International Journal of Algebra, Vol. 14, 2020, no. 3, 139 – 162

HIKARI Ltd, www.m-hikari.com

https://doi.org/10.12988/ija.2020.91250

Enumeration in Ranks of Various Elliptic Curves

𝒚𝟐 = 𝒙𝟑 ± 𝑨𝒙

Shin-Wook Kim

Deokjin-gu, Songcheon 54823

101-703, I-Park Apt

Jeonju, Jeonbuk, Korea

This article is distributed under the Creative Commons by-nc-nd Attribution License.

Copyright © 2020 Hikari Ltd.

Abstract

We appoint that 𝐸𝑝𝑞 and 𝐸−2𝑝 are elliptic curves 𝑦2 = 𝑥3 + 𝑝𝑞𝑥 and 𝑦2 = 𝑥3 −

2𝑝𝑥 then, we will calculate the ranks of these curves. Denote 𝐸±𝑝𝑞 as elliptic

curves 𝑦2 = 𝑥3 ± 𝑝𝑞𝑥 then, we will research the ranks of it and compare the results with that of previous curves. In addition, we shall treat the rank of

𝐸−𝑝: 𝑦2 = 𝑥3 − 𝑝𝑥.

Mathematics Subject Classification: 11A41, 11G05

Keywords: Odd prime, Elliptic curves

1 Introduction

Until now, the known biggest rank in particular elliptic curve is at least 28([3]).

Whereas in generalized elliptic curve, the consequence of rank is small relatively.

If it is gotten as 3 or 4 then, it is not a little rank. And in many cases systematized

ranks exist between from 0 to 3. This is general range of rank in elliptic curve. In

[2], the authors numerated that rank of 𝑦2 = 𝑥(𝑥 − 2𝑚 )(𝑥 + 𝑞 − 2𝑚) where 𝑞 −2𝑚 = 𝑝 are odd primes is 𝑟𝑞,𝑝,3 = 0 if 𝑝 ≡ 3(𝑚𝑜𝑑 4) and is 𝑟𝑞,𝑝,4 ≤1. In [14], the

authors proved that rank of 𝑦2 = 𝑥(𝑥2 − 𝑘2) is at most 2 where 𝑘 = 𝑝𝑞𝑟 and 𝑝 ≡

𝑞 ≡ 5(𝑚𝑜𝑑 8) and 𝑟 ≡ 3(𝑚𝑜𝑑 4) are primes such that (𝑝

𝑞) = (

𝑝

𝑟) = (

𝑞

𝑟) = +1.

In [6], the authors calculated that rank of congruent number elliptic curve

𝑦2 = 𝑥(𝑥2 − 𝑛2) is at least 3 where the positive integer 𝑛 satisfies that 𝑛 =

6(𝑢4 + 2𝑢2𝑣2 + 4𝑣4)(𝑢4 + 8𝑢2𝑣2 + 4𝑣4) with 𝑡 =𝑢

𝑣 and 𝑢 , 𝑣 are integers as

gcd(𝑢, 𝑣) = 1 and (𝑡, 𝑤)(𝑡 ≠ 0) is one of the curve 𝑤2 = 𝑡4 + 14𝑡2 + 4 where

140 Shin-Wook Kim

this curve has infinitely many points. In [7], the author computed that rank of

curve 𝑦2 = 𝑥3 + 2𝑝𝑥 with prime 𝑝 as 𝑝 ≡ 11(𝑚𝑜𝑑 16) is 0. In [8], the rank of elliptic curve 𝑦2 = 𝑥3 + 𝑝(𝑝 − 2)𝑥 with twin primes 𝑝, 𝑝 − 2 as 𝑝 ≡ 5(𝑚𝑜𝑑 16) was computed as 1. In [11], the author enumerated that rank of elliptic curve

