integer solutions to $y^2+2=x^3$

 
Theorem 3.5. The only integer solutions to $y^2+2=x^3$ are $x=3, y=±5.$ Proof. Factor the equation as \[\tag{3.2} (y+\sqrt{-2})(y-\sqrt{-2})=x^3 . \] We claim that the two factors on the left are coprime (the only integers in $\mathbf{Z}[\sqrt{-2}]$ dividing both of them are units). Suppose, to the contrary, that some irreducible $\alpha$ divides both factors. Then $\alpha$ divides $(y+\sqrt{-2})-(y-\sqrt{-2})=2 \sqrt{-2}=-(\sqrt{-2})^3$. Now $\sqrt{-2}$ is irreducible in $\mathbf{Z}[\sqrt{-2}]$, since it has norm 2, so if it factors into two elements of $\mathbf{Z}[\sqrt{-2}]$, one of them must have norm 1 and hence be a unit. Therefore, by unique factorisation into irreducibles, $\alpha$ is an associate of $\sqrt{-2}$. Modifying $\alpha$ by a unit, we can assume that $\alpha=\sqrt{-2}$. Thus $\sqrt{-2} \mid(y+\sqrt{-2})$, and so $\sqrt{-2} \mid y$. Taking norms, we see that $2 \mid y^2$, and hence $2 \mid y$. But then, returning to the original equation $y^2+2=x^3$, we see that $2 \mid x$, and hence $y^2 \equiv 6\pmod{8}$. This is impossible, and so indeed the two factors on the left in (3.2) are coprime. Using unique factorisation again, it follows that both $y \pm \sqrt{-2}$ are associates of cubes in $\mathbf{Z}[\sqrt{-2}]$. Since the only units in $\mathbf{Z}[\sqrt{-2}]$ are $\pm 1$, and $-1$ is a cube, both $y \pm \sqrt{-2}$ are cubes. Suppose that \[ y+\sqrt{-2}=(a+b \sqrt{-2})^3, \] where $a, b \in \mathbf{Z}$. Expanding out and comparing coefficients of $\sqrt{-2}$, we obtain \[ 1=b(3 a^2-2 b^2) . \] This is a very easy equation to solve over the integers. We must have either $b=-1$, in which case $3 a^2-2=-1$, which is impossible, or else $b=1$, in which case $3 a^2-2=1$ and so $a= \pm 1$. This leads to $y+\sqrt{-2}=(\pm 1+\sqrt{-2})^3$ and so $y= \pm 5$.