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It is known that if GRH holds there does not exist additional Idoneal numbers. (see www.mast.queensu.ca/~kani/papers/idoneal.pdf this paper puts on the question of correctnes for Wikipedia and Wolfram MathWorld since they state there could only be ONE additional Idoneal number)

What I am interested in is whether there are any certain properties that an additional idoneal number X should satisfy. I am trying to prove a theorem which would only work if X has three odd divisors. Is anything like that known or easy to derive?

I would like to find a precise reference to the fact that any idoneal number not in the currently known finite list has to have at least three odd prime factors (this should hold, as long as the answer by Pete is valid. Pete also suggested looking into the book primes of the form x^2 + ny^2 but my knowledge of number theory is too limited to derive the stated fact from there)

EDIT: Removed misinterpreted sentence about GRH and idoneal numbers. Added request for reference. I will give 100 bounty points to the first concise reference of this fact.

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    $\begingroup$ Are you sure about your first sentence? It is known that if GRH holds, there are no further idoneal numbers. You say the converse, which sounds unlikely: you're saying we can prove GRH by proving there are no further idoneal numbers. $\endgroup$ Apr 10, 2010 at 22:16
  • $\begingroup$ Yeah you are obviously right. I don't know how I reasoned that out. $\endgroup$
    – Jernej
    Apr 12, 2010 at 11:24
  • $\begingroup$ I supplemented Pete Clark's claims with detailed references, see my response below. $\endgroup$
    – GH from MO
    Jul 13, 2011 at 15:53

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If X is an idoneal number, then the class group of discriminant -4X has exponent dividing 2, so the class number is equal to the number of genera (Theorem 6 in Kani's paper: for me, this is the most convenient definition), which is given by an explicit recipe in terms of the number of prime factors of X and its congruence class mod 32 (formula (3) of Kani's paper).

In particular, if X is idoneal and is a prime or twice a prime, then its class number is at most 4. But all discriminants of class number 4 have been calculated. Indeed, all discriminants of class number up to 100 have been calculated (work of M. Watkins), so a new idoneal number should have at least $6$ odd prime divisors, or something like that.

Also see Cox's book Primes of the Form x^2 + ny^2 for treatment of idoneal numbers.

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The reference

Weinberger, P. J.: Exponents of the class groups of complex quadratic fields, Acta Arith. 22 (1973), 117–124.

was what Matthias Schütt and I cited for the fact that "there is at most one further imaginary quadratic field with class group exponent 2". (The context is constructing at least one K3 surface for each of the 65 [known] idoneal numbers; see http://arxiv.org/pdf/0809.0830.)

For the specific question about 3 or more primes, Pete Clark's argument reduces this to the solution of the class number $h$ problem for $h=1,2,4$. Mark Watkins' paper

Watkins, M.: Class numbers of imaginary quadratic fields, Math. of Comp. 73 (2003) #246, 907-938

which does all $h \leq 100$, has been mentioned already. He starts by reviewing earlier work, ending with "Arno's thesis [2] and subsequent work with Robinson and Wheeler [3] and the work of Wagner [44], which together complete the classication for all $N \leq 7$ and odd $N \leq 23$." On page 5 of Watkins' paper (immediately following the statement of Lemma 3) he notes that "From Gauss's theory of genera [10], [8], we know that $2^{\omega(d)-1}$ divides $h(-d)$ where $\omega(d)$ is the number of distinct prime factors of d...", which corresponds to Pete's observation.

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Let's assume $n$ is an idoneal number with at most 2 prime factors. By Theorem 3 and Corollary 8 of Kani's paper we can assume that $n$ is not a square (hence it is square-free) and it is congruent to 1 or 2 modulo 4. Then $-4n$ is a fundamental discriminant, hence the integral binary quadratic forms of discriminant $-4n$ correspond bijectively to the ideal classes of $\mathbb{Q}(\sqrt{-n})$, see the Appendix here. By Theorem 6 of Kani's paper, each genus of integral binary quadratic forms of discriminant $-4n$ consists of a single class, hence the class number of $\mathbb{Q}(\sqrt{-n})$ equals the number of genera of integral binary quadratic forms of discriminant $-4n$. The number of genera is $2^t$ where $t$ is the number of prime factors of $n$, see the note after Theorem 2.9 in Section 10.2 of Rose's book. We infer that the class number of $\mathbb{Q}(\sqrt{-n})$ equals 1 or 2 or 4. The $n$'s with class number 1 and 2 were found by Heegner-Baker-Stark, those with class number 4 were found by Arno. See Arno's paper and the references there.

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The paper you have quoted says that if the generalized Riemann hypothesis holds then there are only 65 idoneal numbers(see corollary 23). This agrees with the first comment to your answer. According to the paper the error made that allowed only one more idoneal number is the assumption that it has been proved that there are no idoneal even numbers greater than 1848(see remark 24). So if the generalized Riemann hypothesis does not hold that does not imply that there are additional idoneal numbers. Even if there were more there still could be one more odd idoneal number. It is only if the idoneal number after 1848 is even that 4 times that number will also be idoneal.

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Wait, if the next idoneal number after 1848 is even, and 4 times that number is also idoneal, that would mean there are two additional idoneal numbers. It has been proven that at most only one additional idoneal number exists. Therefore, a 66th idoneal number must be odd.

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  • $\begingroup$ "It has been proven that at most only one additional idoneal number exists" - why not cite the exact work where this has been done? $\endgroup$
    – Alex M.
    Jun 5, 2018 at 18:43

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