𝑦2 = 𝑥3 − 2𝑝𝑥 where prime 𝑝 is 𝑝 = 838𝑢4 + 260𝑢2𝑣2 + 25𝑣4 with integers 𝑢 and 𝑣 and (𝑢,𝑣)=1 and 𝑝 ≡ 3(𝑚𝑜𝑑 16) and the case 𝑝 = 481𝑢4 + 930𝑢2𝑣2 +450𝑣4 with integers 𝑢 and 𝑣 and (𝑢, 𝑣)=1 and 𝑝 ≡ 5(𝑚𝑜𝑑 16) are 1 and that of 𝑦2 = 𝑥3 − 𝑝𝑥 where prime 𝑝 is the form 𝑝 = 85𝑢4 + 48𝑢2𝑣2 + 144𝑣4 with integers 𝑢 and 𝑣 and (𝑢 ,𝑣 )=1 and 𝑝 ≡ 5(𝑚𝑜𝑑 16) and the case 𝑝 = 122𝑢4 +242𝑢2𝑣2 + 121𝑣4 with integers 𝑢 and 𝑣 and (𝑢 ,𝑣)=1 and 𝑝 ≡ 5(𝑚𝑜𝑑 16) are also 1 in both cases. In [12], the author showed that rank of 𝑦2 = 𝑥3 − 𝑝𝑥 which satisfies that 𝑝 is a prime 𝑝 = 26𝑢4 + 10𝑢2𝑣2 + 𝑣4 with integers 𝑢 and 𝑣 and (𝑢,𝑣)=1 and 𝑝 ≡ 5(𝑚𝑜𝑑 16) and the case 𝑝 = 37𝑢4 + 24𝑢2𝑣2 + 4𝑣4 with two integers 𝑢 and 𝑣 and (𝑢, 𝑣)=1 and 𝑝 ≡ 5(𝑚𝑜𝑑 16) are both 1. In [13], the author enumerated that rank of elliptic curve 𝑦2 = 𝑥3 − 𝑝𝑞𝑥 that satisfies different odd primes 𝑝 and 𝑞 are 𝑝 = 25𝑢4 + 20𝑢2𝑣2 + 6𝑣4 with two integers 𝑢 and 𝑣 and (𝑢,𝑣)=1 and 𝑝 ≡ 3(𝑚𝑜𝑑 16) and 𝑞 = 25𝑢4 + 20𝑢2𝑣2 + 2𝑣4 with two integers 𝑢 and 𝑣 and (𝑢, 𝑣)=1 and 𝑞 ≡ 15(𝑚𝑜𝑑 16) is 2. In [5], Husem�̈�ller presented that rank of curve 𝑦2 = 𝑥3 − 𝑝𝑥 with prime 𝑝 as 𝑝 ≡ 3(𝑚𝑜𝑑 8) is 0. In this article, we will numerate ranks of various forms of elliptic curves 𝑦2 = 𝑥3 ± 𝐴𝑥 as the change of supposition of number 𝐴. In section 2, the rank of elliptic curve 𝐸𝑝𝑞: 𝑦

2 = 𝑥3 + 𝑝𝑞𝑥 will be managed.

In section 3, we shall search the rank of 𝐸−2𝑝: 𝑦2 = 𝑥3 − 2𝑝𝑥.

In section 4, we will calculate the ranks of elliptic curves 𝐸±𝑝𝑞: 𝑦2 = 𝑥3 ± 𝑝𝑞𝑥

and compare the consequences with previous calculations.

In section 5, the rank of 𝐸−𝑝: 𝑦2 = 𝑥3 − 𝑝𝑥 will be surveyed and compared with

the results in 𝐸±𝑝𝑞. In section 6, we shall present examples of results from section 2 to 5.

To start with, the notations in [16] ought to be considered.

Define 𝐸 as an elliptic curve 𝑦2 = 𝑥3 + 𝑎𝑥2 + 𝑏𝑥 and assign Γ as the set of rational points on 𝐸. Then, we acquire that Γ is a finitely generated abelian group owing to 𝑀𝑜𝑟𝑑𝑒𝑙𝑙′𝑠 Theorem. Furthermore, there derived the relation Γ ≅𝐸(𝑄)𝑡𝑜𝑟𝑠 ⊕ 𝑍

𝑟 where 𝐸(𝑄)𝑡𝑜𝑟𝑠 is torsion subgroup and 𝑟 is Mordell-Weil rank. Moreover, we appoint that 𝑄× is a multiplicative group which is comprised of non-zero rational numbers. Let 𝑄×2 be the subgroup of squares of elements of 𝑄×. Denote 𝛼 as a homomorphism given as 𝛼: Γ → 𝑄×/𝑄×2 where

𝛼(𝑂) = 1(𝑚𝑜𝑑 𝑄×2) and 𝛼(0, 0)= 𝑏(𝑚𝑜𝑑 𝑄×2) and 𝛼(𝑥, 𝑦)= 𝑥(𝑚𝑜𝑑 𝑄×2)

with point at infinity 𝑂 and non-zero 𝑥. Put �̅� as the curve 𝑦2 = 𝑥(𝑥2 − 2𝑎𝑥 + 𝑎2 − 4𝑏) and assume that Γ̅ is the set of

rational points on �̅�. Take �̅� as a homomorphism 𝛼:̅ Γ̅ → 𝑄×/𝑄×2 where

Enumeration in ranks of various elliptic curves 141

�̅�(𝑂) = 1(𝑚𝑜𝑑 𝑄×2) and

�̅�(0, 0)= 𝑎2 − 4𝑏(𝑚𝑜𝑑 𝑄×2) and

�̅�(𝑥, 𝑦)= 𝑥(𝑚𝑜𝑑 𝑄×2)

with point at infinity 𝑂 and 𝑥 ≠ 0. Set 𝑁2 = 𝑏1𝑀

4 + 𝑎𝑀2𝑒2 + 𝑏2𝑒4 as relating equation for Γ where 𝑏1and 𝑏2 are

divisors of 𝑏 as 𝑏 = 𝑏1𝑏2 with 𝑏1 ≢ 1, 𝑏(𝑚𝑜𝑑 𝑄×2).

