Proof

[David W. Wilson]

This appears to be a proof that

    p = 4x^2-4xy+7y^2 => p == 7 (mod 24). (x,y >= 0).

This is, however, only half of the proof (the easy half).  Everyone has
proved this half.

The other, more difficult half, is:

    p == 7 (mod 24) => p = 4x^2-4xy+7y^2 (x,y >= 0).

In an earlier note, I showed that this follows if the rational primes in
Q(sqrt(-6)) are the primes == 1 or 7 (mod 24), which empirically seems to be
the case. I suspect this is a result in quadratic field theory, but I can't
prove it myself, I sold my Harvey Cohn.

> -----Original Message-----
> From: Graeme McRae [mailto:g_m at mcraefamily.com]
> Sent: Sunday, April 27, 2008 4:39 PM
> To: grafix at csl.pl; N. J. A. Sloane
> Cc: seqfan
> Subject: Re: Proof
> 
> A simpler proof...  I considered the 576 possible combinations of x,y
> (mod
> 24), and found that possible values of 4x^2-4xy+7y^2 are 0, 4, 7, 12,
> 15,
> and 16.  Of these, only 7 (mod 24) can be prime.