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Equivalently, integers k such that Z[sqrt(k)] = Z[x]/(x^2 - k) is a principal ideal domain.

-2, -1, together with k such that 4*k is in A003656.

All terms are squarefree and congruent to 2 or 3 modulo 4. It appears that the terms > 2 are of the form p or 2*p, where p is a prime congruent to 3 modulo 4. [This is correct; see Theorem 1 and Theorem 2 of Ezra Brown's link. - Jianing Song, Feb 24 2021]

The smallest prime p == 3 (mod 4) that is not a term is p = 79. The smallest prime p == 3 (mod 4) such that 2*p is not a term is p = 71.

Composite terms of A342368.

For a positive fundamental discriminant d, the class number of the real quadratic field of discriminant d is odd if and only if d = 8 or is of one of the three following forms: (i) p, where p is a prime congruent to 1 modulo 4; (ii) 4p or 8p, where p is a prime congruent to 3 modulo 4; (iii) pq, where p, q are distinct primes congruent to 3 modulo 4. See Theorem 1 and Theorem 2 of Ezra Brown's link. This sequence gives values for d in the cases (ii) and (iii) and that the real quadratic number field with discriminant d has odd class number > 1.

Previous name was: Primes of the form 3*x^2 + 5*x*y - 6*y^2 (as well as of the form 6*x^2 + 11*x*y + y^2).

Discriminant = 97. Class = 1. Binary quadratic forms a*x^2 + b*x*y + c*y^2 have discriminant d = b^2-4ac.

Is this the same as A038987? - R. J. Mathar, Oct 12 2013

From Don Reble, Oct 06 2014: (Start)

G. B. Mathews ("Theory of Numbers" by Chelsea publishing) might have an answer to the relation with A038987. In point 59 on page 65 he claims that

- if X is a non-residue of a discriminant of a quadratic form, then X is not representable; and

- if X is a residue of D, then there is a quadratic form of determinant D which represents X.

If all forms of discriminant 97 are equivalent, then that might suffice. (Indeed, either +97 or -97 has class number 1; but I am not sure which sign matters, A003656 vs. A003173.)

(End)

From Jianing Song, Feb 24 2021: (Start)

Also primes of the form u^2 + u*v - 24*v^2. Substitute u, v by u = 9*x+22*y, v = 2*x+5*y gives 3*x^2 + 5*x*y - 6*y^2.

Yes, this is the same as A038987. For primes p being a (coprime) square modulo 97, they split in the ring Z[(1+sqrt(97))/2]. Since Z[(1+sqrt(97))/2] is a UFD, they are reducible in Z[(1+sqrt(97))/2], so we have p = e*(u + v*(1+sqrt(97))/2)*(u + v*(1-sqrt(97))/2) = e*(u^2 + u*v - 24*v^2), e = +-1. WLOG we can suppose e = 1, otherwise substitute u, v by 5035*u+27312*v and 1138*u+6173*v, then p = u^2 + u*v - 24*v^2. On the other hand, if p is a quadratic nonresidue modulo 97, then they remain inert in Z[(1+sqrt(97))/2] and hence cannot be represented as u^2 + u*v - 24*v^2. (End)