqrt{2}$ , and so $N(z)$ must divide $8$ . If $N(z) = 2$ we must have $z = \pm i\sqrt{2}$ . This would imply that $y$ was even, and hence that $x^3 = y^2 + 2 \equiv 2 \pmod{4}$ , which is impossible. If $N(z) = 4$ we must have $z = \pm2$ , which is not possible. If $N(z) = 8$ we must have $z = \pm2i\sqrt{2}$ , which is not possible either. Thus $N(z) = 1$ , and so $z$ is a unit. Thus $y+i\sqrt{2}$ and $y-i\sqrt{2}$ are coprime in $\mathbb{Z}[i \sqrt{2}]$ . Since their product $x^3$ is a cube, it follows that each of $y + i\sqrt{2}$ and $y - i\sqrt{2}$ is a cube in $\mathbb{Z}[i \sqrt{2}]$ .

Thus we can find $u,v \in \mathbb{Z}$ such that $y + i\sqrt{2} \; = \; (u + iv\sqrt{2})^3 \; = \; u(u^2 - 6v^2) + v(3u^2 - 2v^2)i\sqrt{2}$ so that $y \; = \; u(u^2 - 6v^2) \hspace{2cm} v(3u^2 - 2v^2) \; = \; 1$ From the second equation we see that $v = \pm1$ and that $3u^2 - 2v^2 = \pm1$ , and so $3u^2 = 2 \pm 1$ . Thus it follows that $v=1$ and $u^2 = 1$ , so that $u = \pm1$ , and hence $y = \mp5$ . For each of these values of $y$ , we have $x=3$ .

Thus $A \,=\, \big\{(3,5),(3,-5)\big\}$ , and so the answer is $3+5 + 3 - 5 = \boxed{6}$ .