It's Just A Mo_del_

If $x$ is even, then $y$ is even, and $x^2+2\equiv 0\;\; (\text{mod }8)$ , implying that $x^2\equiv 6\;\; (\text{mod }8)$ , a contradiction. Hence, $x$ is odd. Rewrite the equation in $\mathbb{Z}\left[\sqrt{-2}\right]$ as

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

Note that any common divisor $d$ of $x-\sqrt{-2}$ and $x+\sqrt{-2}$ must divide $x+\sqrt{-2}-\left(x-\sqrt{-2}\right)=2\sqrt{-2}$ , meaning that the norm of $d$ divides the norm of $2\sqrt{-2}=8$ . But the norm of $d$ also divides $x^2+2$ , which is odd since $x$ is odd. Thus the norm of $d$ is $1$ , implying that $d$ is a unit in $\mathbb{Z}\left[\sqrt{-2}\right]$ (namely $d=\pm 1$ ). Hence, $x-\sqrt{-2}$ and $x+\sqrt{-2}$ are relatively prime in $\mathbb{Z}\left[\sqrt{-2}\right]$ , and since $d=-1$ and $d=1$ are both cubes, we can write

$x-\sqrt{-2}=\left(a+b\sqrt{-2}\right)^3$

for some $a,b\in \mathbb{Z}$ . Expanding and equating coefficients yields $x=a^3-6ab^2$ and $3a^2b-2b^3=b(3a^2-2b^2)=-1$ , implying that $b=\pm 1$ . If $b=1$ , then $a\not\in \mathbb{Z}$ , so $b=-1$ , implying $a=\pm 1$ . Hence, the only solutions are $x=\pm 5$ , and thus $y=3$ . Since we seek positive integer solutions, we are left with the single solution $(x,y)=(5,3)$ . Hence, the answer is $\boxed{1}$ .

Note that this is just one case of the more general Mordell's equation .

There is only 1 intersecting point of the graph of $y=x^3$ and $y=x^2 + 2$