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From: Andrew Usher on 13 Jan 2010 16:39
On Jan 13, 9:37 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:

> 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 don't see this ring as being any more a UFD than Z[sqrt(-3)] irself
- the famous factorisation 4 = 2*2 = (1+sqrt(-3))(1-sqrt(-3)) still
works, and shows that your norm is not multiplicative besides.

However, the reduced structure obtained by dividing out all factors of
2, though not a ring, does have unique factorisation (use ideals ...),
and this is enough to prove that x^2+3 = y^3 has no solution by the
argument you gave.

Andrew Usher
From: Achava Nakhash, the Loving Snake on 13 Jan 2010 16:51
On Jan 13, 12:43 pm, Leonid Lenov <leonidle...(a)gmail.com> wrote:
> On Jan 13, 8:34 pm, "Achava Nakhash, the Loving Snake"
>
>
>>
> > As you state
> > it, is it actually multiplicative?  I suspect not.  
>
> I think it is...

.... and your are so right.

Also, your set appears to me to be closed under multiplication, but
maybe not under addition. After all (2 + sqrt(-2)) = alpha and (1 +
2*sqrt(-3)) = beta are both in your set, but alpha + beta = 3 + 3*sqrt
(-3) is not in your set since both coefficients are odd. Unless I
have your definition wrong?

I will start the proof I was talking about, but my time is limited, so
I might not get to finish it.

We want to solve y^2=x^3-3 or x^3 = y^2 + 3 in integers x, y. Notice
that y has to be even, since all odd squares are 1 mod 8, but then we
have 2 divides x, so we would have 4 congruent to 0 mod 8 which it
isn't.

Now move to the ring Q{sqrt(-3)). As mentioned earlier, the elements
of this ring are all numbers of the form (a + b*sqrt(-3))/2 where a, b
in Z and either both even or both odd. This ring is a unique
factorization domain (even Euclidean, actually) and the only units are
+/- 1, +/- z and +/- z^2 where z = (-1 + sqrt(-3))/2 and we see that
z^3 = 1. All of this is very standard stuff, easily proved, and
easily looked up.

Now we factor in this ring. (y + sqrt(-3))( y - sqrt(-3)) = x^3.
Any common factor of these two terms has to divide there difference
which is 2*sqrt(-3). Now 2 and -3 are both primes in this ring (norm
considerations) and 2 can't divide both factors since y cannot be
odd. Also sqrt(-3) cannot divide either term, since then 3 divides y
and the right hand side of the originial equation is divisible by 27
but the left hand side is not.

Thus y + sqrt(-3) must be a cube in the ring times a unit, but unique
factorization. Since 1 and -1 are cubes, we need concern ourselves
only with the unit being absent or being z or z^2.

To start off, assume no unit. Then y + sqrt(-3) = (a + b*sqrt(-3))
^3 / 8, and so

8 = 3*(a^2)*b - 3*(b^3),

where a, b integers, so in particular 3 must be a factor of 8 which it
isn't. That takes care of this case. Please check my arithmetic /
algebra. At worst, we are going to get b is a factor of 8, and that
limits the possibilities and they can all be checked by hand.

That leaves the case where y + sqrt(-3) = (a + b*sqrt(-3))^3 / 8) * z
or z^2, which will work out quite identically. This is a different
kettle of fish entirely, and I actually doubt that I can push this
through, but I don't have time to try right now.

So until I have another chance ...

Regards,
Achava
From: The Pumpster on 13 Jan 2010 19:59
On Jan 13, 1:51 pm, "Achava Nakhash, the Loving Snake"
<ach...(a)hotmail.com> wrote:
> On Jan 13, 12:43 pm, Leonid Lenov <leonidle...(a)gmail.com> wrote:
>
> > On Jan 13, 8:34 pm, "Achava Nakhash, the Loving Snake"
>
> > > As you state
> > > it, is it actually multiplicative?  I suspect not.  
>
> > I think it is...
>
> ... and your are so right.
>
> Also, your set appears to me to be closed under multiplication, but
> maybe not under addition.  After all (2 + sqrt(-2)) = alpha and (1 +
> 2*sqrt(-3)) = beta are both in your set, but alpha + beta = 3 + 3*sqrt
> (-3) is not in your set since both coefficients are odd.  Unless I
> have your definition wrong?
>
> I will start the proof I was talking about, but my time is limited, so
> I might not get to finish it.
>
> We want to solve y^2=x^3-3 or x^3 = y^2 + 3 in integers x, y.  Notice
> that y has to be even, since all odd squares are 1 mod 8, but then we
> have 2 divides x, so we would have 4 congruent to 0 mod 8 which it
> isn't.
>
> Now move to the ring Q{sqrt(-3)).  As mentioned earlier, the elements
> of this ring are all numbers of the form (a + b*sqrt(-3))/2 where a, b
> in Z and either both even or both odd.  This ring is a unique
> factorization domain (even Euclidean, actually) and the only units are
> +/- 1, +/- z and +/- z^2 where z = (-1 + sqrt(-3))/2 and we see that
> z^3 = 1.  All of this is very standard stuff, easily proved, and
> easily looked up.
>
> Now we factor in this ring.  (y + sqrt(-3))( y - sqrt(-3)) = x^3.
> Any common factor of these two terms has to divide there difference
> which is 2*sqrt(-3).  Now 2 and -3 are both primes in this ring (norm
> considerations) and 2 can't divide both factors since  y cannot be
> odd.  Also sqrt(-3) cannot divide either term, since then 3 divides y
> and the right hand side of the originial equation is divisible by 27
> but the left hand side is not.
>
> Thus y + sqrt(-3) must be a cube in the ring times a unit, but unique
> factorization.  Since 1 and -1 are cubes, we need concern ourselves
> only with the unit being absent or being z or z^2.
>
> To start off, assume no unit.  Then y + sqrt(-3) = (a + b*sqrt(-3))
> ^3 / 8, and so
>
> 8 = 3*(a^2)*b - 3*(b^3),
>
> where a, b integers, so in particular 3 must be a factor of 8 which it
> isn't.  That takes care of this case.  Please check my arithmetic /
> algebra.  At worst, we are going to get b is a factor of 8, and that
> limits the possibilities and they can all be checked by hand.
>
> That leaves the case where y + sqrt(-3) = (a + b*sqrt(-3))^3 / 8) *  z
> or z^2, which will work out quite identically.  This is a different
> kettle of fish entirely, and I actually doubt that I can push this
> through, but I don't have time to try right now.
>
> So  until I have another chance ...
>
> Regards,
> Achava

The remaining cases give easy contradictions modulo 9.

I guess nobody wants to invoke Coates-Wiles :)

de P
From: Andrew Usher on 14 Jan 2010 01:22
On Jan 13, 6:59 pm, The Pumpster <pumpledumplek...(a)gmail.com> wrote:

> The remaining cases give easy contradictions modulo 9.
>
> I guess nobody wants to invoke Coates-Wiles :)

I prefer my method, it generalises more. The fact the unique
factorisation can fail in Z[sqrt(-3)] only for 'even' numbers can be
used in other proofs; it also applies to some others such as (I
believe) Z[sqrt(-5)].

Andrew Usher
From: Larry Hammick on 14 Jan 2010 05:45

"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.

Have faith. :)

Let u be a zero of the polynomial 1 + u + uu.
( y - u + uu)(y + u - uu) = yy +3
and use the fact that Z[u] is a UFD.

btw Fermat was familiar with certain cases of these now fashionable elliptic
curve thingees.

LH


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