On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿

$On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (1)$ Eldar SultanowEldar Sultanow
Capgemini Deutschland GmbH
Nuremberg, Germanyeldar.sultanow@capgemini.com, $On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (2)$ Anja JeschkeAnja Jeschke
Capgemini Deutschland GmbH
Hamburg, Germanyanja.jeschke@capgemini.com,Amir Darwish TfihaAmir Darwish Tfiha
Tishreen University
Science Fuculty
Syriaamirtfiha@tishreen.edu.sy, $On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (3)$ Madjid TehraniMadjid G. Tehrani
George Washington University
Washington, DC 20052, USAmadjid_tehrani@gwu.eduand $On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (4)$ William J BuchananWilliam J Buchanan
Edinburgh Napier University
Edinburgh, UKb.buchanan@napier.ac.uk

Abstract.

We investigate a special family of elliptic curves, namely $E_{p,q}:y^{2}=x^{3}-pqx$ where $p<q$ are odd primes.Fix a line $L_{a.b}:y=\frac{a}{b}x$ where $a\in\mathbb{Z},b\in\mathbb{N}$ and $(a,b)=1$ . We study sufficient conditions that $p$ and $q$ must satisfy so that there are infinitely many elliptic curves $E_{p,q}$ that intersect $L_{a,b}$ .As a result we obtained six conditions for $p$ and $q$ each ensuring that $E_{p,q}$ has rational points. Moreover we could compact these six conditions down to four conditions and provide a visualization of cases up to $p,q\leq 3581$ .

Key words and phrases:

Elliptic Curves, Rational Points

2010 Mathematics Subject Classification:

14H52

1. Introduction

The fact whether an elliptic curve has rational points or not has been occupying mathematicians for a fairly while. There are stringent conditions under which elliptic curves have definitely rational points.

2. Finding 60 cases

Let $L_{a,b}:y=\frac{a}{b}x$ be a linear function of rational slope where $a\in\mathbb{Z},b\in\mathbb{N}$ and $(a,b)=1$ .Let $E_{p,q}:y^{2}=x^{3}-pqx$ be an elliptic curve where $p<q$ are fixed primes.We aim at finding sufficient conditions for $p$ and $q$ so that there are infinitely many elliptic curves $E_{p,q}$ intersecting $L_{a,b}$ .Inserting $L_{a,b}$ into $E_{p,q}$ leads to

x^{3}-\left(\frac{a}{b}\right)^{2}x^{2}-pqx=0.

This cubic function has two solutions that are non-trivial. They are given by

x_{1,2}=\frac{1}{2}\left(\frac{a}{b}\right)^{2}\pm\frac{1}{2}\sqrt{\left(\frac%{a}{b}\right)^{4}+4pq}.

In order for $x_{1,2}$ to be rational, we need $a^{4}+4pqb^{4}$ to be a square:

a^{4}+4pqb^{4}=c^{2}

for some $c\in\mathbb{N}$ . This equation can be factored as $4pqb^{4}=c^{2}-a^{4}=(c-a^{2})(c+a^{2})$ .We want to find all possibilities of how to separate the factors of $4pqb^{4}$ into the two factors $c-a^{2}$ and $c+a^{2}$ .Counting the number of cases boils down to computing the number of divisors of $4pqb^{4}$ . For $\tau(n)$ the divisor function with $n=p_{1}^{e_{1}}\cdots p_{k}^{e_{k}}$ and $p_{k}$ prime, we have the identity $\tau(n)=(e_{1}+1)\cdots(e_{k}+1)$ , see [1, p.34]. If $b$ is prime, the number of cases is

\tau(2^{2}pqb^{4})=(2+1)(1+1)(1+1)(4+1)=60.

If $b$ is not prime, the number of cases is greater.Table 1(a) shows the cases $(c-a^{2},c+a^{2})$ whereas Table 1(b) contains the reversed cases $(c+a^{2},c-a^{2})$ . Both tables present conditions for $p$ and $q$ .

