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[David Wilson]

Yeah, Alzheimer's strikes again. I don't even know why I bother.

----- Original Message ----- 
From: "Richard Guy" <rkg at cpsc.ucalgary.ca>
To: "David W. Wilson" <wilson.d at anseri.com>
Cc: <seqfan at ext.jussieu.fr>
Sent: Thursday, April 24, 2008 6:03 PM
Subject: RE:


> 4x^2-4xy+6y^2  looks very even to me.   R.
>
> On Thu, 24 Apr 2008, David W. Wilson wrote:
>
>>
>> Theorem: Let p be prime, x >= 0, y >= 0.  p = 4x^2-4xy+6y^2 <=> p == 7 
>> (mod
>> 24).
>>
>> Part I:  p = 4x^2-4xy+6y^2 => p == 7 (mod 24).
>>
>> Let p = 4x^2-4xy+6y^2.
>>
>> Then p = a^2+6b^2 with a = 2x-y, b = y.
>>
>> Thus p = (a+sqrt(-6)b)(a-sqrt(-6)b) is reducible in Q(sqrt(-6)).
>>
>> Number theory tells us that rational (integer) prime p is reducible in a
>> quadratic field exactly when p has one of a fixed set of residues modulo 
>> the
>> determinant of the field (crossed fingers here). The determinant of
>> Q(sqrt(-6)) is 24, and we find empirically that p is reducible when p == 
>> 1
>> or 7 (mod 24). In summary,
>>
>>    [1] p = a^2+6b^2  <=>  p == 1 or 7 (mod 24).
>>
>> We can show that 2 is not of form a^2+6b^2, so p is odd => a is odd => y 
>> is
>> odd => b is odd. a and b odd rules out a^2+6b^2 == 1 (mod 24), so p == 7
>> (mod 24). []
>>
>> Part II: p == 7 (mod 24) => p = 4x^2-4xy+6y^2
>>
>> Let p == 7 (mod 24).
>>
>> [1] gives p = a^2+6b^2.
>>
>> p != 2, so p is odd => a is odd.
>>
>> If b is even, then p = a^2+6b^2 != 7 (mod 24) so b is odd.
>>
>> Since a^2+6b^2 = |a|^2+6|b|^2, so we can assume a >= 0, b >= 0.
>>
>> a and b odd >= 0 gives integer (a+b)/2 >= 0.
>>
>> Hence x = (a+b)/2, y = b are positive integers with p = 4x^2-4xy+6y^2. []
>>
>> QED.
>>
>>
>>
>>
>>
>>
>
>
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