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Is there a simple form for prmes of X^2+11Y^2?
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Primes of the form X^2+Y^2 are of the form 4n+1, Primes of the form
X^2+2Y^2 are of the form 8n+1 or 8n+1. Some other examples of
quadratic forms yielding primes of linear form are listed here:
http://mathworld.wolfram.com/PrimeRepresentation.html
Primes of the form X^2+11Y^2 begin 47, 53, 103, 163, 199. There
doesn't appear to be a linear form that separates these primes into
two groups  I've checked through 4000. Is there some simple rule
that characterizes these primes? Is there a simple rule for when a
linear characterization exists? 0 Attachment
I find this all very interesting, it is new to me. What specifically
caught my attention was the linear representation of primes generated
by the equation x^2 + p*y^2, where p is prime. The nature of the
linear representation is not always the same from prime to prime. Is
there are rule as Sleephound queried, I'm not sure.
Below are the representations for p=2, p=3, p=5, p=7 and p=11. The
p=11 result is courtesy of Sleephound's data, not Wolfram's site.
From the Wolfram site,
For p=2 :
All primes that can be expressed as x^2 + 2y^2 can also be
represented as 8n+1 or 8n+3. Conversely, if I understand it
correctly, all primes of the form 8n+1 or 8n+3 can be expressed as
x^2 + 2y^2.
All primes that can be expressed as x^2  2y^2 can also be
represented as 8n+1 or 8n+7.
Observation: There is partial intersection between the two groups.
One quarter of the primes are not represented at all, that is primes
of the form 8n + 5.
For p=3:
All primes that can be expressed as x^2 + 3y^2 can also be
represented as 6n+1.
All primes that can be expressed as x^2  3y^2 can also be
represented as 12n+1.
Observation: One group is entirely a subset of the other. One half of
the primes are not represented, that is primes of the form 6n1.
For p=5:
All primes that can be expressed as x^2 + 5y^2 can be also be
represented as 20n+1 or 20n+9.
All primes that can be expressed as x^2  5y^2 can also be
represented as 10n+1 or 10n+9.
Observation: One group is entirely a subset of the other. One half of
the primes are not represented, that is primes of the form 10n+3 and
10n+7.
For p=7:
All primes that can be expressed as x^2 + 7y^2 can also be
represented as 14n+1 or 14n+9 or 14n+25.
All primes that can be expressed as x^2  7y^2 can also be
represented as 28n+1 or 28n+9 or 28n+25.
Observation: One group is entirely a subset of the other. One half of
the primes are not represented, that is primes of the form 14n+3 and
14n+5 and 14n+9.
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.
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.
Observation: I'm not sure if the two groups overlap even though they
have the same linear representation. And the fact that the converse
is not true in this cases sets it apart from the earlier primes. I
wonder if all the rest of the primes after 11 (13,17,19 ...) are like
11 in this regard?
Final Note:
Also, I noticed from Wolfram's table that all primes of the form x^2
+ y^2 can be written as 4n+1, and (I am supposing) the converse is
true, all primes of the form 4n+1 can be written as x^2 + y^2. That
is half the primes. And as we know all of the primes can be written
as x^2  y^2.
It seems that no expression x^2  by^2 combined with x^2 + by^2 will
yield all the primes, except of course when b = 1. But it appears
that the combined *three* expressions, x^2 + y^2, x^2 + 2y^2 and x^2 
2y^2 will yield all the primes since they cover all the
possibilities: 8n+1, 8n+3, 8n+5 and 8n+7. I doubt that any other
combination of three will yield all the primes. Wait, could it be
that no combination of four will yield all the primes, even if up to
two of the above three expressions are used? Further, no combination
of 5, or 6, or 7 or ....?
