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Is it true that if $$4 x^3 = y^2 +27$$ then $$y=3k$$ I tried the Fermat theorem for $x^3$, but seems I am missing something.

EDITED:

$x$ and $y$ are integers.

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    Try looking up the Mordell curve, see what the internet spits out. This can certainly not be solved by Fermat's Little Theorem, because of the $4$ coefficient in front of the $x$. – Landuros Jan 05 '18 at 01:13
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    The only integral (and rational) solution is $(x,y) = (3, \pm 9)$, according to Magma. – Hw Chu Jan 05 '18 at 01:16
  • Very interesting, I am just wondering about the algorithm it uses. – Gevorg Hmayakyan Jan 05 '18 at 01:17

2 Answers2

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You are correct that you could use Fermat's theorem for $x^3+y^3=z^3$. Starting with your elliptic curve $$4x^3=y^2+27$$ we borrow Yong Hao Ng's answer by multiplying it by $16$ to get $$(4x)^3=(4y)^2+432$$

$$X^3-432=Y^2\tag1$$

This is a special case of a well-known family. The problem of finding two rational cubes equal to $N$ $$p^3+q^3 = N$$ or, $$\left(\frac{36N+v}{6u}\right)^3+\left(\frac{36N-v}{6u}\right)^3 =N$$ simplifies to

$$u^3-432N^2 =v^2\tag2$$

Thus, your curve is just the case $N=1$. Of course, by Fermat's Last Theorem, there are no non-trivial rational solutions to,

$$p^3 + q^3 =1$$

so the only rational point on $(1)$ is its torsion point $X=12$, and no other.

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You can multiply by $16$ to convert this into $$ \begin{align*} (4y)^2 &= (4x)^3 - 432 \\ Y^2 &= X^3 - 432 \end{align*} $$ i.e. a Mordell curve with $n=-432$.

General theory
In the article you will see a mention that the number of integers is finite, but more generally any equation of the form $$ Y^2 = f(X) $$ where $f(X)\in \mathbb Q[X]$ has $\geq 3$ distinct roots in $\mathbb C$ will have only finitely many integral points. This is the Siegel's theorem on integral points. Therefore we find all of them and check which ones are of the form $(4x,4y)$.

From what I have learned there are no easy ways to find these integral points. However it is known that there is an effective method, i.e. we can find an upper bound of $|X|,|Y|$, as referenced here: $$ \begin{align*} \max\{|X|,|Y|\} &< e^{(10^6 H)^{10^6}} \end{align*} $$ where $H$ is the maximum absolute value of the coefficients of the equation. The problem here is that the bounds are too huge for exhaustive search, so often you see adhoc methods for specific problems to find the optimal bounds to try. This paper has computations deriving said bounds for Mordell curves, but the methods are very advanced.

Solving it via Sagemath
The easy ones to find are the torsion points, using Nagell-Lutz theorem. This says that either $Y=0$ or $Y^2|D$, where $D$ is the discriminant. Here $D=-27(-432)^2=-5038848$ so the list of $Y$ to try are $$ \{1,2,3,4,6,8,9,12,16,18,24,27,36,48,54,72,81,108,144,162,216,324,432,648,1296\} $$ Then trying this list shows that only $Y=36$ works, giving $X=12$. The symmetry of Elliptic curves gives you the other solution $(X,Y)=(12,-36)$. Hence the torion points are of the form $(4x,4y)$ that we want, giving $(x,y) = (3,\pm 9)$.

The difficult ones are the non-torsion points. However from Sagemath this curve has rank $0$, so the only rational points are the torsion points and we are done.

Yong Hao Ng
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