How to solve $y^2=x^3+3$ where x,y integer?

Asked Jun 11 '24 at 10:28

Active Jun 11 '24 at 11:35

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In my Number theory class my exam problem was to solve Mordell equation $y^2=x^3+3$.

But I don't know how to solve this. One approach was to subtract 3 and use the fact that $F=Q(\sqrt 3)$ is ED, but what I can do was just transforming this equations to another Thue equations... So I want to know how to solve this.

edited Jun 11 '24 at 11:29

EDX

2,413

asked Jun 11 '24 at 10:28

Ena

y= 2,-2 x = 1 ez – uggupuggu Jun 11 '24 at 11:52
@uggupuggu Did you actually read the question? – Antony Theo. Jun 11 '24 at 11:53
@uggupuggu Proving there are no other solutions is less "ez", though. There is a similar question ( https://math.stackexchange.com/questions/473180/the-diophantine-equation-x2-2-y3 ) but its approach can't be used directly since $\mathbb{Z}[\sqrt{3}]$ is not UFD. – Abstraction Jun 11 '24 at 11:58
3

$\Bbb{Z}[\sqrt{3}]$ is a UFD. – SomeCallMeTim Jun 11 '24 at 12:07
Cf. this question – J. W. Tanner Jun 11 '24 at 12:26
May you use the writing
$$ y^2-4 =x^3-1$$ $$ (y-2)(y+2) =(x-1)(x^2+x+1)$$
– EDX Jun 11 '24 at 12:33
@SomeCallMeTim Indeed it is. I am sorry for the misleading comment. – Abstraction Jun 11 '24 at 13:13
I think the reason why the method for $y^2=x^3-2$ couldn't be applied is that Z[sqrt(3)] is a real quadratic field, so it has infinitely many units. So we need to solve $y+\sqrt 3 =u a^3$, for infinitely many u(you can reduce the possibility of u, but there are more than 1 u, and some of them make this problem hard to solve – Ena Jun 11 '24 at 15:29
1

Does this answer your question? Diophantine equation $n^3+3=m^2$ - found from J. W. Tanner's comment. – John Omielan Jun 11 '24 at 15:58
Oh I checked that and another informations related to that, but what I can know was that the number of solution is finite and can be bounded by incredibly large number(or just Im too stupid so that I missed some important information) – Ena Jun 11 '24 at 16:07
Umm...I think I was a bit stupid. I found some books that has a possibility to solve my problem, but I don't want to buy books to solve this problem.... – Ena Jun 11 '24 at 16:18

How to solve $y^2=x^3+3$ where x,y integer?

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