Denote 𝑁2 = 𝑏1𝑀4 − 2𝑎𝑀2𝑒2 + 𝑏2𝑒

4 as relating equation for Γ̅ where the coefficients 𝑏1 and 𝑏2 are divisors of 𝑎

2 − 4𝑏 such that 𝑏1𝑏2 = 𝑎2 − 4𝑏 with

𝑏1 ≢ 1, 𝑎2 − 4𝑏(𝑚𝑜𝑑 𝑄×2).

Suppose that (𝑀, 𝑒, 𝑁) is an integral solution of above two relating equations that satisfies (𝑀, 𝑁)=(𝑀, 𝑒)=(𝑁, 𝑒) =(𝑏1, 𝑒) = (𝑏2, 𝑀) =1 and 𝑀 ≠ 0, 𝑒 ≠ 0 .

And there is deduced 2𝑟 =#𝛼(Γ)#�̅�(Γ)

4 where 𝑟 denotes the rank of curve 𝐸.

We take several notations as follows:

𝐿𝐷𝑉: Legendre symbols of different values([10]).

𝐿𝑆𝑉: Legendre symbols having same value([10]).

𝑤. 𝑖. 𝑢. 𝑣. 1: with integers 𝑢 and 𝑣 and (𝑢, 𝑣)=1.

𝑟2.4: 2𝑟 = 2∙4

4.

𝑟4.2: 2𝑟 =4∙2

4.

𝑟≥4.4: 2𝑟 ≥4∙4

4.

𝑟4.4: 2𝑟 =4∙4

4.

2 In 𝑬𝒑𝒒

In second section, we calculate the rank of elliptic curve 𝐸𝑝𝑞: 𝑦

2 = 𝑥3 + 𝑝𝑞𝑥.

Examining the solvability of equation 1) for Γ and that of 1) and 5) for Γ̅ will be omitted. See [13] for this.

Proposition 2.1. (1). Define 𝐸𝑝𝑞 as an elliptic curve 𝑦2 = 𝑥3 + 𝑝𝑞𝑥 where 𝑝

and 𝑞 are different odd primes 𝑝 = 𝑢4 + 72𝑢2𝑣2 + 10𝑣4 and 𝑞 = 𝑢4 +74𝑢2𝑣2 + 10𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 3(𝑚𝑜𝑑 16) and 𝑞 ≡ 5(𝑚𝑜𝑑 16) then, we

142 Shin-Wook Kim

gain 𝑟𝑎𝑛𝑘(𝐸(𝑢4+72𝑢2𝑣2+10𝑣4)(𝑢4+74𝑢2𝑣2+10𝑣4)(𝑄)) = 1.

(2). If 𝐸𝑝𝑞 is appointed as an elliptic curve 𝑦2 = 𝑥3 + 𝑝𝑞𝑥 where 𝑝 , 𝑞 are

distinct odd primes as 𝑝 = 𝑢4 − 10𝑢2𝑣2 + 12𝑣4 and 𝑞 = 𝑢4 − 10𝑢2𝑣2 +14𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 3(𝑚𝑜𝑑 16) and 𝑞 ≡ 5(𝑚𝑜𝑑 16) then, the result 𝑟𝑎𝑛𝑘(𝐸(𝑢4−10𝑢2𝑣2+12𝑣4)(𝑢4−10𝑢2𝑣2+14𝑣4)(𝑄)) = 1 is gotten.

(3). We appoint that 𝐸𝑝𝑞 is an elliptic curve 𝑦2 = 𝑥3 + 𝑝𝑞𝑥 where 𝑝 , 𝑞 are

distinct odd primes of the forms 𝑝 = 𝑢4 − 2𝑢2𝑣2 + 4𝑣4 and 𝑞 = 𝑢4 − 2𝑢2𝑣2 +6𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 3(𝑚𝑜𝑑 16) and 𝑞 ≡ 5(𝑚𝑜𝑑 16) then, we obtain that 𝑟𝑎𝑛𝑘(𝐸(𝑢4−2𝑢2𝑣2+4𝑣4)(𝑢4−2𝑢2𝑣2+6𝑣4)(𝑄)) = 1.

