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From: Leonid Lenov on 13 Jan 2010 10:37
On Jan 13, 12:14 am, "Achava Nakhash, the Loving Snake"
<ach...(a)hotmail.com> wrote:
> 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.

Can we just use the ring R={2^k (a+b*sqrt{-3})| a,b,k in Z and either
one of a or b is odd} with the norm N(2^k(a+b*sqrt{-3}))=a^2+3b^2? Now
units will be +/-1 and numbers of the form +/-2^k where k in Z.
If R is a UFD it follows that y-sqrt{-3}=+/-2^k(x+iy)^3 which has no
solutions.

From: Leonid Lenov on 13 Jan 2010 10:40
On Jan 13, 3:51 pm, master1729 <tommy1...(a)gmail.com> wrote:
> what approach ?
If Z[sqrt{-d}] is a UFD and y^2+d=x^3 than if (y-sqrt{-d},y+sqrt{-d})
=1 it follow that y+sqrt{-d}=u(n+m*sqrt{-d})^3 from where we can see
all the solutions if we know what the unit u can be like.
From: The Pumpster on 13 Jan 2010 12:50
On Jan 13, 7:40 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
> On Jan 13, 3:51 pm, master1729 <tommy1...(a)gmail.com> wrote:> what approach ?
>
> If Z[sqrt{-d}] is a UFD and y^2+d=x^3 than if (y-sqrt{-d},y+sqrt{-d})
> =1 it follow that y+sqrt{-d}=u(n+m*sqrt{-d})^3 from where we can see
> all the solutions if we know what the unit u can be like.

I guess one could note that the related L-function has L(1) > 2
(and there is no torsion), but maybe this is cheating!

de P

From: Achava Nakhash, the Loving Snake on 13 Jan 2010 14:34
On Jan 13, 7:37 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
> On Jan 13, 12:14 am, "Achava Nakhash, the Loving Snake"
>
> <ach...(a)hotmail.com> wrote:
> > 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.
>
> Can we just use the ring R={2^k (a+b*sqrt{-3})| a,b,k in Z and either
> one of a or b is odd} with the norm N(2^k(a+b*sqrt{-3}))=a^2+3b^2? Now
> units will be +/-1 and numbers of the form +/-2^k where k in Z.
> If R is a UFD it follows that y-sqrt{-3}=+/-2^k(x+iy)^3 which has no
> solutions.

I am confused by 2 things in your reply. Do you mean to allow k to be
negative? I assume that you are simply pulling out the common factor
of the highest power of 2 dividing both a and b and leaving out cases
where a and b are both odd. If you do mean k to be non-negative, you
will never get unique factorization because you need your ring to be
integrally closed. That means that any element of the quotient field
of your ring (in this case Q(sqrt(-3)) that satisfies a monic
polynomial with coefficients in the ring must already be in the ring.
That is clearly not true in this instance.

The other point of confusion is what you mean by norm. As you state
it, is it actually multiplicative? I suspect not. After all, the
norm of any power of 2 is 1 the way you have it defined.

I will try to present a proof of your contention that there are no
solutions when I get a chance, which I can't promise will be less than
a week from now because it is not clear to me that I will have a
computer to use.

Regards,
Achava
From: Leonid Lenov on 13 Jan 2010 15:43
On Jan 13, 8:34 pm, "Achava Nakhash, the Loving Snake"
<ach...(a)hotmail.com> wrote:
> On Jan 13, 7:37 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
>
> > On Jan 13, 12:14 am, "Achava Nakhash, the Loving Snake"
>
> > <ach...(a)hotmail.com> wrote:
> > > 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.
>
> > Can we just use the ring R={2^k (a+b*sqrt{-3})| a,b,k in Z and either
> > one of a or b is odd} with the norm N(2^k(a+b*sqrt{-3}))=a^2+3b^2? Now
> > units will be +/-1 and numbers of the form +/-2^k where k in Z.
> > If R is a UFD it follows that y-sqrt{-3}=+/-2^k(x+iy)^3 which has no
> > solutions.
>
> I am confused by 2 things in your reply.  Do you mean to allow k to be
> negative?  

Yes.

> I assume that you are simply pulling out the common factor
> of the highest power of 2 dividing both a and b and leaving out cases
> where a and b are both odd.  If you do mean k to be non-negative, you
> will never get unique factorization because you need your ring to be
> integrally closed.  That means that any element of the quotient field
> of your ring (in this case Q(sqrt(-3)) that satisfies a monic
> polynomial with coefficients in the ring must already be in the ring.
> That is clearly not true in this instance.
>
> The other point of confusion is what you mean by norm.

By norm I mean Euclidean norm.

> As you state
> it, is it actually multiplicative?  I suspect not.  

I think it is...

> After all, the
> norm of any power of 2 is 1 the way you have it defined.

Yes, so the units of R are +/-2^k, k in Z.

> I will try to present a proof of your contention that there are no
> solutions when I get a chance, which I can't promise will be less than
> a week from now because it is not clear to me that I will have a
> computer to use.

Thanks in advance.
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