A137389

[N. J. A. Sloane]

------=_Part_587_20406391.1209400988721
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 7bit

The determination of primes produced by the quadratic form x^2 + 6y^2 can be done using an analysis analogous to the one given for Q(sqrt(-30)).
Again, by table lookup, Q(sqrt(-6)) has class number 2.

Two dissimilar ideals in Z[sqrt(-6)] are (1) and (2,sqrt(-6)). So any ideal in Z[sqrt(-6)] similar to (1) will be (x + y*sqrt(-6)) and have norm equal to the absolute value of
| x    y |
| -6y x | = x^2 + 6y^2, which is clearly nonnegative.

Any ideal similar to (2,sqrt(-6)) will be (x + y*sqrt(-6), -3y + x*sqrt(-6)/2 ). Integrality of the generators forces x to be even (write x = 2z) and it forces y to be an integer.
Then the norm of (2z + y*sqrt(-6), -3y + z*sqrt(-6)) is the absolute value of
| 2z   y |
| -3y  z | = 2z^2 + 3y^2, which is clearly nonnegative.

Thus, by using some of the other reasoning in the Q(sqrt(-30)) email, all primes which split in Q(sqrt(-6)) are produced by one of the quadratic forms x^2 + 6y^2 or 2x^2 + 3y^2.
Again, by quadratic reciprocity and the (-1/p) and (2/p) supplements to it, the primes p such that (-6/p) = 1 are those such that p == 1, 5, 7, 11 (mod 24).

If p = 2x^2 + 3y^2 and p is not 2 or 3, then looking mod 24 shows p == 5 or 11 mod 24.

If p = x^2 + 6y^2, then looking mod 24 shows p == 1 or 7 mod 24.

Since there are no residue classes mod 24 produced by more than one of those quadratic forms, the necessary congruential conditions for p to be produced by one of those forms are also sufficient. Therefore any p == 1 or 7 mod 24 has p = x^2 + 6y^2 for some integers x and y. I think this is what David Wilson wanted. In any case, the rest of the argument can proceed using elementary modular arithmetic.

---- David
------=_Part_587_20406391.1209400988721--