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In Question 337879, I conjectured that for any prime $p\equiv3\pmod4$ the equation $$3x^2+4\left(\frac p3\right)=py^2\tag{1}$$ always has integer solutions, where $(\frac p3)$ is the Legendre symbol. Motivated by this, here I pose the following conjectures.

Conjecture 1. For any prime $p\equiv13\pmod{24}$, the equation $$3x^2+1=py^2\tag{2}$$ always has integer solutions.

For example, when $p=829$ the least positive integer solution of $(2)$ is $$(x,\,y)=(1778674,\,106999).$$

Conjecture 2. For any prime $p\equiv3\pmod4$, the equation $$2x^2-py^2=\left(\frac 2p\right)\tag{3}$$ always has integer solutions, where $(-)$ is the Legendre symbol.

For example, when $p=167$ the smallest positive integer solution of $(3)$ is $$(x,\,y)=(3993882,\,437071).$$

Conjecture 3. For any prime $p\equiv3\pmod4$ and $q\in\{7,11,19,43,67,163\}$, the equation $$qx^2+4\left(\frac pq\right)=py^2\tag{4}$$ always has integer solutions.

QUESTION. How to solve the conjectures?

Your comments are welcome!

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  • $\begingroup$ After learning this conjecture from me, Prof. Bo He has verified my above conjecture for all primes $p<72253$ with $p\equiv13\pmod{24}$. $\endgroup$ – Zhi-Wei Sun Aug 14 at 11:55
  • $\begingroup$ Note that those imaginary quadratic fields $\mathbb Q(\sqrt{-q})$ with $q\in\{1,2,3,7,11,19,43,67,163\}$ has class number one! $\endgroup$ – Zhi-Wei Sun Aug 14 at 14:00
  • $\begingroup$ Prof. Ping-Zhi Yuan has just told me that he could prove Conjecture 2. $\endgroup$ – Zhi-Wei Sun Aug 14 at 14:12
  • $\begingroup$ Prof. Bo He has just verified my Conjecture 3 with $q=163$ for all primes $p<5107$ with $p\equiv3\pmod4$. $\endgroup$ – Zhi-Wei Sun Aug 14 at 15:35
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Here is a proof of Conjecture 1. The proof of the other conjectures should be similar. I will use the usual terminology and notations for binary quadratic forms. In particular, I will use the first two pages of Pall: Discriminantal divisors of binary quadratic forms, J. Number Theory (1) 1969, 525–533.

Consider the fundamental discriminant $d=12p$. The generic characters for this discriminant are $\bigl(\tfrac{\cdot}{3}\bigr)$, $\bigl(\frac{\cdot}{p}\bigr)$, $\bigl(\frac{-1}{\cdot}\bigr)$, hence there are $2^3/2=4$ genera. Also, there are $8$ ancipital forms of discriminant $d$ and positive first coefficient, which belong to the various genera as follows: $$[1,0,-3p]\quad\text{and}\quad [p,0,-3]\quad\text{belong to the signs}\quad +++$$ $$[3p,0,-1]\quad\text{and}\quad [3,0,-p]\quad\text{belong to the signs}\quad -+-$$ $$[2,2,(1-3p)/2]\quad\text{and}\quad [2p,2p,(p-3)/2]\quad\text{belong to the signs}\quad --+$$ $$[6p,6p,(3p-1)/2]\quad\text{and}\quad [6,6,(3-p)/2]\quad\text{belong to the signs}\quad +--$$ However, by Theorem 1 of the quoted paper (which is essentially due to Gauss), each ambiguous class of discriminant $d$ contains exactly two ancipital forms with positive first coefficient, hence $[1,0,-3p]$ and $[p,0,-3]$ in the first line must be equivalent. Now $[1,0,-3p]$ trivially represents $1$, hence $[p,0,-3]$ also represents $1$. That is, the OP's equation $(2)$ has an integer solution. Done.

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  • $\begingroup$ Is ancipital a regular English word or a purely mathematical word? I couldn't find it in Word Reference. $\endgroup$ – Sylvain JULIEN Aug 14 at 20:53
  • $\begingroup$ @SylvainJULIEN: The meaning and etimology of "ancipital" is explained on the first page of the quoted paper. $\endgroup$ – GH from MO Aug 14 at 21:53
  • $\begingroup$ Indeed, thank you. $\endgroup$ – Sylvain JULIEN 2 days ago

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