TITLE: Integer solutions of $x^3-19 = y^2$
QUESTION [6 upvotes]: I am asked to find all integer solutions of the elliptic curve $x^3-19 = y^2$.
I've noticed so far that $x=7$ and $y=18$ is a solution and I've also noticed that $x$ and $y$ are coprime and that $x$ must be odd. 
I've been trying to work in $K=\mathbb{Q}(\sqrt{-19})$, but since $19 \equiv 1 \pmod{4}$, I haven't found a way to relate this equation with the ring of integers of $K$.

REPLY [5 votes]: So $x^3=(y+\sqrt{-19})(y-\sqrt{-19})$. One shows in the usual way that
$y+\sqrt{-19}$ and $y-\sqrt{-19}$ are coprime in $R=\Bbb Z[\frac12(1+\sqrt{-19})]$. Then $R$ is a PID ($\Bbb Q(\sqrt{-19})$ has class number $1$) and its only units are $\pm1$. Then we get $y+\sqrt{-19}=\alpha^3$
with $\alpha\in R$. This means that
$$\left(\frac{a+b\sqrt{-19}}2\right)^3=y+\sqrt{-19}$$
where $a$ and $b$ have the same parity. Then $a^3-57ab^2=8y$
and $3a^2b-19b^3=8$. This last equation narrows down $a$ and $b$ to
a finite set.