Mordell : Where is the bug yeah ???

[Nadim Ul Abrar]

Using this method I became able to solve the equation for many ($<10$ :S) values of $k$ .
But I've failed to solve for $k=24$ .

Boro Vaiz , Please check my solution

$y^2=x^3+24$
or $x^3=(y+2 \sqrt6)(y-2 \sqrt6)$

Now $(y+2 \sqrt6),(y-2 \sqrt6)$ are relatively prime in $\mathbb {Z}[\sqrt(6)]$

let$ d|(y+2 \sqrt6),(y-2 \sqrt6)$
then $d|4 \sqrt6$
So $N(d)|96$.
Again $N(d)|x^3+24$
With $mod8 $we have $x^3+24 $is odd .
So $N(d)=1,3$

If $N(d)=3 $then
$3|x^3+24$
or $3|x$ . And this property gives contradiction .
So $N(d)=1$ and $d$ is Unit .
And $(y+2 \sqrt6),(y-2 \sqrt6)$ are relatively prime in $\mathbb {Z}[\sqrt(6)]$

So that we can write ,
$y+2\sqrt6=(m+n\sqrt6)^3$ ($m,n $are integer)
using this relation we can say
$3(m^2n+2n^3)=2$ which leads there is no such $n,m$ .

(But google says the equation has solution for $x=-2,1,10,8158 $:( )

[Tahmid Hasan]

$y^2=x^3+24$
or $x^3=(y+2 \sqrt6)(y-2 \sqrt6)$

can you really conclude that?

[nayel]

Using this method I became able to solve the equation for many ($<10$ :S) values of $k$ .
But I've failed to solve for $k=24$ .

Boro Vaiz , Please check my solution

$y^2=x^3+24$
or $x^3=(y+2 \sqrt6)(y-2 \sqrt6)$

Now $(y+2 \sqrt6),(y-2 \sqrt6)$ are relatively prime in $\mathbb {Z}[\sqrt(6)]$

let$ d|(y+2 \sqrt6),(y-2 \sqrt6)$
then $d|4 \sqrt6$
So $N(d)|96$.
Again $N(d)|x^3+24$
With $mod8 $we have $x^3+24 $is odd .
So $N(d)=1,3$

If $N(d)=3 $then
$3|x^3+24$
or $3|x$ . And this property gives contradiction .
So $N(d)=1$ and $d$ is Unit .
And $(y+2 \sqrt6),(y-2 \sqrt6)$ are relatively prime in $\mathbb {Z}[\sqrt(6)]$

So that we can write ,
$y+2\sqrt6=(m+n\sqrt6)^3$ ($m,n $are integer)
using this relation we can say
$3(m^2n+2n^3)=2$ which leads there is no such $n,m$ .

(But google says the equation has solution for $x=-2,1,10,8158 $:( )

Your method fails because $\mathbb Z[\sqrt 6]$ is not a Unique Factorisation Domain. As an example, we have $-2=(2+\sqrt 6)(2-\sqrt 6)$ and it can be shown that $-2$ and $2\pm\sqrt 6$ are irreducibles.