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

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• In a message dated 06/07/03 06:57:13 GMT Daylight Time, ... In fact, the fraction of primes p congruent to 1 or 3 or 5 or 9 or 15 mod 22 which can be so
Message 1 of 12 , Jul 6, 2003
In a message dated 06/07/03 06:57:13 GMT Daylight Time,
mark.underwood@... writes:

> For p=11:
>
> All primes that can be expressed as x^2 + 11y^2 can also be
> represented as 22n+1, 22n+3, 22n+5, 22n+9 or 22n + 15. But the
> converse is not true; not all primes of the form 22n+1, 22n+3, 22n+9
> or 22n+15 can be written as x^2 + 11y^2.
>

In fact, the fraction of primes p congruent to 1 or 3 or 5 or 9 or 15 mod 22
which can be so expressed is certainly 1/3.

Experimental evidence:-
p_max fraction
1000 0.29762
10000 0.3271
100000 0.32840
1000000 0.33270
10000000 0.33283
100000000 0.33322
(execution time 35 GHz-mins.)

Theoretical evidence:-
David Cox's "Primes of the Form x^2 + ny^2" (Wiley, 1989) proves the following
Theorem 9.2: Let n > 0 be an integer. Then there is a monic irreducible
polynomial f(x) of degree h(-4n) such that if an odd prime p divides neither n nor
the discriminant of f(x) then
p = x^2 + ny^2 iff
(-n|p) = 1 and f(x) = 0 mod p has an integer solution.

[In the above, (-n|p) = is the Legendre symbol, and h(-4n) is the class
number of the quadratic field Q(sqrt(-4n)).]

Now (-11|p) = 1 precisely for the above set of 5 residues.

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
it's reasonable to assume that it has a root mod p for 1 in 3 values of p, by
analogy with cubic residuocity.

> All primes that can be expressed as x^2 - 11y^2 can be also
> represented as 22n+1 or 22n+3 or 22n+9 or 22n+15. But again, I don't
> think the converse is true.
>

This case is much easier:-
p = x^2 - 11y^2 iff p = 1 or 5 or 9 or 25 or 37 mod 44.

Mike Oakes

[Non-text portions of this message have been removed]
• ... haven t been able to determine the polynomial explicitly By searching a subspace of the space of all cubics, I have found a suitable one (it is not
Message 2 of 12 , Jul 6, 2003
> So the extra condition is that a cubic polynomial has a root mod p; I
haven't been > able to determine the polynomial explicitly

By searching a subspace of the space of all cubics, I have found a suitable
one (it is not unique), so can state, finally, the following Theorem:-

If p is not 2 or 11,
then p = x^2 + 11*y^2 iff
(i) p = 1 or 3 or 5 or 9 or 15 mod 22, and
(ii) there is an integer solution x to the equation x^3 - 4*x + 4 = 0 mod p.

The exclusion of 2 and 11 is because the cubic has discriminant -176 =
-11*2^4.

It's interesting that all the candidate cubics had discriminants = 0 mod 44.

Mike Oakes

[Non-text portions of this message have been removed]
• Thanks Mike. Your findings are starting to go over my head but are very appreciated. I shall have to look into this further. Based on your conclusion that p
Message 3 of 12 , Jul 6, 2003
Thanks Mike. Your findings are starting to go over my head but are
very appreciated. I shall have to look into this further.

Based on your conclusion that "p = x^2 - 11y^2 iff p = 1 or 5 or 9 or
25 or 37 mod 44", that would mean it covers 1/4 of the primes, since
these are 5 of the possible twenty residues mod 44. It turns out then
that the table at Wolfram's site was never meant to be complete but
just a small sample.

I found your data that x^2 + 11y^2 covers 1/3 of the primes very

If p is not 2 or 11, then p = x^2 + 11*y^2 iff
(i) p = 1 or 3 or 5 or 9 or 15 mod 22, and
(ii) there is an integer solution x to the equation x^3 - 4*x + 4 = 0
mod p.

is simply amazing (and beyond my reach right now!).

Incidentally, a clever and seasoned veteran in these matters informed
me via email that the three quadratic forms x^2+y^2, x^2+2y^2 and x^2-
2y^2 are not unique in their ability to cover all the primes as I had
supposed. As a hint, he showed that x^2+y^2 and x^2+3y^2 cover the
12n+1, 12n+5, and 12n+7 residues, leaving only the 12n+11 residue
uncovered. Then he said there is indeed a quadratic form with a small
b that does cover all the primes of 12n+11 but left it to me to find
it (!) I wonder if the b is positive. Another hint was to do a seach
on Bernstein and Atkin's sieve, which I shall do as time allows.

Mark

> > So the extra condition is that a cubic polynomial has a root mod
p; I
> haven't been > able to determine the polynomial explicitly
>
> By searching a subspace of the space of all cubics, I have found a
suitable
> one (it is not unique), so can state, finally, the following
Theorem:-
>
> If p is not 2 or 11,
> then p = x^2 + 11*y^2 iff
> (i) p = 1 or 3 or 5 or 9 or 15 mod 22, and
> (ii) there is an integer solution x to the equation x^3 - 4*x + 4 =
0 mod p.
>
> The exclusion of 2 and 11 is because the cubic has discriminant -
176 =
> -11*2^4.
>
> It's interesting that all the candidate cubics had discriminants =
0 mod 44.
>
> Mike Oakes
>
>
> [Non-text portions of this message have been removed]
• In a message dated 05/07/03 15:52:25 GMT Daylight Time, sleephound@yahoo.com ... For there to be a characterisation of the form p = x^2 + n*y2 iff p = r_1 or
Message 4 of 12 , Jul 6, 2003
In a message dated 05/07/03 15:52:25 GMT Daylight Time, sleephound@...
writes:

> ? Is there a simple rule for when a
> linear characterization exists?
>

For there to be a characterisation of the form
"p = x^2 + n*y2 iff p = r_1 or r_2 or .. or r_k mod q"
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 (see my earlier post).

[Note: h(-4*n) = 2 is ok because it will give a subsidiary equation of the
form x^2 - b*x + c = 0 mod p, i.e. (b^2 - 4*c)/4 is to be a quadratic residue
mod p, which, by Gauss's law of quadratic reciprocity, will be true iff p is a
quadratic residue or nonresidue mod some integer s, which in turn will give a
condition of the form p = r_1 or r_2 or .. or r_k mod s.]

The /only/ values of (square-free) n for which this class number condition
holds are (again from Cohen, Appendix B):-
h(-4*n) = 1 for n=1,2,3,7.
h(-4*n) = 2 for n=5,6,10,13,15,22,37,58.

This explains why tables like that at
http://mathworld.wolfram.com/PrimeRepresentation.html
only have entries which are a subset of these particular values of n.

Mike Oakes

[Non-text portions of this message have been removed]
• 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 5 of 12 , Jul 6, 2003
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 6 of 12 , Jul 6, 2003

> 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 7 of 12 , Jul 7, 2003
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 8 of 12 , Jul 7, 2003
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 9 of 12 , Jul 7, 2003
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 10 of 12 , Jul 8, 2003
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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