6
$\begingroup$

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 $K$.

$\endgroup$
2
  • $\begingroup$ Ummm, $19 \equiv 3 \mod 4$. $\endgroup$ Dec 5 '17 at 2:55
  • 1
    $\begingroup$ But $-19\equiv1\bmod4$, which is probably creating some difficulty. $\endgroup$ Dec 5 '17 at 3:32
5
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ But $y\pm\sqrt{-19}$ are not coprime, are they? Aren't both divisible by $2$, if $y$ is even? $\endgroup$ Dec 5 '17 at 3:34
  • 1
    $\begingroup$ @Gerry These elements are divisible by $2$ if $y$ is odd. $\endgroup$ Dec 5 '17 at 6:01
  • $\begingroup$ @GerryMyerson The OP points out that $x$ is odd, therefore $y$ is even, and that means that $y+\sqrt{-19}$ is not divisible by $2$. $\endgroup$ Dec 5 '17 at 7:37
  • $\begingroup$ @franz, of course. I don't know what I was thinking. $\endgroup$ Dec 5 '17 at 8:10
  • $\begingroup$ Quite right, LStU (though neither you nor OP explicitly pointed out that $y$ is even, but that's a tiny omission). $\endgroup$ Dec 5 '17 at 8:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.