

A330191


For d == 0, 1 (mod 4), let E(d) to be the exponent of the class group of binary quadratic forms with discriminant d, b(d) to be the smallest prime p such that Kronecker(d,p) = 1, then sequence gives d such that E(d) > 2 and b(d) > sqrt(d/4).


0



76, 108, 172, 252, 268, 387, 400, 540, 588, 592, 603, 652, 988, 1068, 1072, 1332, 1467, 2088, 2608, 2832, 2907, 3712, 4075, 5868
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OFFSET

1,1


COMMENTS

The exponent of a group G is the smallest e > 0 such that x^e = I for all x in G, where I is the group identity.
d is in this sequence if and only d == 0, 1 (mod 4), d is not in A003171 and b(d) > sqrt(d/4). It seems that 5868 is the largest term. In general, it seems that for any t > 0, b(d) = o(d^t) as d > oo.
For d == 0, 1 (mod 4), we want to determine the size of b(d). Let R = Z[sqrt(d)/2] if d == 0 (mod 4), R = Z[(1+sqrt(d))/2)] otherwise, note that R is not necessarily a Dedekind domain. It is conjectured that if Kronecker(d,p) = 1, then p*R is the product of two distinct prime ideals of R (this is obviously true if d is a fundamental discriminant). It also seems that if p*R = I*I', then I^(k*e) must be principal, e = E(d) (again, this is true if d is fundamental). If these statements are indeed true, let t*O_k = I^(k*e), then t/p is not in R, and the norm of t over R is p^e. Define f(x,y) = x^2 + (d/4)*y^2 if d == 0 (mod 4), x^2 + x*y + ((d+1)/4)*y^2 otherwise, it is easy to see f(x,y) = p^(k*e) has integral solutions (x,y) such that gcd(x,y) = 1.
If f(x,y) = p^(k*e) < d, then y = 1, so it seems that 4*p^(k*e)  d must be a (positive) square. Setting k = 1 gives b(d) > (d/4)^(1/e) (and furthermore we have: if Kronecker(d,p) = 1 and p^(k*e) < d, then k = 1, or (p,k,e,d) = (2,2,1,7), (3,2,1,11)).
We also have the following observations (not proved):
(b) if e > 2, then b(d) < sqrt(d/4) unless d is in this sequence.
Note that 7392 is conjectured to be the largest term in A003171. Therefore, it seems that b(d) < sqrt(d/4) for all d > 7392.
Although these terms satisfy b(d) > sqrt(d/4), for each d it is fairly simple to find a prime p such that Kronecker(d,p) = 1 and f(x,y) = p^2 has no coprime solution (x,y). In contrast, if (a) is true, for d in A003171 (i.e., d such that E(d) <= 2) we have b(d) > sqrt(d/4); if Kronecker(d,p) = 1 then there always exists coprime (x,y) such that f(x,y) = p^2.


LINKS



EXAMPLE

76 is in this sequence because the class group of binary quadratic forms with discriminant 76 is isomorphic to C_3 (generated by 4x^2  x*y + 5y^2), and the smallest prime p such that Kronecker(76,p) = 1 is p = 5 > sqrt(76/4).
387 is in this sequence because the class group of binary quadratic forms with discriminant 387 is isomorphic to C_4 (generated by 9x^2  3x*y + 11y^2), and the smallest prime p such that Kronecker(387,p) = 1 is p = 11 > sqrt(387/4).
5868 is in this sequence because the class group of binary quadratic forms with discriminant 5868 is isomorphic to C_12 (generated by 36x^2  6x*y + 41y^2), and the smallest prime p such that Kronecker(5868,p) = 1 is p = 41 > sqrt(5868/4).


PROG

(PARI) isok(d) = (d>0) && (d)%4<=1 && (quadclassunit(d)[2]!=[]&&quadclassunit(d)[2][1]!=2) && !sum(p=1, sqrt(d/4), isprime(p)&&kronecker(d, p)==1)


CROSSREFS



KEYWORD

nonn,more


AUTHOR



STATUS

approved