Case	$c-a^{2}$	$c+a^{2}$	Condition
1	$4pqb^{4}$	1	$pq=\frac{1-2a^{2}}{4b^{4}}$
2	$2pqb^{4}$	2	$pq=\frac{1-a^{2}}{b^{4}}$
3	$pqb^{4}$	$4$	$pq=\frac{4-2a^{2}}{b^{4}}$
4	$b^{4}$	$4pq$	$pq=\frac{2a^{2}+b^{4}}{4}$
5	$b^{3}$	$4pqb$	$pq=\frac{2a^{2}+b^{3}}{4b}$
6	$b^{2}$	$4pqb^{2}$	$pq=\frac{2a^{2}+b^{2}}{4b^{2}}$
7	$b$	$4pqb^{3}$	$pq=\frac{2a^{2}+b}{4b^{3}}$
8	$2pb$	$2qb^{3}$	$p=qb^{2}-\frac{a^{2}}{b}$
9	$4qb^{4}$	$p$	$p=2a^{2}+4qb^{4}$
10	$2qb^{4}$	$2p$	$p=a^{2}+qb^{4}$
11	$qb^{4}$	$4p$	$p=\frac{2a^{2}+qb^{4}}{4}$
12	$qb^{3}$	$4pb$	$p=\frac{2a^{2}+qb^{3}}{4b}$
13	$qb^{2}$	$4pb^{2}$	$p=\frac{2a^{2}+qb^{2}}{4b^{2}}$
14	$qb$	$4pb^{3}$	$p=\frac{2a^{2}+qb}{4b^{3}}$
15	$2qb$	$2pb^{3}$	$q=pb^{2}-\frac{a^{2}}{b}$
16	$4pb^{4}$	$q$	$q=2a^{2}+4pb^{4}$
17	$2pb^{4}$	$2q$	$q=a^{2}+pb^{4}$
18	$pb^{4}$	$4q$	$q=\frac{2a^{2}+pb^{4}}{4}$
19	$pb^{3}$	$4qb$	$q=\frac{2a^{2}+pb^{3}}{4b}$
20	$pb^{2}$	$4qb^{2}$	$q=\frac{2a^{2}+pb^{2}}{4b^{2}}$
21	$pb$	$4qb^{3}$	$q=\frac{2a^{2}+pb}{4b^{3}}$
22	$2qb^{2}$	$2pb^{2}$	$p=q+\frac{a^{2}}{b^{2}}$
23	$2b$	$2pqb^{3}$	$pq=\frac{a^{2}+b}{b^{3}}$
24	$2b^{2}$	$2pqb^{2}$	$pq=\frac{a^{2}+b^{2}}{b^{2}}$
25	$2b^{3}$	$2pqb$	$pq=\frac{a^{2}+b^{3}}{b}$
26	$2b^{4}$	$2pq$	$pq=a^{2}+b^{4}$
27	$4b$	$pqb^{3}$	$pq=\frac{2a^{2}+4b}{b^{3}}$
28	$4b^{2}$	$pqb^{2}$	$pq=\frac{2a^{2}+4b^{2}}{b^{2}}$
29	$4b^{3}$	$pqb$	$pq=\frac{2a^{2}+4b^{3}}{b}$
30	$4b^{4}$	$pq$	$pq=2a^{2}+4b^{4}$

