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Recently I met a problem when I'm studying algebraic number theory.

Problem. Find all positive integer solutions of $y^3=x^2+20$.

I solved the situation when $x$ is an odd because the two ideals $(x+2\sqrt{-5})\text{ and }(x-2\sqrt{-5})$ are relatively prime, so it must be a cube of a principal ideal for the ideal class number of $\mathbb{Z}[\sqrt{-5}]$ is $2$. But when $x$ is an even, let $x=2k$, then I know that $(k+\sqrt{-5})=IJ^3$, where $I=(2,1+\sqrt{-5})$, and $J$ is also a ideal. I don't know how to deal with it for they aren't principal ideals anymore. I believe that there is a unique solution $(x,y)=(14,6)$, and wish that someone could solve it.

I'm also interested in whether there is a method by elementary number theory, thanks.

Fate Lie
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    This is the elliptic curve $y^2=x^3-20$. The integral points have been determined, see K. Conrad's notes about it, here, section 3, and references. The only integral points are $(6,\pm 14)$. – Dietrich Burde Dec 07 '23 at 13:23
  • The question has been asked already here. – Dietrich Burde Dec 07 '23 at 13:38
  • @DietrichBurde Sorry, but this paper seems not to give the answer when $k=-20$. – Fate Lie Dec 07 '23 at 14:04
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    The answer is known, see the other links to elliptic curves. But it is not clear that there is an elementary proof. Nevertheless, Conrad's notes are the best place to look for such methods, also from algebraic number theory. – Dietrich Burde Dec 07 '23 at 15:19
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    Note this is cross-posted on MathOverflow at A Mordell Equation $y^3=x^2+20$. Note that cross-posting is generally frowned on, especially if it's done within less than about a week's time. Also, if you do cross-post, please provide a link within each post to the other one, to help avoid people duplicating efforts. – John Omielan Dec 07 '23 at 17:19
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    Given the gazillion of trivial questions that seem to be appropriate here, this looks like a reasonable question. Those looking down upon it are welcome to provide an answer. –  Dec 09 '23 at 10:14
  • That was Lemmermeyer's Last Answer. –  Dec 09 '23 at 22:14

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Let me see if I can get it right. Consider the case $y^2 = x^3 - 20$ with $x$ and $y$ even; then $x^3 = y^2 + 20 = (y+2\sqrt{-5})(y-2\sqrt{-5})$ shows that $y + 2\sqrt{-5} = 2{\mathfrak a} {\mathfrak b}^3$ and $y - 2\sqrt{-5} = 2{\mathfrak a} {\mathfrak c}^3$, where ${\mathfrak a} = (2, 1+\sqrt{-5})$ is the prime ideal above $2$ and where ${\mathfrak c}$ is the conjugate of ${\mathfrak b}$. This is because, with $y = 2k$ and $k$ necessarily odd, the elements $k + \sqrt{-5}$ and $k - \sqrt{-5}$ have gcd ${\mathfrak a}$.

Now clearly ${\mathfrak b}$ is not principal since ${\mathfrak a}$ is not; since the class group of $K = {\mathbb Q}(\sqrt{-5})$ is cyclic of order $2$, this implies that $ {\mathfrak a} {\mathfrak b} = (r+s\sqrt{-5})$ is principal. Since $(2) = {\mathfrak a}^2$, our equations now become $$ y + 2\sqrt{-5} = (r+s\sqrt{-5})^3, \quad y - 2\sqrt{-5} = (r-s\sqrt{-5})^3. $$ Subtracting them from each other gives, after simplification, $$ 2 = s(3r^2 - 5s^2). $$ This implies $s \in \{\pm 1, \pm 2\}$. Only $s = -1$ leads to a solvable equation, namely $2 = -(3 - 5)$ with $r = \pm 1$. But now $(\pm 1 - \sqrt{-5})^3 = \pm 14 + 2 \sqrt{-5}$ shows that $x = \pm 14$ and thus $y = 6$.