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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 $p\equiv13\pmod{24}$. $\endgroup$ Aug 14 '19 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! $\endgroup$ Aug 14 '19 at 14:00
  • $\begingroup$ Prof. Ping-Zhi Yuan has just told me that he could prove Conjecture 2. $\endgroup$ Aug 14 '19 at 14:12
  • $\begingroup$ Prof. Bo He has just verified my Conjecture 3 with $q=163$ for all primes $p<5107$ with $p\equiv3\pmod4$. $\endgroup$ Aug 14 '19 at 15:35
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Updated on 2019/08/21: I prove Conjectures 1-3 below. 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.

Proof of Conjecture 1. 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.

Proof of Conjecture 2. Consider the fundamental discriminant $d=8p$. As $p\equiv 3\pmod{4}$, the generic characters for this discriminant are $\bigl(\tfrac{\cdot}{3}\bigr)$ and $\bigl(\frac{-2}{\cdot}\bigr)$, hence there are $2^2/2=2$ genera. Also, there are $4$ ancipital forms of discriminant $d$ and positive first coefficient, which belong to the various genera as follows.

If $p\equiv 3\pmod{8}$, then: $$[1,0,-2p]\quad\text{and}\quad[p,0,-2]\quad\text{belong to the signs}\quad ++$$ $$[2p,0,-1]\quad\text{and}\quad[2,0,-p]\quad\text{belong to the signs}\quad --$$

If $p\equiv 7\pmod{8}$, then: $$[1,0,-2p]\quad\text{and}\quad[2,0,-p]\quad\text{belong to the signs}\quad ++$$ $$[2p,0,-1]\quad\text{and}\quad[p,0,-2]\quad\text{belong to the signs}\quad --$$ As in the proof of Conjecture 1, each ambiguous class of discriminant $d$ contains exactly two ancipital forms with positive first coefficient, hence $[1,0,-2p]$ must be equivalent to $[p,0,-2]$ (resp. $[2,0,-p]$) when $p\equiv 3\pmod{8}$ (resp. $p\equiv 7\pmod{8}$). Now $[1,0,-2p]$ trivially represents $1$, hence $[p,0,-2]$ (resp. $[2,0,-p]$) also represents $1$ when $p\equiv 3\pmod{8}$ (resp. $p\equiv 7\pmod{8}$). As $\left(\frac{2}{p}\right)=-1$ when $p\equiv 3\pmod{8}$, and $\left(\frac{2}{p}\right)=+1$ when $p\equiv 7\pmod{8}$, we conclude that the OP's equation $(3)$ has an integer solution.

Proof of Conjecture 3. I will only use that $p,q\equiv 3\pmod{4}$. Note that if we switch $p$ and $q$, the quadratic residue symbol $\left(\frac{p}{q}\right)$ changes to its negative, hence the solvability of the OP's equation $(4)$ remains unchanged. Therefore, without loss of generality, $\left(\frac{p}{q}\right)=1$, and we need to show that $[p,0,-q]$ represents $4$. Equivalently, after a simple change of variables, $[p,p,(p-q)/4]$ represents $1$. Consider the fundamental discriminant $d=pq$. The generic characters for the discriminant $d$ are $\bigl(\frac{\cdot}{p}\bigr)$ and $\bigl(\frac{\cdot}{q}\bigr)$, hence there are $2^2/2=2$ genera. Also, there are $4$ ancipital forms of discriminant $d$ and positive first coefficient, which belong to the various genera as follows: $$[1,1,(1-pq)/4]\quad\text{and}\quad[p,p,(p-q)/4]\quad\text{belong to the signs}\quad ++$$ $$[pq,pq,(pq-1)/4]\quad\text{and}\quad[q,q,(q-p)/4]\quad\text{belong to the signs}\quad --$$ As in the proof of Conjecture 1, each ambiguous class of discriminant $d$ contains exactly two ancipital forms with positive first coefficient, hence $[1,1,(1-pq)/4]$ and $[p,p,(p-q)/4]$ in the first line must be equivalent. Now $[1,1,(1-pq)/4]$ trivially represents $1$, hence $[p,p,(p-q)/4]$ also represents $1$. That is, the OP's equation $(4)$ has an integer solution.