Case	$c+a^{2}$	$c-a^{2}$	Condition
31	1	$4pqb^{4}$	$pq=\frac{2a^{2}+1}{4b^{4}}$
32	2	$2pqb^{4}$	$pq=\frac{a^{2}+1}{b^{4}}$
33	4	$pqb^{4}$	$pq=\frac{2a^{2}+4}{b^{4}}$
34	$4pq$	$b^{4}$	$pq=\frac{b^{4}-2a^{2}}{4}$
35	$4pqb$	$b^{3}$	$pq=\frac{b^{3}-2a^{2}}{4b}$
36	$4pqb^{2}$	$b^{2}$	$pq=\frac{b^{2}-2a^{2}}{4b^{2}}$
37	$4pqb^{3}$	$b$	$pq=\frac{b-2a^{2}}{4b^{3}}$
38	$2qb^{3}$	$2pb$	$p=qb^{2}+\frac{a^{2}}{b}$
39	$p$	$4qb^{4}$	$p=4qb^{4}-2a^{2}$
40	$2p$	$2qb^{4}$	$p=qb^{4}-a^{2}$
41	$4p$	$qb^{4}$	$p=\frac{qb^{4}-2a^{2}}{4}$
42	$4pb$	$qb^{3}$	$p=\frac{qb^{3}-2a^{2}}{4b}$
43	$4pb^{2}$	$qb^{2}$	$p=\frac{qb^{2}-2a^{2}}{4b^{2}}$
44	$4pb^{3}$	$qb$	$p=\frac{qb-2a^{2}}{4b^{3}}$
45	$2pb^{3}$	$2qb$	$q=pb^{2}+\frac{a^{2}}{b}$
46	$q$	$4pb^{4}$	$q=4pb^{4}-2a^{2}$
47	$2q$	$2pb^{4}$	$q=pb^{4}-a^{2}$
48	$4q$	$pb^{4}$	$q=\frac{pb^{4}-2a^{2}}{4}$
49	$4qb$	$pb^{3}$	$q=\frac{pb^{3}-2a^{2}}{4b}$
50	$4qb^{2}$	$pb^{2}$	$q=\frac{pb^{2}-2a^{2}}{4b^{2}}$
51	$4qb^{3}$	$pb$	$q=\frac{pb-2a^{2}}{4b^{3}}$
52	$2qb^{2}$	$2pb^{2}$	$p=q-\frac{a^{2}}{b^{2}}$
53	$2pqb^{3}$	$2b$	$pq=\frac{b-a^{2}}{b^{3}}$
54	$2pqb^{2}$	$2b^{2}$	$pq=\frac{b^{2}-a^{2}}{b^{2}}$
55	$2pqb$	$2b^{3}$	$pq=\frac{b^{3}-a^{2}}{b}$
56	$2pq$	$2b^{4}$	$pq=b^{4}-a^{2}$
57	$pqb^{3}$	$4b$	$pq=\frac{4b-2a^{2}}{b^{3}}$
58	$pqb^{2}$	$4b^{2}$	$pq=\frac{4b^{2}-2a^{2}}{b^{2}}$
59	$pqb$	$4b^{3}$	$pq=\frac{4b^{3}-2a^{2}}{b}$
60	$pq$	$4b^{4}$	$pq=4b^{4}-2a^{2}$

3. Six conditions for $E_{p,q}$ to have rational points

The 60 cases in Table 1 reduce to only six different cases with solutions. These cases are shown in Table 2 and described in more detail in the following.All other cases are treated in the Appendix. Appendix A lists the cases that have no solution and Appendix B explains the cases exhibiting redundancy to one of these six cases.All elliptic curves will have the same structure. As an illustration, Figure 1 shows the sample elliptic curves of case 40 and 56 including their rational points.

Case	Condition	Samples	Resulting sample	Resulting rational points
		a, b, p, q	elliptic curve	on sample elliptic curve
17	$q=a^{2}+pb^{4}$	$8,3,3,307$	$y^{2}=x^{3}-921x$	$\left(\frac{307}{9},\frac{2456}{27}\right),(-27,-72)$
26	$pq=a^{2}+b^{4}$	$7,2,p,\frac{65}{p}$	$y^{2}=x^{3}-65x$	$\left(\frac{65}{4},\frac{455}{8}\right),(-4,-14)$
32	$pq=\frac{a^{2}+1}{b^{4}}$	$1432,5,p,\frac{3281}{p}$	$y^{2}=x^{3}-3281x$	$\left(-\frac{1}{25},-\frac{1432}{125}\right),(82025,23491960)$
40	$p=qb^{4}-a^{2}$	$2,1,3,7$	$y^{2}=x^{3}-21x$	$(7,14),(-3,-6)$
47	$q=pb^{4}-a^{2}$	$5,2,3,23$	$y^{2}=x^{3}-69x$	$(12,30),\left(-\frac{23}{4},-\frac{115}{8}\right)$
56	$pq=b^{4}-a^{2}$	$1,2,p,\frac{15}{p}$	$y^{2}=x^{3}-15x$	$(4,2),\left(-\frac{15}{4},-\frac{15}{8}\right)$