Mark
 In primenumbers@yahoogroups.com, "sleephound" <sleephound@y...>
wrote:> Primes of the form X^2+Y^2 are of the form 4n+1, Primes of the form
> X^2+2Y^2 are of the form 8n+1 or 8n+1. Some other examples of
> quadratic forms yielding primes of linear form are listed here:
>
> http://mathworld.wolfram.com/PrimeRepresentation.html
>
> Primes of the form X^2+11Y^2 begin 47, 53, 103, 163, 199. There
> doesn't appear to be a linear form that separates these primes into
> two groups  I've checked through 4000. Is there some simple rule
> that characterizes these primes? Is there a simple rule for when a
> linear characterization exists? 0 Attachment
In a message dated 06/07/03 06:57:13 GMT Daylight Time,
mark.underwood@... writes:
> For p=11:
In fact, the fraction of primes p congruent to 1 or 3 or 5 or 9 or 15 mod 22
>
> 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.
>
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 GHzmins.)
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
(np) = 1 and f(x) = 0 mod p has an integer solution.
[In the above, (np) = is the Legendre symbol, and h(4n) is the class
number of the quadratic field Q(sqrt(4n)).]
Now (11p) = 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
This case is much easier:
> represented as 22n+1 or 22n+3 or 22n+9 or 22n+15. But again, I don't
> think the converse is true.
>
p = x^2  11y^2 iff p = 1 or 5 or 9 or 25 or 37 mod 44.
Mike Oakes
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> 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
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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
interesting. And your finding that
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
 In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
> > 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
>
>
> [Nontext portions of this message have been removed] 0 Attachment
In a message dated 05/07/03 15:52:25 GMT Daylight Time, sleephound@...
writes:
> ? Is there a simple rule for when a
For there to be a characterisation of the form
> linear characterization exists?
>
"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 (squarefree) 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
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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
Hilbert polynomial of the quadratic field with discriminant 44 is
> 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
x^31122662608*x^2+270413882112*x653249011576832
Satoshi Tomabechi 0 Attachment
 In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
> And h(44) = 3  see e.g. H.Cohen "A Course in Computational
I've found two places on the web where I think h(44)=3 ought to show
>Algebraic Number Theory" (Springer, 1996), Appendix B.
up, and it isn't in either place:
http://mathworld.wolfram.com/ClassNumber.html
http://www.research.att.com/cgibin/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.
12 and 28 don't show up for h(d)=1.
> h(4*n) = 2 for n=5,6,10,13,15,22,37,58.
60 doesn't show up for h(d)=2.
Am I doing something wrong, or so these sources disagree with your
source? 0 Attachment
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(12) = h(28) = 1 give n = 3 resp. 7.
>
> > 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(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
Further reading has made it clear that not only the great Gauss (1801) but
> degree of the polynomial in the subsidiary condition
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 noone 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
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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
Thanks!
> x^31122662608*x^2+270413882112*x653249011576832
>
Can you reduce it? (I'm no pari expert)  it ought to be equivalent to my
cubic.
Mike Oakes
[Nontext portions of this message have been removed] 0 Attachment
On Mon, 7 Jul 2003 04:31:45 EDT
mikeoakes2@... wrote:
> > Hilbert polynomial of the quadratic field with discriminant 44 is
x^3  x^2 + x + 1
> > x^31122662608*x^2+270413882112*x653249011576832
> >
>
> Thanks!
> Can you reduce it? (I'm no pari expert)  it ought to be equivalent to my
> cubic.
David Broadhurst told me:>? polred(x^31122662608*x^2+270413882112*x653249011576832)
It says that two polynomials define the same (class) filed.
>%1 = [x  1, x^3  x^2  x  1, x^3  x^2 + x + 1]
>so x^3x^2x+1 seems to do the same job?
You can use another Weber polynomial x^3  2*x^2 + 2*x  2,
which also defines the same field.
Satoshi Tomabechi 0 Attachment
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
Thank you.
>> cubic.
>
> x^3  x^2 + x + 1
Meanwhile, (pari Grand Master) David Broadhursthas emailed me with> henri cohen's "minimal height" form for your d=44 cubic is
So that's good !  there's complete agreement between your theoretical
> x^3x^2+x+1
Hilbert polynomial and my experimentally discovered cubic.
Mike Oakes
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