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[Richard Guy]

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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