(4). Denote 𝐸𝑝𝑞 as an elliptic curve 𝑦2 = 𝑥3 + 𝑝𝑞𝑥 that satisfies 𝑝 and 𝑞 are

different odd primes such that 𝑝 = 𝑢4 − 4𝑢2𝑣2 + 6𝑣4 and 𝑞 = 𝑢4 − 4𝑢2𝑣2 +8𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 3(𝑚𝑜𝑑 16) and 𝑞 ≡ 5(𝑚𝑜𝑑 16) then, we take that 𝑟𝑎𝑛𝑘(𝐸(𝑢4−4𝑢2𝑣2+6𝑣4)(𝑢4−4𝑢2𝑣2+8𝑣4)(𝑄)) = 1.

Proo𝑓. (1). We appoint that 𝐸𝑝𝑞 is an elliptic curve 𝑦2 = 𝑥3 + 𝑝𝑞𝑥 where 𝑝 and

𝑞 are different odd primes as the forms 𝑝 = 𝑢4 + 72𝑢2𝑣2 + 10𝑣4 and 𝑞 = 𝑢4 +74𝑢2𝑣2 + 10𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 3(𝑚𝑜𝑑 16) and 𝑞 ≡ 5(𝑚𝑜𝑑 16) . Take 𝑝 and 𝑞 as 𝑝 = 16𝑘 + 3 and 𝑞 = 16𝑘′ + 5 with integers 𝑘 , 𝑘′ then, we attain relating equations for Γ as follows:

1)𝑁2 = 𝑀4 + (16𝑘 + 3)(16𝑘′ + 5)𝑒4 and

2)𝑁2 = (16𝑘 + 3)𝑀4 + (16𝑘′ + 5)𝑒4.

Cutting down on 2) by 16 educes the congruence 0 , 1 , 4 , 9 ≡ 𝑁2 ≡ 3𝑀4 +5𝑒4 ≡ 8, 3, 5(𝑚𝑜𝑑 16). Due to unmatched numeration we obtain a contradiction. Thereby, we have that #α(Γ) = 2. In the next step, the curve 𝐸𝑝𝑞̅̅ ̅̅ ̅ is 𝑦

2 = 𝑥3 − 4(16𝑘 + 3)(16𝑘′ + 5)𝑥.

Whence, there exist relating equations for Γ̅:

1)𝑁2 = 𝑀4 − 4(16𝑘 + 3)(16𝑘′ + 5)𝑒4 and

2)𝑁2 = −𝑀4 + 4(16𝑘 + 3)(16𝑘′ + 5)𝑒4 and

3)𝑁2 = 2𝑀4 − 2(16𝑘 + 3)(16𝑘′ + 5)𝑒4 and

4)𝑁2 = −2𝑀4 + 2(16𝑘 + 3)(16𝑘′ + 5)𝑒4 and

5)𝑁2 = 4𝑀4 − (16𝑘 + 3)(16𝑘′ + 5)𝑒4 and

Enumeration in ranks of various elliptic curves 143

6)𝑁2 = −4𝑀4 + (16𝑘 + 3)(16𝑘′ + 5)𝑒4 and

7)𝑁2 = (16𝑘 + 3)𝑀4 − 4(16𝑘′ + 5)𝑒4 and

8)𝑁2 = −(16𝑘 + 3)𝑀4 + 4(16𝑘′ + 5)𝑒4 and

9)𝑁2 = 2(16𝑘 + 3)𝑀4 − 2(16𝑘′ + 5)𝑒4 and

10)𝑁2 = −2(16𝑘 + 3)𝑀4 + 2(16𝑘′ + 5)𝑒4 and

11)𝑁2 = 4(16𝑘 + 3)𝑀4 − (16𝑘′ + 5)𝑒4 and

12)𝑁2 = −4(16𝑘 + 3)𝑀4 + (16𝑘′ + 5)𝑒4.

Doing a reduction modulo 𝑝 in 2) deduces that 𝑁2 ≡ −𝑀4(𝑚𝑜𝑑 𝑝) but there

also derived (−𝑀4

𝑝) = −1. These two things cannot exist simultaneously and so

we attain a contradiction.

Treating a reduction modulo prime 𝑞 in equations 3) and 4) implies that 𝑁2 ≡

2𝑀4(𝑚𝑜𝑑 𝑞) and 𝑁2 ≡ −2𝑀4(𝑚𝑜𝑑 𝑞) but we also obtain that (2𝑀4

𝑞) =

(−2𝑀4

𝑞) = − 1. Pairing these with previous results in each case educes a

contradiction.

Cut down on equation 6) by 4 shows that 1 ≡ 𝑁2 ≡ 15𝑒4 ≡ 3(𝑚𝑜𝑑 4) and two sides are unmatched. Hence, 6) cannot have a solution.

After cutting down on relating equation 7) by 16 there comes 1, 9 ≡ 𝑁2 ≡ 3 +12𝑒4 ≡ 15, 3(𝑚𝑜𝑑 16) and this is unmatched congruence. Henceforth, we gain a contradiction.

