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From: Leonid Lenov on 12 Jan 2010 16:48
Hello,
Diophantine equation y^2=x^3-3 can have only finitely many solutions.
Unfortunately, Z[sqrt{-3}] is not UFD so one cannot use unique
factorization to obtain it's solutions. Nevertheless, I got that it
has no solutions but am unsure about the correctness of my method and
therefor the result. Can someone tell me is it true that it has no
solution?
Thanks in advance.
From: preedmont on 12 Jan 2010 17:11

"Leonid Lenov" <leonidlenov(a)gmail.com> wrote in message
news:1219fc00-5d64-48ef-911c-05d0a4a9ec2b(a)j24g2000yqa.googlegroups.com...
> Hello,
> Diophantine equation y^2=x^3-3 can have only finitely many solutions.
> Unfortunately, Z[sqrt{-3}] is not UFD so one cannot use unique
> factorization to obtain it's solutions. Nevertheless, I got that it
> has no solutions but am unsure about the correctness of my method and
> therefor the result. Can someone tell me is it true that it has no
> solution?
> Thanks in advance.

go to

http://www.alpertron.com.ar/QUAD.HTM


From: master1729 on 12 Jan 2010 07:43
Helpful wrote :

>
> "Leonid Lenov" <leonidlenov(a)gmail.com> wrote in
> message
> news:1219fc00-5d64-48ef-911c-05d0a4a9ec2b(a)j24g2000yqa.
> googlegroups.com...
> > Hello,
> > Diophantine equation y^2=x^3-3 can have only
> finitely many solutions.
> > Unfortunately, Z[sqrt{-3}] is not UFD so one cannot
> use unique
> > factorization to obtain it's solutions.
> Nevertheless, I got that it
> > has no solutions but am unsure about the
> correctness of my method and
> > therefor the result. Can someone tell me is it true
> that it has no
> > solution?
> > Thanks in advance.
>
> go to
>
> http://www.alpertron.com.ar/QUAD.HTM
>
>

your mad.

how does a degree 2 diophantine equation applet help to solve a degree 3 diophantine equation ?

not very " helpful " , mr helpful.

this just shows how bad things are on sci.math.

most questions are answered by links or book references and such.

instead of really helping and doing math.

helpful has rediculized JSH number theory posts.

now lets see how well he can answer number theory questions himself.

tommy1729
From: Achava Nakhash, the Loving Snake on 12 Jan 2010 18:14
On Jan 12, 1:48 pm, Leonid Lenov <leonidle...(a)gmail.com> wrote:
> Hello,
> Diophantine equation y^2=x^3-3 can have only finitely many solutions.
> Unfortunately, Z[sqrt{-3}] is not UFD so one cannot use unique
> factorization to obtain it's solutions. Nevertheless, I got that it
> has no solutions but am unsure about the correctness of my method and
> therefor the result. Can someone tell me is it true that it has no
> solution?
> Thanks in advance.

The ring of algebraic integers in Q[sqrt(-3)] is a unique
factorization domain, and the only units are plus and minus 1 and the
2 imaginary cube roots of unity. The integers in this ring are of the
form (a + b*sqrt(-3))/2 where a and b are either both odd or both
even. So it is not the same as Z[sqrt(-3)] which isn't even
integrally closed.

Now the approach you appear to have had in mind should work without
any difficulty.


Regards,
Achava
From: master1729 on 12 Jan 2010 23:51
> On Jan 12, 1:48 pm, Leonid Lenov
> <leonidle...(a)gmail.com> wrote:
> > Hello,
> > Diophantine equation y^2=x^3-3 can have only
> finitely many solutions.
> > Unfortunately, Z[sqrt{-3}] is not UFD so one cannot
> use unique
> > factorization to obtain it's solutions.
> Nevertheless, I got that it
> > has no solutions but am unsure about the
> correctness of my method and
> > therefor the result. Can someone tell me is it true
> that it has no
> > solution?
> > Thanks in advance.
>
> The ring of algebraic integers in Q[sqrt(-3)] is a
> unique
> factorization domain, and the only units are plus and
> minus 1 and the
> 2 imaginary cube roots of unity. The integers in
> this ring are of the
> form (a + b*sqrt(-3))/2 where a and b are either both
> odd or both
> even. So it is not the same as Z[sqrt(-3)] which
> isn't even
> integrally closed.
>
> Now the approach you appear to have had in mind
> should work without
> any difficulty.
>
>
> Regards,
> Achava

what approach ?
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