In the book "Elliptic curves, number theory and cryptography" by Washington in section 2.5.2 it says that it is somewhat not trivial to show that the only rational solutions to the curve $y^2=x^3-432$ are $(12,\pm36)$ and $\infty$. I am curious as to how one would show this.

  • $\begingroup$ You may want to look at this: math.stackexchange.com/questions/1346512/… $\endgroup$ – user25406 Feb 20 at 0:05
  • $\begingroup$ And also this: math.stackexchange.com/questions/485079/… $\endgroup$ – user25406 Feb 20 at 0:06
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    $\begingroup$ Let $(x,y)= (X/Z,Y/Z)$ for some integers $X,Y,Z$. One way is to clear denominators to get $$Y^2Z = X^3 - 432Z^3,$$ and writing it as $$ (36 Z + Y)^3 + (36 Z - Y)^3 = (6 X)^3 $$ Then using Fermat's Last Theorem we get $36Z+Y=0, 36Z-Y=0$ or $6X=0$. $X=0$ has no integer solutions $Y,Z$, so $Y= \pm 36Z\implies y=Y/Z = \pm 36$, but it's probably not as good as factorization over number fields. $\endgroup$ – Yong Hao Ng Feb 20 at 2:43

The following is a reduction to $4(x')^3 = (y')^2 + 27$ and $(x'')^3 = 3(y'')^2+16$ (these might be marginally easier?)

Edit: This is for integer solutions (sorry, I misread the question).

Write the equation as $x^3 = y^2 + 432 = (y - \sqrt{-423})(y+\sqrt{-432}) = (y - 12\sqrt{-3})(y + 12\sqrt{-3})$.

The ring of integers of $K = \mathbb{Q}(\sqrt{-3})$ is $\mathcal{O}_K = \mathbb{Z}\big[\frac{1+\sqrt{-3}}{2}\big]$ and it is a UFD.

A prime $\pi$ in $\mathcal{O}_K$ that divides both $y+12\sqrt{-3}$ and $y-12\sqrt{-3}$ divides their difference $24\sqrt{-3} = 2^3 \sqrt{-3}^3$.

So, up to units $\pi = 2$ or $\pi = \sqrt{-3}$ or $\pi$ does not exist.

If $\pi=2$ then $y$ is even, hence $x$ is even, hence $y^2$ is divisible by 8, hence $y$ is divisible by $4$, hence $x^3$ is divisible by $16$, hence $x$ is divisible by $4$. Substitutions $x=4x', y=4y'$ yield $4(x')^3 = (y')^2 + 27$. This has solutions $x'=3$ and $y'=\pm 9$; hence $x = 12$ and $y = \pm 36$.

If $\pi=\sqrt{-3}$ then $y$ is threeven, hence $y^2$ is divisible by $9$, hence $x$ is threeven, hence $y^2$ is divisible by $27$, hence $y$ is divisible by $9$. Substitutions $x=3x'', y=9y''$ yield $(x'')^3 = 3(y'')^2+16$. This has solutions $x''= 4$ and $y'' = \pm 4$; hence $x = 12$ and $y = \pm 36$ again.

If $\pi$ does not exist then the factors $y\pm 12\sqrt{-3} = (y\mp 12) \pm 24\big(\frac{1+\sqrt{-3}}{2}\big)$ are coprime and hence they are both cubes, say $$\begin{align*}(y-12) + 24\bigg(\frac{1+\sqrt{-3}}{2}\bigg) &= \bigg(a + b\frac{1+\sqrt{-3}}{2}\bigg)^3 \\ &= a^3 + 3ab^2\bigg(\frac{1+\sqrt{-3}}{2}\bigg)^2 + 3a^2b\bigg(\frac{1+\sqrt{-3}}{2}\bigg) + b^3\bigg(\frac{1+\sqrt{-3}}{2}\bigg)^3\\ &= a^3 + 3ab^2\bigg(\frac{1+\sqrt{-3}}{2} - 1\bigg) + 3a^2b\bigg(\frac{1+\sqrt{-3}}{2}\bigg) - b^3\\ &= (a^3 - 3ab^2 -b^3) + (3ab^2+3a^2b)\bigg(\frac{1+\sqrt{-3}}{2}\bigg)\end{align*}$$

This implies $3ab(a+b) = 24$, i.e. $ab(a+b) = 8$. Hence both $a$ and $b$ are $\pm$ powers of $2$, say $a = 2^n, b = 2^m$ with $n \leq m$ w.l.o.g. (the equation is symmetric, and $\pm$ won't affect the argument). Then $a+b = 2^n + 2^m = 2^n(1+2^{m-n})$ is also a power of $2$, so $m-n=0$, so $8 = 2^n \cdot 2^n \cdot 2 \cdot 2^n = 2^{3n + 1}$ is a contradiction.


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