How to prove that $x^3=6y^2+2$ has only solution $(2,1)$

Question

How to prove that diophantine equation: $x^3=6y^2+2$ has only solution $(2,1)$

Answer 1

This is $$\left(\frac x2\right)^3=\left(\frac{1+3y\sqrt{-3}}2\right) \left(\frac{1-3y\sqrt{-3}}2\right).$$ We can use unique factorisation in $R=\Bbb Z[\frac12(1+\sqrt{-3})]$ to write $$\frac{1+3y\sqrt{-3}}2=\omega^j\alpha^3$$ where $\omega=\frac12(-1+\sqrt{-3})$, $j\in\{0,1,2\}$ and $\alpha\in R$. Then $1=\omega^j\alpha^3+\omega^{-j}\overline{\alpha}^3$ and one can proceed from there.

Answer 2

I've tried an unfinished approach by elementary conclusions, maybe it helps you. Firstly: $$x^3=6y^2+2$$

So, you can conclude, that $x$ is even because right always even. Put $x=2k, (k\in \mathbb{N})$. Then $$4k^3=3y^2+1$$

From here we conclude that $y$ is odd. Put $y=2m-1, (m\in \mathbb{N})$, we have: $$4k^3=12m^2-12m+4$$ $$k^3-1=3m(m-1)$$

As, $k^3-1$ must divide $3$, then $k=3l+1, (l\in \mathbb{N_0})$. Then, we obtain: $$3l(3l^2+3l+1)=m(m-1)$$

In the right side we have products 0f consecutive pairs like: $(0;1), (1;2), (2;3)$ and so on. All divisors of $3l^2+3l+1$ are congruent to $1$ modulo $6$ see proof here. Not including trivial solutions $m=1, l=0$, we proceed further... By above congruence and by that $m(m-1)$ divides $3$, we restrict our consecutive pairs to these: $(6;7), (12;13), (18,19), (24; 25),...=(6j;6j+1), (j\in \mathbb{N})$

And, I left with proving: $$\frac{l(3l^2+3l+1)}{j(6j+1)}\neq2$$

So, I don't know how proceed further, maybe you or someone can finish this...

Answer 3

Facts about Mordell equations can be used.

$x^3=6y^2+2$ is equivalent to

$(6x)^3=(36y)^2+432$.

In this case, we're lucky that $-10000<432<10000$, otherwise another method would be needed, as in this case I've answered before.

See http://oeis.org/search?q=mordell&language=english&go=Search

in particular (this link is given inside the link), e.g., http://oeis.org/A081120/b081120.txt

You'll find that $a^3=b^2+432$ has exactly two integer solutions, so indeed $(x,y)=(2,\pm 1)$ are the only integer solutions.

Unfortunately, this link only gives integer solutions $(x,y)$ to $y^2=x^3+k$, where $1\le k\le 10000$, $k\in\mathbb Z$, not when $-10000\le k\le -1$.

Apparently, OEIS claims that there exists only a cached copy, after the original web site tnt.math.se.tmu.ac.jp was shut down in 2017.