3.1. Case 17: $\boldsymbol{q=a^{2}+pb^{4}}$

With $b=1$ , case 8 is a special case of case 17. The square $a^{2}$ is even, because $p$ and $q$ are odd. Because $a\in\mathbb{Z}$ , a has to be even. In other words, $(p-q)\mod{4}=0$ .According to Polignac’s conjecture (which isn’t proved), there are infinitely many cases of two consecutive prime numbers with difference n if n is even. [2, p. 295].Hence, Case 17 with $b=1$ describes a subset of Polignac’s conjecture.

3.2. Case 26: $\boldsymbol{pq=a^{2}+b^{4}}$

Proving that this case has infinitely solutions is a difficult endeavor. A possible direction is to verify the conjecture that $pq=a^{2}+b^{4}$ has infinitely many solutions, which corresponds to the theorem of Iwaniec and Frielander [3] that there are infinitely many primes of the form $p=a^{2}+b^{4}$ . Up to a big integer $X$ we have $Y$ of the primes whose product is a semiprime of the form $a^{2}+b^{4}$ .One example is the curve $y^{2}=x^{3}-65x$ , which is also listed in the LMFDB [4]. The value $c=81$ and the discriminant is $\frac{6561}{64}$ .

In the case that $p\equiv 3\pmod{4}$ or $q\equiv 3\pmod{4}$ no solution exist [5, p.21]. Let us set $p=r^{2}+s^{2}$ and $q=u^{2}+v^{2}$ . If $p\equiv 1\pmod{4}$ and $q\equiv 1\pmod{4}$ we exactly obtain one solution $r>s>0$ for $p$ and one solution $u>v>0$ for $q$ [5, p.21]. If we now have these unique solutions $p=r^{2}+s^{2}$ and $q=u^{2}+v^{2}$ , then the product of both primes is $pq=(r^{2}+s^{2})(u^{2}+v^{2})=(ru+sv)^{2}+(rv-su)^{2}=(ru-sv)^{2}+(rv+su)^{2}$ .Consider $b^{2}=c$ , other integer solutions for $pq=a^{2}+c^{2}=a^{2}+b^{4}$ do not exist, unless one of the four integers $ru+sv$ , $|rv-su|$ , $|ru-sv|$ and $rv+su$ is a perfect square.

3.3. Case 32: $\boldsymbol{pq=\frac{a^{2}+1}{b^{4}}}$

If $b=0,2[4]$ then $a=1,3[4]$ . If $b=1,3[4]$ then $a=0,2[4]$ .

3.4. Case 40: $\boldsymbol{p=qb^{4}-a^{2}}$

We note that if $b$ is odd then $a$ must be even. This is a similar problem as case 8 (which is case 17 with $b=1$ ), but not the same.If $b$ is even, then $2^{4}|p+a^{2}$ .This case can be transformed to the problem describing primes of the form $x^{2}+ny^{2}$ which is extensively elaborated by David A. Cox [6].

3.5. Case 47: $\boldsymbol{q=pb^{4}-a^{2}}$

Let us set $b=2c$ and thus consider $a^{2}=16pc^{3}-qc$ . It follows $a^{2}\equiv 16pc^{3}\bmod{q}$ and $a^{2}\equiv-qc\bmod{p}$ . Using these two congruences we can approach a solution using Quadratic residue and the Chinese remainder theorem.