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  • $\begingroup$ Is ancipital a regular English word or a purely mathematical word? I couldn't find it in Word Reference. $\endgroup$ Aug 14 '19 at 20:53
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    $\begingroup$ @SylvainJULIEN: The meaning and etimology of "ancipital" is explained on the first page of the quoted paper. $\endgroup$
    – GH from MO
    Aug 14 '19 at 21:53
  • $\begingroup$ Indeed, thank you. $\endgroup$ Aug 15 '19 at 8:35
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    $\begingroup$ To avoid confusion, for anyone without access: "In the Gaussian theory of integral binary quadratic forms, a form $[a, b, c] = ax^2 + bxy+cy^2$ is called ambiguous if $a \mid b$; and an ambiguous class is one which contains an ambiguous form. As is well known, the primitive ambiguous classes are those which are self-inverse under composition. For the sake of brevity we will write class and form hereafter to mean primitive class and form. The term ambiguous was a translation of Gauss’s term anceps. We will borrow this word and call the forms $[a, 0, c]$ or $[a, a, c]$ ancipital." $\endgroup$
    – LSpice
    Oct 21 '19 at 1:07
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    $\begingroup$ The paper itself is at Pall - Discriminantal divisors of binary quadratic forms (MSN). $\endgroup$
    – LSpice
    Oct 21 '19 at 1:08
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Here's a second proof of Claim 1 (I guess the others can be taken care of similarly) that perhaps explains a little bit better what is going on. Assume first that $3x^2 + 1 = py^2$ has an integral solution. Then $\eta = x\sqrt{3} + y \sqrt{p}$ is a unit in ${\mathbb Q}(\sqrt{3},\sqrt{p})$ satisfying $\eta^2 = t + u\sqrt{3p}$; thus $\eta^2$ is an odd power of the fundamental unit $\varepsilon$ of ${\mathbb Q}(\sqrt{3p})$. For $p = 829$ we have $\varepsilon = 18982087189657 + 380632678652\sqrt{3p}$, and using the observation that $(2 \cdot 18982087189657 - 2)/3 = 4 \cdot 1778674^2$ (see the proof below) we find $\eta = 1778674 \sqrt{3} + 106999 \sqrt{829}$.

For proving the existence of a solution we simply work backwards (essentially this a classical descent on Pell conics). We start with the fundamental solution $(t, u)$ of $t^2 - 3pu^2 = 1$ and write this equation in the form $(t-1)(t+1) = t^2 - 1 = 3pu^2$. The fact that the fundamental unit has norm $+1$ (the discriminant is divisible by $3$) and that $(t,u)$ is fundamental implies that $3$ and $p$ divide different factors. Using elementary congruences and the fact that $(2/p) = -1$ and $(3/p) = +1$ it is easy to show that the only possibility is $$ t-1 = 6a^2, \quad t+1 = 2pb^2, $$ which implies $1 = pb^2 - 3a^2$.

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  • $\begingroup$ Your $\eta$ does not lie in $\mathbb{Q}(\sqrt{3p})$. It lies in the biquadratic field $\mathbb{Q}(\sqrt{3},\sqrt{p})$. $\endgroup$
    – GH from MO
    Oct 24 '21 at 11:39
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    $\begingroup$ @GH: thanks - corrected. $\endgroup$ Oct 24 '21 at 11:56
  • $\begingroup$ Nice proof! Let me add more detail. First assume that $t$ is even. Then $t-1$ and $t+1$ are coprime, and their product is $3pu^2$. This leads to four cases. If $t-1=a^2$ and $t+1=3pb^2$, then $a^2-3pb^2=-2$, so $(-2/p)=1$, contradiction. If $t-1=3a^2$ and $t+1=pb^2$, then $pb^2-3a^2=2$, so $(2/3)=1$, contradiction. If $t-1=pa^2$ and $t+1=3b^2$, then $3b^2-pa^2=2$, so $(6/p)=1$, contradiction. If $t-1=3pa^2$ and $t+1=b^2$, then $b^2-3pa^2=2$, so $(2/3)=1$, contradiction. I continue in the next remark. $\endgroup$
    – GH from MO
    Oct 24 '21 at 13:30
  • $\begingroup$ Now assume that $t$ is odd. Then $(t-1)/2$ and $(t+1)/2$ are coprime, and their product is $3p(u/2)^2$. This leads to four cases. If $(t-1)/2=a^2$ and $(t+1)/2=3pb^2$, then $a^2-3pb^2=-1$, so $(-1/3)=1$, contradiction. If $(t-1)/2=3a^2$ and $(t+1)/2=pb^2$, then $pb^2-3a^2=1$, which is the conclusion we want. If $(t-1)/2=pa^2$ and $(t+1)/2=3b^2$, then $pa^2-3b^2=-1$, so $(-1/3)=1$, contradiction. If $(t-1)/2=3pa^2$ and $(t+1)/2=b^2$, then $b^2-3pa^2=1$, which contradicts the minimality of $(t,u)$. $\endgroup$
    – GH from MO
    Oct 24 '21 at 13:39

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