- 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

>

>

> [Non-text portions of this message have been removed] - 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^3-x^2+x+1

Hilbert polynomial and my experimentally discovered cubic.

Mike Oakes

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