3.6. Case 56: $\boldsymbol{pq=b^{4}-a^{2}}$

We can write $pq=b^{4}-a^{2}=(b^{2}-a)(b^{2}+a)$ which leads with $p<q$ to $p=b^{2}-a$ and $q=b^{2}+a$ . This is a special case of the prime gap problem. Setting $a=1$ , this is well known as the twin prime conjecture [7, 8].

On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (5)

4. Considering $\boldsymbol{b}$ being not prime

The six cases in section 3 are derived under the condition that $b$ is prime. If we consider that $b$ is not necessarily prime, we can reduce these six cases to four cases.The case 32 with $b$ prime will be a special case of case 26 with $b$ not prime.The case 40 with $b$ prime will be a special case of case 17 with $b$ not prime.

4.1. Case 17 with $\boldsymbol{b}$ not prime: $\boldsymbol{q=a^{2}+pb^{4}}$

Writing $b=b_{1}b_{2}$ with $b_{1},b_{2}$ prime and setting $c-a^{2}=2b_{1}^{4}p$ and $c+a^{2}=2b_{2}^{4}q$ leads to the modified case $b_{2}^{4}q=a^{2}+pb_{1}^{4}$ . The special case $b_{2}=1$ corresponds to the original case 17. The special case $b_{1}=1$ corresponds to case 40. Therefore this modified case makes the case 40 obsolete.

4.2. Case 26 with $\boldsymbol{b}$ not prime: $\boldsymbol{pq=a^{2}+b^{4}}$

Writing $b=b_{1}b_{2}$ with $b_{1},b_{2}$ prime and setting $c-a^{2}=2b_{1}^{4}$ and $c+a^{2}=2b_{2}^{4}pq$ leads to the modified case $b_{2}^{4}pq=a^{2}+b_{1}^{4}$ . The special case $b_{2}=1$ corresponds to the original case. The special case $b_{1}=1$ corresponds to case 32.

4.3. Case 47 with $\boldsymbol{b}$ not prime: $\boldsymbol{q=pb^{4}-a^{2}}$

Writing $b=b_{1}b_{2}$ with $b_{1},b_{2}$ prime and setting $c+a^{2}=2b_{1}^{4}p$ and $c-a^{2}=2b_{2}^{4}q$ leads to the modified case $b_{2}^{4}q=pb_{1}^{4}-a^{2}$ . The special case $b_{2}=1$ corresponds to the original case. The special case $b_{1}=1$ corresponds to case 10 which is impossible to satisfy. Moreover $b_{1}$ must be larger than $b_{2}$ .

4.4. Case 56 with $\boldsymbol{b}$ not prime: $\boldsymbol{pq=b^{4}-a^{2}}$

Writing $b=b_{1}b_{2}$ with $b_{1},b_{2}$ prime and setting $c+a^{2}=2b_{1}^{4}$ and $c-a^{2}=2b_{2}^{4}pq$ leads to the modified case $b_{2}^{4}pq=b_{1}^{4}-a^{2}$ . The special case $b_{2}=1$ corresponds to the original case. The special case $b_{1}=1$ corresponds to case 2 which is impossible to satisfy. Moreover $b_{1}$ must be larger than $b_{2}$ .

5. Visualize Patterns

Using the four cases presented in Section 4 we cover any curve $E_{p,q}:y^{2}=x^{3}-pqx$ where $p<q$ are odd primes.In Figure 2 we colorize each cell by red for case 4.1, green for case 4.2, blue for case 4.3 and yellow for case 4.4. The more curves covered by case 4.1 exist, the more intense is the color red. Analogously for the other colors. We unveiled the following patterns:

On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (6)

On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (7)

On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (8)

On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (9)

On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (10)

On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (11)

6. Conclusion and Outlook

So far, we have inferred the conditions that two distinct odd primes $p,q$ must satisfy so that the elliptic curve $y^{2}=x^{3}-pqx$ can have rational points. There exist many curves that fulfill at least one of these conditions and therefore have rational points. However, the case $p=5,q=7$ , namely the elliptic curve $y^{2}=x^{3}-35x$ has no other solution than $(x,y)=(0,0)$ .