In the next step, if 32 is used in taking a reduction modulo in 9) then, there is

attained that 0, 4, 16 ≡ 𝑁2 ≡ 6𝑀4 + 22𝑒4 ≡ 28(𝑚𝑜𝑑 32) and both sides do not match and so there deduced a contradiction.

No solution exists in equation 11) since reducing this by 4 derives that 1 ≡ 𝑁2 ≡−5𝑒4 ≡ 3(𝑚𝑜𝑑 4) and 𝐿𝐻𝑆, 𝑅𝐻𝑆 are unmatched. In equation 10) replacing 1 into both 𝑀 and 𝑒 gives that

−2(𝑢4 + 72𝑢2𝑣2 + 10𝑣4) + 2(𝑢4 + 74𝑢2𝑣2 + 10𝑣4)

= −144𝑢2𝑣2 + 148𝑢2𝑣2 = 4𝑢2𝑣2.

Therefore, the triple (1, 1, 2𝑢𝑣) is gotten as the solution of equation 10). We appoint that there exists a solution in 8) then, the algebraic structure

−(16𝑘 + 3) ∙ (−2(16𝑘 + 3)) ≡ 2 ∈ �̅�(Γ̅)(𝑚𝑜𝑑 𝑄×2) is deduced but as we

examined in the previous, relating equation 3)𝑁2 = 2𝑀4 − 2(16𝑘 + 3)(16𝑘′ +5)𝑒4 cannot take a solution and thus we acquire a contradiction. Moreover, this is

144 Shin-Wook Kim

also applied to equation 12). In conclusion, none of equations 8) and 12) possess a solution.

Eventually, we reach the conclusion #�̅�(Γ̅) = 4. Consequentially, the result 𝑟𝑎𝑛𝑘(𝐸(𝑢4+72𝑢2𝑣2+10𝑣4)(𝑢4+74𝑢2𝑣2+10𝑣4)(𝑄)) = 1 is

given from 𝑟2.4. (2). From (1) in above, it is sufficient that we only search the solution of

equation

10)𝑁2 = −2(𝑢4 − 10𝑢2𝑣2 + 12𝑣4)𝑀4 + 2(𝑢4 − 10𝑢2𝑣2 + 14𝑣4)𝑒4 for Γ̅.

Taking 𝑀 and 𝑒 as 1 shows that

−2(𝑢4 − 10𝑢2𝑣2 + 12𝑣4) + 2(𝑢4 − 10𝑢2𝑣2 + 14𝑣4)

= −24𝑣4 + 28𝑣4 = 4𝑣4.

For this reason, the triple (1, 1, 2𝑣2) is derived as the solution of 10). Wherefore, we take that #�̅�(Γ̅) = 4. Consequently, there is acquired 𝑟𝑎𝑛𝑘(𝐸(𝑢4−10𝑢2𝑣2+12𝑣4)(𝑢4−10𝑢2𝑣2+14𝑣4)(𝑄)) =

1 because of 𝑟2.4. (3). On account of (1) in the above, it is enough that we only examine the

solvability of next equation

10)𝑁2 = −2(𝑢4 − 2𝑢2𝑣2 + 4𝑣4)𝑀4 + 2(𝑢4 − 2𝑢2𝑣2 + 6𝑣4)𝑒4 for Γ̅.

Put 1 into both 𝑀 and 𝑒 gives that

−2(𝑢4 − 2𝑢2𝑣2 + 4𝑣4) + 2(𝑢4 − 2𝑢2𝑣2 + 6𝑣4)

= −8𝑣4 + 12𝑣4 = 4𝑣4.

Whence, the solution of 10) is gotten as (1, 1, 2𝑣2). And so we confront to #�̅�(Γ̅) = 4. As a result, we have that 𝑟𝑎𝑛𝑘(𝐸(𝑢4−2𝑢2𝑣2+4𝑣4)(𝑢4−2𝑢2𝑣2+6𝑣4)(𝑄)) = 1 from

𝑟2.4. (4). Because of (1) in above, it is remained only equation

10)𝑁2 = −2(𝑢4 − 4𝑢2𝑣2 + 6𝑣4)𝑀4 + 2(𝑢4 − 4𝑢2𝑣2 + 8𝑣4)𝑒4 for Γ̅

that is necessary to consider the solvability.

Replace 1 into 𝑀 and 𝑒 educes that

−2(𝑢4 − 4𝑢2𝑣2 + 6𝑣4) + 2(𝑢4 − 4𝑢2𝑣2 + 8𝑣4)

Enumeration in ranks of various elliptic curves 145

= −12𝑣4 + 16𝑣4 = 4𝑣4.

