The adjoined element is not a pure imaginary number, hence the vector that is based at the origin is not perpendicular to the x axis. In other words, the cells of the lattice are no longer rectangles. They are pushed over rectangles, or parallelograms. It is tempting to cut each parallelogram in half, making an infinite lattice of perfect equilateral triangles, but that disguises some of the algebra. There are really two vectors, 1 and (1+q)/2, and they span a lattice of parallelograms, just like any other pair of independent vectors in the plane.
The norm, a2+3b2, is still a valid multiplicative map into the positive integers, but this time a and b can be half integers. Watch the lattice though; they are either both half integers or they are both whole integers. Verify that the norm is still an integer, even if a and b are half integers. Then show there are six different units, namely the six roots of 1 in the complex plane.
Next consider the gcd algorithm. Let the shorter vector establish a lattice of parallelograms, (in this case rhombuses), find the cell that contains the longer vector, and measure its distance to the four corners. The distance is always less than the side of the rhombus. The gcd algorithm runs to completion, and we have a unique factorization domain (ufd).
If q is the square root of -7 or -11, and we adjoin (1+q)/2, the parallelograms grow taller, but the distance between the center and the nearest corner is still less than the short side of the parallelogram. The gcd algorithm works, and these are all ufds.
Conversely, any prime other than 2 has associates with half integer coefficients. The norm is odd, so a and b have opposite parity. Multiply by (1±q)/2 and you get half integers. The exception is 2, with norm 4, and associates 1±q.
To keep things simple, let the order of a and b be significant, and let's allow a and b to be negative. Any solution can be multiplied by any power of q to get a new solution. Only 0 stays put when multiplied by q, so each solution expands into 6. And how many solutions are there? Review the primes over primes theorem, and apply this to primes over 3. If n is divisible by a prime p that is 2 mod 3, p does not split. p has to divide a and b, and that means p2 divides n. Pull all these primes out; they don't contribute to the combinatorics.
If 3 divides n, then 1+q or its conjugate is part of a+qb. It doesn't matter which one, as they are associates. So these can come out.
That leaves the primes that are 1 mod 3. If pi is raised to the ki, there are ki+1 possibilities. None, some, or all of the upper primes can be folded into a+qb. Take the product over ki+1, then multiply by 6 as described above. This is the number of a,b pairs that yield n.
If signs don't matter, then multiplying by q3 = -1 gives the same solution. Thus the product is multiplied by 3, not 6.
Draw a+qb in the complex plane as a vector a along the x axis, then a vector of length b pointing up and to the right. Conjugate, and b points down and to the right. Rotate this 60 degrees counterclockwise and find b+qa. The variables have been swapped. This rotation is the same as multiplying by q, which is a unit in our ufd. The conjugate solution merely swaps variables, up to associates. Something similar can be done when a or be is negative. And this process can be reversed, so that swapping variables equalss conjugation. Thus, like the sum of squares problem, we should clump conjugate solutions together. Divide the product over ki+1 by 2, then multiply by 3.
This works when the product is even, i.e. when any prime is raised to an odd exponent. But there is one case, when every exponent is even, where we can get into trouble. Step back and ask when conjugation produces the same solution, up to associates. Certainly this is the case when a+qb is real, i.e. when b = 0. This is the half and half solution, the odd member of the product. What about other associates? Reflect through the x axis, then rotate by q. If this gives the same solution then the angle was 30 degrees. This is a+qb, where a is an irrational multiple (2/sqrt(3)) times b. That is impossible, so set this aside. If conjugation, followed by q2, yields the same point, then the angle is 60 degrees, and it is an associate of a real number. Other cases are handled similarly. So the only solution overcounted is a = sqrt(n). This one is lost if we simply divide an odd number by 2, so bring it back in (as one solution) if the product is odd. This is illustrated in the example below.
Here is an example that does not waste time with factors of 3, or factors that are 2 mod 3. Let n = 72×132. Increment the exponents and multiply them together, giving 9. Divide by 2, giving 4. There are 4 base solutions, and each one expands into 3. Since 9 is odd, the last solution stands alone. Use 2+q and 3+q for the upper primes over 7 and 13 respectively. Start with 2 copies of the first and 0 1 or 2 of the second, then 1 of the first and 0 or 1 of the second. Expand each into three solutions as you go along, except for the last. Verify that a2 + ab + b2 = 8281.
80, 19 -19, 99 -99, 80 39, 65 -65, 104 -104, 39 -11, 96 -96, 85 -58, -11 105, -49 49, 56 -56, 105 91, 0
When n is a square with no primes equal to 2 mod 3, we have a direct connection to a2 + ab + b2 = c2, (similar to coprime pythagorean triples), and these in turn become the rational points on the ellips x2 + xy + y2 = 1.