Interesting directions for future research may include:

•
fixing a line and search for a family of curves where $pq$ is a congruent number. Plot the $(p,q)$ pairs and explore a structure
•
fixing a curve and search for a family of lines that intersect this curve in rational points
•
taking a closer look at the visualized patterns and interpret these by drawing some conclusions about the distribution of prime numbers

7. Acknowledgements

We are grateful for the kind assistance and useful input from mathematics communities like the Stack Exchange Network or the MatheBoard Community. As an example let us mention Servaes and HAL 9000, for whose instant help in matters of number theory we are very grateful as for the help by many other pleasant people from the math communities.

We owe the graphical illustration of the patterns to Arty. He implemented the C++ programs that generate the needed data, which can be checked out on GitHub [9].

Appendix A Reasons for some cases to be unsatisfiable

In the following we provide (numbered) reasons for the unsatisfiability of conditions in Table 1:

a)
Cases 1, 2, 3, 37, 53 and 54 are impossible to satisfy. This is due to the fact that $pq$ is an integer requires the fraction’s numerator to be larger than the denominator.
b)
These cases require $b$ to be even and as a consequence $a$ to be odd (since $a\in\mathbb{Z}$ , $b\in\mathbb{N}$ and the fraction $\frac{a}{b}$ is reduced). Let us take for example case 4 with the equation $4pq=2a^{2}+b^{4}$ that leads to $2pq=a^{2}+\frac{b^{4}}{2}$ when dividing it by $2$ and thus to the contradictory requirement $a$ is even. Another example is case 27 with $pqb^{3}=2a^{2}+4b$ . Also here $b$ must be even and therefore $a$ odd. But $\frac{pqb^{3}}{2}=a^{2}+2b$ requires $a$ must be even.
c)
Case 7 is unsolvable. We know that $b$ must be even and by substituting $b$ with $2t$ we have to solve $16t^{3}pq=a^{2}+t$ which is $(16t^{2}pq-1)t=a^{2}$ . Since $t$ is coprime with $16t^{2}pq-1$ , we conclude that $16t^{2}pq-1$ is a perfect square, which is impossible by an argument $\bmod 4$ . Recall that if $x$ is a perfect square then $x\equiv 0\bmod 4$ or $x\equiv 1\bmod 4$ [5, p.21].
d)
Case 31 is impossible to satisfy because the equation $4pqb^{4}=2a^{2}+1$ has no solution. The reason for this is that on the left side of the equation is an even number and on the right side is an odd number.
e)
Cases 9, 10, 22 and 38 are unsatisfiable due to the assumed inequality $p<q$ .
f)
Case 15 is unsatisfiable, since it requires $b=1$ leading to $q=p-a^{2}$ which violates the assumption $p<q$ .
g)
Cases 9, 16, 30, 39, 46, 60 are unsatisfiable, since $p$ and $q$ are odd primes and the difference or sum of two even integers cannot be odd.
h)
Case 55 is unsatisfiable, because it requires $b=1$ leading to the equation $pq=1-a^{2}$ which has no solution.

Appendix B Reasons for some cases to be redundant

B.1. Case 8: $\boldsymbol{p=qb^{2}-\frac{a^{2}}{b}}$

The integer $b$ can only accept the value $1$ , because p is an integer. The result is the equation $p=q-a^{2}$ , which is case 17 with $b=1$ .

B.2. Case 14: $\boldsymbol{p=\frac{2a^{2}+qb}{4b^{3}}}$

We can rewrite this condition as $q=4b^{2}p-a^{2}\cdot\frac{2}{b}$ . There are only two possible solutions $b=1$ and $b=2$ . Setting $b=1$ leads to $q=4p-2a^{2}$ which is a special case of case 46 and is impossible to satisfy due to reason g (Appendix A). Setting $b=2$ leads to $q=2^{4}p-a^{2}$ which is a special case of case 47.