Thus, the solution of 10) is given as (1, 1, 2𝑣2). Wherefore, it is derived that #�̅�(Γ̅) = 4. For that reason, there is educed 𝑟𝑎𝑛𝑘(𝐸(𝑢4−4𝑢2𝑣2+6𝑣4)(𝑢4−4𝑢2𝑣2+8𝑣4)(𝑄)) = 1

since we get 𝑟2.4. □

In above curve 𝐸𝑝𝑞 the ranks are all 1 and primes 𝑝 and 𝑞 are the forms 𝑝 ≡

3 (𝑚𝑜𝑑 16) and 𝑞 ≡ 5 (𝑚𝑜𝑑 16). In other forms of primes 𝑝 and 𝑞, there also deduced the rank of 𝐸𝑝𝑞 as 1.

Remark 2.2. The form 𝐸𝑝𝑞 is considerable curve in elliptic curve. Maximal rank in this curve is 4 and it is not big value in generally but in systematized rank in

elliptic curve, it is not a small rank. According to conditions of 𝑝 and 𝑞 the consequence of generalized rank is decided. The bigger in rank, more valuable to

notice.

Remark 2.3. The result (1) in the above, the solvability of equation 7)𝑁2 =(16𝑘 + 3)𝑀4 − 4(16𝑘′ + 5)𝑒4 for Γ̅ can also be shown by 𝐿𝑆𝑉 and 𝐿𝐷𝑉. If we cut down on 7) by 𝑞 and 𝑝 then, we are confronted with 𝑁2 ≡ (16𝑘 +3)𝑀4(𝑚𝑜𝑑 𝑞) , 𝑁2 ≡ −4(16𝑘′ + 5)𝑒4(𝑚𝑜𝑑 𝑝) and so it should be derived that

1 = ((16𝑘+3)𝑀4

𝑞) = (

𝑝

𝑞) and 1= (

−4(16𝑘′+5)𝑒4

𝑝) = − (

𝑞

𝑝). Whence, we gain 𝐿𝐷𝑉

but between 𝑝 and 𝑞 it must be gotten 𝐿𝑆𝑉 and so there is appeared a contradiction.

Remark 2.4. In 𝐸−𝑝 if prime 𝑝 is 𝑝 ≡ 5 (𝑚𝑜𝑑 16) then, there were induced many

cases whose rank is 1 and in curve 𝐸−2𝑝, if primes are 𝑝 ≡ 5,11,13 (𝑚𝑜𝑑 16) then, we can search many cases that rank is 1(generalized).

3 Curves 𝑬−𝟐𝒑

In third section, we will survey the rank of curve 𝐸−2𝑝: 𝑦

2 = 𝑥3 − 2𝑝𝑥.

Lemma 3.1. (1). If a hypothesis is given that prime 𝑝 is 16603𝑢4 +3068𝑢2𝑣2 + 676𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 11(𝑚𝑜𝑑 16) in 𝐸−2𝑝 then, there comes

that 𝑟𝑎𝑛𝑘(𝐸−2(16603𝑢4+3068𝑢2𝑣2+676𝑣4)(𝑄)) = 1.

(2). Suppose that prime 𝑝 is 421𝑢4 + 464𝑢2𝑣2 + 128𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡5(𝑚𝑜𝑑 16) in 𝐸−2𝑝 then, the result 𝑟𝑎𝑛𝑘(𝐸−2(421𝑢4+464𝑢2𝑣2+128𝑣4)(𝑄)) = 1 is

gotten.

146 Shin-Wook Kim

(3). We appoint that prime 𝑝 is 405𝑢4 + 216𝑢2𝑣2 + 32𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡13(𝑚𝑜𝑑 16) in 𝐸−2𝑝 then, we get 𝑟𝑎𝑛𝑘(𝐸−2(405𝑢4+216𝑢2𝑣2+32𝑣4)(𝑄)) = 1.

(4). Take prime 𝑝 is 313𝑢4 + 100𝑢2𝑣2 + 8𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 5(𝑚𝑜𝑑 16) in 𝐸−2𝑝 then, the result 𝑟𝑎𝑛𝑘(𝐸−2(313𝑢4+100𝑢2𝑣2+8𝑣4)(𝑄)) = 1 is gotten.

Proo𝑓. (1). From [9], the remanent relating equation which should be examined the solvability is

4)𝑁2 = −2𝑀4 + (16603𝑢4 + 3068𝑢2𝑣2 + 676𝑣4)𝑒4 for Γ.

Now 676 is a square and 3068 is factored as 2 ∙ 1534 = 2 ∙ 26 ∙ 59. If we consider the relation of the term for 𝑢4 and 𝑢2𝑣2, there must be emerged

the term 592𝑢4. If we take a computation

16603𝑢4 − 592𝑢4 then,

it is 13122𝑢4. Therefore, in 2𝑀4 = 13122𝑢4 the value 𝑀 is educed as 9𝑢. Henceforth, the pair (𝑒, 𝑀)=(1, 9𝑢) is induced as a part of solution. And from

−13122𝑢4 + 16603𝑢4 + 3068𝑢2𝑣2 + 676𝑣4

= 3481𝑢4 + 3068𝑢2𝑣2 + 676𝑣4

the integer 𝑁 is deduced as 59𝑢2 + 26𝑣2. Whence, the triple (9𝑢, 1, 59𝑢2 + 26𝑣2) is produced as the solution of equation

4). Accordingly, we reach that #α(Γ) = 4. Resultantly, 𝑟𝑎𝑛𝑘(𝐸−2(16603𝑢4+3068𝑢2𝑣2+676𝑣4)(𝑄)) = 1 is induced due to 𝑟4.2.

