OFFSET
1,219
COMMENTS
2-rank of the form class group of discriminant D = A079896(n) quotient by {I,-I}, where I is the principal class. (So I corresponds to the form x^2 - (D/4)*y^2 for 4|D and x^2 - x*y - ((D-1)/4)*y^2 for D == 1 (mod 4), and -I corresponds to the form (D/4)*x^2 - y^2 for 4|D and ((D-1)/4)*x^2 - x*y - y^2 for D == 1 (mod 4)).
Let Cl+(D) and Cl(D) be the narrow class group and the class group of the quadratic order of discriminant D. Then Cl+(D)/Cl(D) = 1 if D < 0 or D > 0 and the fundamental unit has norm -1, C_2 if D > 0 and the fundamental unit has norm 1.
Claim. Suppose that the fundamental unit has norm 1.
(a) If 16|D or D has a prime factor congruent to 3 modulo 4, then Cl+(D) = Cl(D) X C_2;
(b) Otherwise, there exists a subgroup G of Cl+(D) such that Cl+(D) = G X C_{2^(r+1)} and Cl(D) = G X C_{2^r}.
Proof. We prove that -I is not a square in Cl+(D) if 16|D or D has a prime factor congruent to 3 modulo 4, and -I is a square otherwise. ((Thankfully) almost nothing more than factorization properties in Z[i] is used). Let f_0(x,y) = (D/4)*x^2 - y^2 for 4|D and ((D-1)/4)*x^2 - x*y - y^2 for D == 1 (mod 4) be in the class -I.
(a) Assume that 16|D or D has a prime factor congruent to 3 modulo 4. Suppose that f is a form such that f^2 is equivalent to f_0. Take a prime p not dividing D represented by f, which exists by W. E. Briggs link. Then p^2 is represented by f^2 ~ f_0, and thus -p^2 is represented by the principal form, which means that -p^2 is a quadratic residue modulo 4 or a prime congruent to 3 modulo 4, impossible.
(b) Suppose first that 4|D. There exists gcd(r,s) = 1 with odd s such that r^2 + s^2 = D/4. Then f_0 is equivalent to f_0(x+y,r*x+(r+1)*y) = s^2*x^2 + 2*(s^2-2*r)*x*y + (s^2-2*r-1)*y^2. Take f(x,y) = s*x^2 + 2*(s^2-2*r)*x*y + s*(s^2-2*r-1)*y^2, then f is a primitive form since gcd(s,4*r) = 1, and f^2 ~ f_0 by Definition 3.1.4 in Rick L. Shepherd link.
For D == 1 (mod 4), there exists gcd(r,s) = 1 with odd r and even s such that r^2 + s^2 = D. Then f_0 is equivalent to f_0(x+y,((r-1)/2)*x+((r+1)/2)*y) = (s^2/4)*x^2 + (s^2/2-r)*x*y + (s^2/4-r-1)*y^2. Take f(x,y) = (s/2)*x^2 + (s^2/2-r)*x*y + (s/2)*(s^2/4-r-1)*y^2, then f is a primitive form, and f^2 ~ f_0.
This result can be used in the other direction: if 16 does not divide D, D has no prime factor congruent to 3 modulo 4, and Cl+(D) has no cyclic subgroup of order 4, then -I = I, namely the norm of the fundamental unit has norm -1. This proves for example that
- For primes p == 1 (mod 4), x^2 - p*y^2 = -1 has solutions, since |Cl+(p)| is odd; (Note that for D == 1 (mod 4), the norm of the fundamental unit of discriminant D is -1 if and only if the norm of the fundamental unit of discriminant 4*D is -1: if (x_0,y_0) is the fundamental solution to x^2 - D*y^2 = -4, then the fundamental solution to x^2 - D*y^2 = -1 is (x_0/2,y_0/2) for even x_0,y_0 and ((x_0^3+3*x_0*y_0^2*D)/8,(3*x_0*y_0^2+y_0^3*D)/8) for odd x_0,y_0).
- For primes p == 5 (mod 8), x^2 - 2*p*y^2 = -1 has solutions, since |Cl+(8*p)| == 2 (mod 4);
- For different primes p, q == 1 (mod 4) such that each is not a quadratic residue modulo the other, x^2 - p*q*y^2 = -1 has solutions, since |Cl+(p*q)| == 2 (mod 4).
See Theorem 1 and Theorem 2 of Ezra Brown's link for these assertions on Cl+(D), and the subsequent theorems for more results like them.
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000
W. E. Briggs, An Elementary Proof of a Theorem About the Representation of Primes by Quadratic Forms, Canadian Journal of Mathematics, 1954;6:353-363.
Ezra Brown, Class numbers of real quadratic number fields, Trans. Amer. Math. Soc. 190 (1974), 99-107.
Rick L. Shepherd, Binary quadratic forms and genus theory, Master of Arts Thesis, University of North Carolina at Greensboro, 2013.
FORMULA
EXAMPLE
Let D = A079896(23) = 60. It is not divisible by 32 nor congruent to 4 modulo 16, so A391441(23) = omega(60) - 1 = 2. Since 60 has a factor (3) congruent to 3 modulo 4, a(23) = A391441(23) - 1 = 1. Indeed, Cl(60) = C_2.
PROG
(PARI) r(D) = omega(D) - if(D%32==0, 0, if(D%16==4, 2, 1)) \\ gives 2-rank of Cl+(D)
s(D) = if(D%16==0, return(r(D)-1)); my(f = factor(D)[, 1]~); for(i=1, #f, if(f[i]%4==3, return(r(D)-1))); return(r(D)) \\ gives 2-rank of Cl(D)
for(D=1, 1000, if(D%4<=1 && !issquare(D), print1(s(D), ", ")))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Dec 09 2025
STATUS
approved