B.3. Case 21 and 51: $\boldsymbol{q=\frac{pb\pm 2a^{2}}{4b^{3}}}$

We can rewrite condition of case 21 as $p=4b^{2}q-a^{2}\cdot\frac{2}{b}$ . There are only two possible solutions $b=1$ and $b=2$ . Setting $b=1$ leads to $p=4q-2a^{2}$ which is a special sub case of case 39. Setting $b=2$ leads to $p=2^{4}q-a^{2}$ which is a special sub case of case 40.

Analoguously, we can rewrite condition of case 51 as $p=4b^{2}q+a^{2}\cdot\frac{2}{b}$ . There are only two possible solutions $b=1$ and $b=2$ . Setting $b=1$ leads to $p=4q+2a^{2}$ which is a special sub case of case 9. Setting $b=2$ leads to $p=2^{4}q+a^{2}$ which is a special sub case of case 10. Both cases are impossible to satisfy due to reason e (Appendix A).

B.4. Case 52: $\boldsymbol{p=q-\frac{a^{2}}{b^{2}}}$

The value of $b$ can only be $1$ to ensure that $p$ remains an integer. This leads to $p=q-a^{2}$ , which simultaneously goes for case 8, which is a special case of case 17.

B.5. Case 23: $\boldsymbol{pq=\frac{a^{2}+b}{b^{3}}}$

It must $b$ be odd and $a$ must be even. Only solutions for $b=1$ exist, since the right-hand side of $\frac{a^{2}}{b}=pqb^{2}-1$ is an integer and the left-hand side is a fraction unless $b$ is $1$ . Setting $b=1$ boils the condition down to $pq=a^{2}+1$ which is a special sub case of case 26.

B.6. Case 24: $\boldsymbol{pq=\frac{a^{2}+b^{2}}{b^{2}}}$

Also here no solution exist for $b>1$ . The product $pq=\left(\frac{a}{b}\right)^{2}+1$ can only be an integer when $b=1$ because the fraction $\frac{a}{b}$ is reduced (by assumption $a$ and $b$ are coprime). For this reason, case 24 is the same special case of case 26, just as case 23 does.

B.7. Case 25: $\boldsymbol{pq=\frac{a^{2}+b^{3}}{b}}$

This case is identical with cases 23 and 24, since it provides only solutions for $b=1$ as well. The reason for this is analogous to both previous cases. Here $\frac{a^{2}}{b}$ is a fraction unless $b=1$ . For this reason, case 25 is the same special case of case 26, just as case 23 does.

B.8. Case 29 and Case 59: $\boldsymbol{pq=\frac{4b^{3}\pm 2a^{2}}{b}}$

In this case we get solutions if $b$ divides $2a^{2}$ . Therefore only solutions exist if $b=2$ , as per assumption $a$ and $b$ are coprime. For Case 29, setting $b=2$ boils the condition down to $pq=a^{2}+2^{4}$ which is a special sub case of case 26. For case 59, setting $b=2$ boils the condition down to $pq=2^{4}-a^{2}$ which is a special sub case of case 56.

B.9. Case 44: $\boldsymbol{p=\frac{qb-2a^{2}}{4b^{3}}}$

We can rewrite this condition as $q=4b^{2}p+a^{2}\frac{2}{b}$ . There are only two possible solutions $b=1$ and $b=2$ . Setting $b=1$ leads to $q=4p+2a^{2}$ which is a special case of case 16 and is impossible to satisfy due to reason g (Appendix A). Setting $b=2$ leads to $q=2^{4}p+a^{2}$ which is a special case of case 17.

B.10. Case 45: $\boldsymbol{q=pb^{2}+\frac{a^{2}}{b}}$

See argumentation in B.4.

References

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On families of elliptic curves 𝐸_{𝑝,𝑞}:𝑦²=𝑥³-𝑝⁢𝑞⁢𝑥 that intersect the same line 𝐿_{𝑎,𝑏}:𝑦={𝑎/𝑏}⁢𝑥 of rational slope (2024)