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• ## Re: [PrimeNumbers] Re: Is there a simple form for prmes of X^2+11Y^2?

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• On Sun, 6 Jul 2003 08:19:29 EDT ... Hilbert polynomial of the quadratic field with discriminant -44 is x^3-1122662608*x^2+270413882112*x-653249011576832
Message 1 of 12 , Jul 6, 2003
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On Sun, 6 Jul 2003 08:19:29 EDT
mikeoakes2@... wrote:
> And h(-44) = 3 - see e.g. H.Cohen "A Course in Computational Algebraic Number
> Theory" (Springer, 1996), Appendix B.
> So the extra condition is that a cubic polynomial has a root mod p; I haven't
> been able to determine the polynomial explicitly (this stuff is /hard/), but

Hilbert polynomial of the quadratic field with discriminant -44 is
x^3-1122662608*x^2+270413882112*x-653249011576832

Satoshi Tomabechi
• ... I ve found two places on the web where I think h(-44)=3 ought to show up, and it isn t in either place: http://mathworld.wolfram.com/ClassNumber.html
Message 1 of 12 , Jul 6, 2003
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> And h(-44) = 3 - see e.g. H.Cohen "A Course in Computational
>Algebraic Number Theory" (Springer, 1996), Appendix B.

I've found two places on the web where I think h(-44)=3 ought to show
up, and it isn't in either place:

http://mathworld.wolfram.com/ClassNumber.html

http://www.research.att.com/cgi-bin/access.cgi/as/
njas/sequences/eisA.cgi?Anum=A006203

I also see some differences from your list here:

> h(-4*n) = 1 for n=1,2,3,7.
> h(-4*n) = 2 for n=5,6,10,13,15,22,37,58.

12 and 28 don't show up for h(-d)=1.
60 doesn't show up for h(-d)=2.

Am I doing something wrong, or so these sources disagree with your
source?
• In a message dated 07/07/03 04:23:08 GMT Daylight Time, sleephound@yahoo.com ... h(-12) = h(-28) = 1 give n = 3 resp. 7. h(-60) = 2 gives n = 15. Ok? Notice
Message 1 of 12 , Jul 7, 2003
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In a message dated 07/07/03 04:23:08 GMT Daylight Time, sleephound@...
writes:

> I also see some differences from your list here:
>
> > h(-4*n) = 1 for n=1,2,3,7.
> > h(-4*n) = 2 for n=5,6,10,13,15,22,37,58.
>
> 12 and 28 don't show up for h(-d)=1.
> 60 doesn't show up for h(-d)=2.
>
> Am I doing something wrong, or so these sources disagree with your
> source?
>

h(-12) = h(-28) = 1 give n = 3 resp. 7.
h(-60) = 2 gives n = 15.
Ok?

Notice that I said "..(squarefree) n...". The list for /all/ n is a bit
longer:-
h(-4*n) = 1 for n=1,2,3,4,7.
h(-4*n) = 2 for n=5,6,8,9,10,12,13,15,16,18,22,25,28,37,58.

In fact, there's quite a serious error in my earlier claim that
> it is necessary that the class number h(-4*n) = 1 or 2, since that is the
> degree of the polynomial in the subsidiary condition

Further reading has made it clear that not only the great Gauss (1801) but
even before him the mighty Euler (c. 1750) were in posession of /65/ values of n
which have a "linear" characterisation of the type we are discussing.
To the above must be added (using our modern notation of class numbers of
fields):-
h(-4*n) = 4 for n=21,24,30,33,40,42,45,48,57,60,70,72,
78,85,88,93,102,112,130,133,177,190,232,253
h(-4*n) = 8 for n=105,120,165,168,210,240,273,280,312,330,
345,357,385,408,462,520,760.
h(-4*n) = 16 for n=840,1320,1365,1848.
[These values come from David Cox's superb book.]

Apparently it has since been proved that there is at most one other value of
n, but no-one has been able to show whether or not it exists!
This is one of the deepest and richest areas of number theory, still not
exhausted by centuries of mining.

Mike Oakes

[Non-text portions of this message have been removed]
• In a message dated 07/07/03 02:08:44 GMT Daylight Time, ... Thanks! Can you reduce it? (I m no pari expert) - it ought to be equivalent to my cubic. Mike Oakes
Message 1 of 12 , Jul 7, 2003
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In a message dated 07/07/03 02:08:44 GMT Daylight Time,
tomabeti@... writes:

> Hilbert polynomial of the quadratic field with discriminant -44 is
> x^3-1122662608*x^2+270413882112*x-653249011576832
>

Thanks!
Can you reduce it? (I'm no pari expert) - it ought to be equivalent to my
cubic.

Mike Oakes

[Non-text portions of this message have been removed]
• On Mon, 7 Jul 2003 04:31:45 EDT ... x^3 - x^2 + x + 1 ... It says that two polynomials define the same (class) filed. You can use another Weber polynomial x^3
Message 1 of 12 , Jul 7, 2003
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On Mon, 7 Jul 2003 04:31:45 EDT
mikeoakes2@... wrote:

> > Hilbert polynomial of the quadratic field with discriminant -44 is
> > x^3-1122662608*x^2+270413882112*x-653249011576832
> >
>
> Thanks!
> Can you reduce it? (I'm no pari expert) - it ought to be equivalent to my
> cubic.

x^3 - x^2 + x + 1

>? polred(x^3-1122662608*x^2+270413882112*x-653249011576832)
>%1 = [x - 1, x^3 - x^2 - x - 1, x^3 - x^2 + x + 1]
>so x^3-x^2-x+1 seems to do the same job?
It says that two polynomials define the same (class) filed.

You can use another Weber polynomial x^3 - 2*x^2 + 2*x - 2,
which also defines the same field.

Satoshi Tomabechi
• In a message dated 08/07/03 02:14:47 GMT Daylight Time, ... Thank you. Meanwhile, (pari Grand Master) David Broadhursthas emailed me with ... So that s good !
Message 1 of 12 , Jul 8, 2003
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In a message dated 08/07/03 02:14:47 GMT Daylight Time,
tomabeti@... writes:

>> Can you reduce it? (I'm no pari expert) - it ought to be equivalent to my
>> cubic.
>
> x^3 - x^2 + x + 1

Thank you.
Meanwhile, (pari Grand Master) David Broadhursthas emailed me with
> henri cohen's "minimal height" form for your d=-44 cubic is
> x^3-x^2+x+1

So that's good ! - there's complete agreement between your theoretical
Hilbert polynomial and my experimentally discovered cubic.

Mike Oakes

[Non-text portions of this message have been removed]
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