(2). Due to [13], we only have to look into the solvability of equation

2)𝑁2 = −𝑀4 + 2(421𝑢4 + 464𝑢2𝑣2 + 128𝑣4)𝑒4 for Γ.

Substitute 𝑢 and 1 into 𝑀 and 𝑒 derives that

−𝑢4 + 2(421𝑢4 + 464𝑢2𝑣2 + 128𝑣4)

= 841𝑢4 + 2 ∙ 464𝑢2𝑣2 + 256𝑣4.

Thereby, we attain the integer 𝑁 as 29𝑢2 + 16𝑣2. Thus, the solution of 2) is produced as (𝑢, 1, 29𝑢2 + 16𝑣2).

Enumeration in ranks of various elliptic curves 147

Hence, educed conclusion is #α(Γ) = 4. On this account, we arrive at 𝑟4.2. For this reason, we obtain that 𝑟𝑎𝑛𝑘(𝐸−2(421𝑢4+464𝑢2𝑣2+128𝑣4)(𝑄)) = 1.

(3). The only equation which requires to check the solvability is

2)𝑁2 = −𝑀4 + 2(405𝑢4 + 216𝑢2𝑣2 + 32𝑣4)𝑒4 for Γ

from [13].

Choose 𝑀 and 𝑒 as 3𝑢 and 1 then, deduced calculation is

−81𝑢4 + 2 ∙ 405𝑢4 + 432𝑢2𝑣2 + 64𝑣4

= 729𝑢4 + 432𝑢2𝑣2 + 64𝑣4.

Hence, the value 𝑁 is gotten as 27𝑢2 + 8𝑣2. Eventually, we obtain the solution of 2) as (3𝑢, 1, 27𝑢2 + 8𝑣2). Consequently, we are faced with conclusion #α(Γ) = 4. Therefore, there is educed 𝑟4.2. Now we confront to 𝑟𝑎𝑛𝑘(𝐸−2(405𝑢4+216𝑢2𝑣2+32𝑣4)(𝑄)) = 1.

(4). We only find the solution of equation

2)𝑁2 = −𝑀4 + 2(313𝑢4 + 100𝑢2𝑣2 + 8𝑣4)𝑒4 for Γ

due to [13].

Let 𝑀 and 𝑒 be 𝑢 and 1 then, derived numeration is

−𝑢4 + 626𝑢4 + 200𝑢2𝑣2 + 16𝑣4

= 625𝑢4 + 200𝑢2𝑣2 + 16𝑣4.

Thereby, we gain the integer 𝑁 as 25𝑢2 + 4𝑣2. Resultantly, the triple (𝑢, 1, 25𝑢2 + 4𝑣2) is given as the solution of equation 2). On that account, it is induced that #α(Γ) = 4. Eventually, we acquire the result 𝑟𝑎𝑛𝑘(𝐸−2(313𝑢4+100𝑢2𝑣2+8𝑣4)(𝑄)) = 1 from

𝑟4.2. □

Remark 3.2. Compared with other forms in elliptic curves 𝑦2 = 𝑥3 ± 𝐴𝑥 in form 𝐸−2𝑝 we can search easily the case of generalized rank 1. It is matter of enumeration. Above results are also that ranks are all 1.

Remark 3.3. If prime 𝑝 is 𝑝 ≡ 1(𝑚𝑜𝑑 8) in 𝐸−2𝑝 then, rank can be deduced as 3. But as the other case of finding maximal rank in some form of curve, in this curve

searching the maximal rank is also difficult relatively.

148 Shin-Wook Kim

4 In 𝑬±𝒑𝒒 with Correlated to Rank 2

In fourth section, we will treat the ranks of curves 𝐸±𝑝𝑞: 𝑦

2 = 𝑥3 ± 𝑝𝑞𝑥 . In next

proof of theorem, we will not manage the solvability of relating equation 1) for Γ and that of 1) and 5) for Γ̅(in 𝐸𝑝𝑞 ) and 1) for Γ and 1) and 3) for Γ̅(in 𝐸−𝑝𝑞 ).

Refer [13] for this. 𝐿𝑆𝑉, 𝐿𝐷𝑉 are in [10].

