Deleted.
2 or 3 days ago, the question What's the connection between the Heegner numbers and 37? was asked.
(1). The first version of that question, was not eligible as a question to be asked in MSE, and I was the fifth user, who voted to close the question.
(2). In the comments:
- User @David wrote that:
37 is the largest prime for which -163 is a quadratic non-residue.
- Then I pointed out that his comment is not right by this comment:
Consider this Legendre symbol: (−163/137)=−1.
- Then he answered that:
The statement is more like 37 is largest prime p s.t. All primes, q: 2<q<=p have (−163/q)=−1. I would like to state this correctly and succinctly.
(3). Why I was convinced that his question is a good question?
- His comment was interesting to me, because I've realized that the phenomenon, which he pointed out, is not accidental and is closely related to the properties of Heegner numbers. [Not important: As you can see in the comments; at first, I wrongly thought that it is related to Rabinowitz theorem. For proof of the Rabinowitz theorem see this answer by Will Jagy.]
- Definition: A Heegner number is a square-free positive integer $d$ such that the imaginary quadratic field $\mathbb{Q}(\sqrt{-d})$ has class number $1$. I wrote my answer, in the comments [see these six consecutive comments]. If we set this definition as the definition of Heegner Numbers, then the First case is just the immediate consequence of the definition. I mean whether we know the total set of Heegner numbers is finite or not, we can do the first case without any use of powerful theorems like Baker–Stark–Heegner theorem, just by the knowledge of $19^{\text{th}}$ century. [Very Important: Unfortunately, in the Second case, I strongly used the Baker–Stark–Heegner theorem. Not important: If we break the second case in some new cases, then the case "$(\dfrac{-H}{q})=0$ and $q^2 \nmid (H)$" can be done in a quite elementary manner, but yet I don't have any idea to do the case "$(\dfrac{-H}{q})=0$ and $q^2 \mid (H)$".]
(4). If a curious person sees that question, then probably he/ she would click on the show N more comments to see the comments. So Why do I insist to reopen that question? An unsatisfactory answer would be something like this: (A) Suppose that a random student who does not have enough curiosity about this special question, will see this question. (+) Since this question is closed for the lack of details, (++) and since there is not an answer for it; the most probable scenario would be that he/ she will be taught there is nothing special behind this question. (B) Another unsatisfactory answer would be something like this: From a pedagogical point of view: Someone else can write a better answer, maybe someone can be able to answer it without using any powerful theorems. Finally, reading a solution as a single answer is better to follow the separated comments. At last, I should confess whether this question reopens or not, it would not cause important changes.
(5): TOTALLY IRRELEVANT: As a math student, I should be careful about my counting: In my last comment, I wrote My 7th comment, but it was my eighth comment.
Added after the edit:
Let's set this definition as the definition of a Heegner number. As I stated in the comments, the question can be considered as
- Q($1$): Let $H$ be a Heegner number, then for any prime $2 \neq q \leq \dfrac{1+H}{4}-2$ we have $(\dfrac{-H}{q})=-1$,
Or equivalently:
- Q($2$): Let $H$ be a positve square-free integer, and let $q$ be a prime number $2 \neq q \leq \dfrac{1+H}{4}-2$, such that $(\dfrac{-H}{q})=+1$. Then $H$ is not a Heegner number.
Both of these equivalent questions are very surprising to me. Now consider this question:
- Q($3$): Let $200 \leq H$ be a positve square-free integer, then there exists a prime number $q$ such that: $2 \neq q \leq \dfrac{1+H}{4}-2$ and $(\dfrac{-H}{q})=+1$.
If I was able to prove the above question, then the Baker–Heegner–Stark theorem, would follow immediately. So Q($3$) is not an easy question.
- Reamrk($4$): Let $\dfrac{49}{3} \leq H \stackrel{8}{\equiv} 3$, then the Gauss's upper bound [Gauss's bound, is a better bound, than Minkowski's bound, for the norm of ideals to be checked in order to determine the class number of the imaginary quadratic field $\mathbb{Q}(\sqrt{-H})$], tells us: By considering Baker–Heegner–Stark theorem, if $H \notin \{ 11, 19, 43, 67, 163 \}$, then there is a prime $q$ such that: $2 \neq q \leq \sqrt{\dfrac{H}{3}}$ and $(\dfrac{-H}{q})=+1$. Note that for $\dfrac{49}{3} \leq H$ this new bound is surprisingly smaller than $\dfrac{1+H}{4}-2$. These were some parts of my unsuccessful attempts to prove Q($3$), without use of Baker–Heegner–Stark theorem.
