An imaginary quadratic field Q(sqrt(-d)) has class number 1 when its integers factor uniquely -- just like ordinary integers. There are exactly nine such values of d. Conjectured by Gauss, proved by Stark and Heegner in 1967. The first five are {1, 2, 3, 7, 11} -- the primes of Z/2,310 except 5. The remaining four are generated by 3 through the Cunningham map c(n) = 2n+1.
Every prime of Z/2,310 except 5 is a Heegner number. Why? Because h(-5) = 2. The field Q(sqrt(-5)) has two ideal classes -- factorization is ambiguous. 5 is also the only prime of the ring that splits in Z[i]: 5 = (2+i)(2-i). Two ideal classes, two Gaussian factors -- the pattern is consistent.
| p | 3*p | c(3*p) | Heegner? |
|---|---|---|---|
| 1 | 3 | 7 | YES |
| 2 | 6 | 13 | NO (h(-52)=2) |
| 3 | 9 | 19 | YES |
| 5 | 15 | 31 | NO (h(-31)=3) |
| 7 | 21 | 43 | YES |
| 11 | 33 | 67 | YES |
The failures have small prime class numbers: h(-52)=2, h(-31)=3. Exactly 2 and 5 are the elements whose Cunningham lift fails -- the same two primes excluded by different Heegner mechanisms.
Powers of 3 through the Cunningham map: c(3^n) = 2*3^n + 1.
| n | c(3^n) | Heegner? |
|---|---|---|
| 1 | c(3) = 7 | YES |
| 2 | c(9) = 19 | YES |
| 3 | c(27) = 55 | NO (5*11) |
| 4 | c(81) = 163 | YES (!) |
| 5 | c(243) = 487 | NO (prime, not Heegner) |
Heegner at n in {1, 2, 4}. The class numbers stay 11-smooth (factors only from {2,3,5,7,11}) for n=0..8 (9 consecutive). First non-smooth failure at n=9. At n=7: c(3^7) = 4375 = 5^4 * 7.
Subtract 6 (= 2*3) from each odd prime squared:
Three consecutive odd primes squared minus 6 yield three consecutive Heegner primes: 19, 43, 67.
All 5 splitting Heegner d's are Eisenstein norms N(a,b) = a^2 + ab + b^2 with small-prime coordinates:
| Heegner d | Eisenstein norm | Pair |
|---|---|---|
| 7 | 1 + (-2) + 4 | (1, 2) |
| 19 | 4 + (-6) + 9 | (2, 3) |
| 43 | 1 + (-6) + 36 | (1, 6) |
| 67 | 4 + (-14) + 49 | (2, 7) |
| 163 | 9 + (-33) + 121 | (3, 11) |
The Eisenstein lattice Z[omega] (where omega = e^{2*pi*i/3}) reflects the prime chain. Inert: {2, 11} (both 2 mod 3). Ramified: 3. Unit: 1.
| h | (a, b) | sqrt(disc) |
|---|---|---|
| 1 | (0, 1) | 2 |
| 3 | (1, 2) | 3 |
| 7 | (1, 3) | 5 |
| 19 | (2, 5) | 8 (= 2^3) |
| 43 | (1, 7) | 13 |
| 67 | (2, 9) | 16 (= 2^4) |
| 163 | (3, 14) | 25 (= 5^2) |
163 Triangle: E3(3, 14) = E3(11, 14) = 163. Both share b-value 14 (= 2*7). 14 - 3 = 11. 3 + 11 = 14. Also: E3(3, 11) = 97. 163 - 97 = 66 = 2*3*11.
Enter d. Is Q(sqrt(-d)) a unique factorization domain? The nine Heegner numbers are {1,2,3,7,11,19,43,67,163} -- the primes of Z/2,310 (except 5) plus four extensions generated by 3 via c(n) = 2n+1.
Check discriminant d:
Try: 1, 2, 3, 7, 11 (all Heegner), 5 (excluded! h(-5)=2), 19 (c(9)=19), 163 (c(81)=163), 23 (h(-23)=3).
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