Quadratic forms vs congruences

[T. D. Noe]

similar to the  "15 theorem", "33 theorem", or "290 theorem" (see A030050-1
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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.