Notes on A107006 ---------------- Date: Wed, 23 Apr 2008 21:20:23 -0600 From: Don Reble <djr(AT)nk.ca> A107006 is primes of the form 4x^2-4xy+7y^2, with x and y nonnegative. We ponder whether this could be renamed, primes of the form 24k+7. Call this second set A2. It's easy to show that A107006 is a subset of A2. (This is essentially what Artur did.) Simply observe that, modulo 24, 4x^2-4xy+7y^2 has the range {0, 4, 7, 12, 15, 16}. Of these, only 24k+7 numbers can be prime. The hard part is, to show that A2 is a subset of A107006. First, observe that 4x^2-4xy+7y^2 = (2x-y)^2 + 6y^2. Let w = 2x-y. Let p be any 24k+7 prime. Then (by quadratic reciprocity), -6 is a quadratic residue, modulo p. Let r^2 == -6 mod p. Using Farey sequences (see Hardy&Wright, theorem 36), one can show that (for any real s, integer n>1) there is an integer z, and an integer y, 0<y<n, such that such that abs(s - z/y) <= 1/(yn). Letting n = ceil(sqrt(p)), s=r/p, we have 0 < y < sqrt(p), abs(r/p - z/y) <= 1/(yn) < 1/(y sqrt(p)). Multiply by py: abs(ry - zp) < p/sqrt(p) = sqrt(p). Let w = ry - zp. w^2 = (ry - zp)^2 == (r^2 y^2) mod p == -6 y^2 mod p. w^2 + 6y^2 = 0 mod p. So some multiple of p is representable as w^2 + 6y^2. Also, abs(ry - zp) < sqrt(p), so 0 <= w^2 < p; 0 < y < sqrt(p), so 0 < y^2 < p, 0 < w^2 + 6y^2 < 7p. --- One of {p, 2p, 3p, 4p, 5p, 6p} is representable as w^2 + 6y^2. --- Modulo 24, the range of w^2 + 6y^2 is {0, 1, 4, 6, 7, 9, 10, 12, 15, 16, 18, 22}. But (2p, 3p, 5p) == (14, 21, 11) mod 24; these numbers have no representation. --- One of {p, 4p, 6p} is representable as w^2 + 6y^2. --- If 4p = w^2 + 6y^2, then w is even; w^2 is a multiple of 4; 6y^2 is a multiple of 4; y^2 is even; y is even. Since both w and y are even, p = (w/2)^2 + 6(y/2)^2. If 6p = w^2 + 6y^2, then w^2 is a multiple of 6; and w is a multiple of 6, say 6v. So p = y^2 + 6v^2. In all three cases: --- p is representable as w^2 + 6y^2. --- We can choose w and y to be non-negative. w is not zero, because p is not divisible by 6; y is not zero, because p is not a square. Therefore w and y are positive. Since p = w^2 + 6y^2, w is odd; and w^2 is 1 mod 4. p is 3 mod 4, so 6y^2 = 2 mod 4, not a multiple of 4. Therefore y is odd. Since w and y are odd positives, (w+y)/2 = x is a positive integer; and p = 4x^2-4xy+7y^2. That shows that A2 is a subset of A107006, and the sequences are equal.