# Integer solutions of $x^3-19 = y^2$

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 $K$.

• Ummm, $19 \equiv 3 \mod 4$. Dec 5 '17 at 2:55
• But $-19\equiv1\bmod4$, which is probably creating some difficulty. Dec 5 '17 at 3:32

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.
• But $y\pm\sqrt{-19}$ are not coprime, are they? Aren't both divisible by $2$, if $y$ is even? Dec 5 '17 at 3:34
• @Gerry These elements are divisible by $2$ if $y$ is odd. Dec 5 '17 at 6:01
• @GerryMyerson The OP points out that $x$ is odd, therefore $y$ is even, and that means that $y+\sqrt{-19}$ is not divisible by $2$. Dec 5 '17 at 7:37
• Quite right, LStU (though neither you nor OP explicitly pointed out that $y$ is even, but that's a tiny omission). Dec 5 '17 at 8:12