Theorem 4.1. (1). If 𝐸𝑝𝑞 is assigned as an elliptic curve 𝑦2 = 𝑥3 + 𝑝𝑞𝑥 where

𝑝 and 𝑞 are different odd primes 𝑝 = 2𝑢4 − 4𝑢2𝑣2 + 𝑣4 and 𝑞 = 2𝑢4 − 4𝑢2𝑣2 +3𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 1(𝑚𝑜𝑑 8) and 𝑞 ≡ 3(𝑚𝑜𝑑 16) then, we acquire that

𝑟𝑎𝑛𝑘(𝐸(2𝑢4−4𝑢2𝑣2+𝑣4)(2𝑢4−4𝑢2𝑣2+3𝑣4)(𝑄)) ≥

𝑟𝑎𝑛𝑘(𝐸(𝑢4+72𝑢2𝑣2+10𝑣4)(𝑢4+74𝑢2𝑣2+10𝑣4)(𝑄))

+ 𝑟𝑎𝑛𝑘(𝐸−2(16603𝑢4+3068𝑢2𝑣2+676𝑣4)(𝑄)).

(2). Denote 𝑝 and 𝑞 as two different odd primes 𝑝 = 4𝑢4 + 20𝑢2𝑣2 + 27𝑣4 and 𝑞 = 4𝑢4 + 20𝑢2𝑣2 + 23𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1, 𝑝 ≡ 3(𝑚𝑜𝑑 16) and 𝑞 ≡ 15(𝑚𝑜𝑑 16) in curve 𝐸−𝑝𝑞 then, derived consequence is

𝑟𝑎𝑛𝑘(𝐸−(4𝑢4+20𝑢2𝑣2+27𝑣4)(4𝑢4+20𝑢2𝑣2+23𝑣4)(𝑄)) =

𝑟𝑎𝑛𝑘(𝐸(𝑢4−10𝑢2𝑣2+12𝑣4)(𝑢4−10𝑢2𝑣2+14𝑣4)(𝑄))

+𝑟𝑎𝑛𝑘(𝐸−2(421𝑢4+464𝑢2𝑣2+128𝑣4)(𝑄)).

(3). Let 𝑝 and 𝑞 be two different odd primes 𝑝 = 36𝑢4 + 36𝑢2𝑣2 + 11𝑣4 and 𝑞 = 36𝑢4 + 36𝑢2𝑣2 + 7𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1, 𝑝 ≡ 11(𝑚𝑜𝑑 16) and 𝑞 ≡ 7(𝑚𝑜𝑑 16) in elliptic curve 𝐸−𝑝𝑞 then, we obtain the result

𝑟𝑎𝑛𝑘(𝐸−(36𝑢4+36𝑢2𝑣2+11𝑣4)(36𝑢4+36𝑢2𝑣2+7𝑣4)(𝑄)) =

𝑟𝑎𝑛𝑘(𝐸(𝑢4−2𝑢2𝑣2+4𝑣4)(𝑢4−2𝑢2𝑣2+6𝑣4)(𝑄))

+𝑟𝑎𝑛𝑘(𝐸−2(405𝑢4+216𝑢2𝑣2+32𝑣4)(𝑄)).

(4). Define 𝐸−𝑝𝑞 as an elliptic curve 𝑦2 = 𝑥3 − 𝑝𝑞𝑥 where two distinct odd

primes 𝑝 and 𝑞 are 𝑝 ≡ 7(𝑚𝑜𝑑 16) and 𝑞 ≡ 7(𝑚𝑜𝑑 16) as −𝑝 + 𝑞 = 𝑡2 with integer 𝑡 and 2𝑢4 + 2𝑝𝑞 = 𝑣2 with integers 𝑢 and 𝑣 then, we gain the result

Enumeration in ranks of various elliptic curves 149

𝑟𝑎𝑛𝑘(𝐸−(−𝑝+𝑞,2𝑢4+2𝑝𝑞)(𝑄)) =

𝑟𝑎𝑛𝑘(𝐸(𝑢4−4𝑢2𝑣2+6𝑣4)(𝑢4−4𝑢2𝑣2+8𝑣4)(𝑄))

+𝑟𝑎𝑛𝑘(𝐸−2(313𝑢4+100𝑢2𝑣2+8𝑣4)(𝑄)).

Proo𝑓. (1). Denote 𝐸𝑝𝑞 as an elliptic curve 𝑦2 = 𝑥3 + 𝑝𝑞𝑥 where two different

odd primes 𝑝 and 𝑞 are gotten as 𝑝 = 2𝑢4 − 4𝑢2𝑣2 + 𝑣4 and 𝑞 = 2𝑢4 −4𝑢2𝑣2 + 3𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 1(𝑚𝑜𝑑 8) and 𝑞 ≡ 3(𝑚𝑜𝑑 16) . Suppose that 𝑝 = 8𝑘 + 1 and 𝑞 = 16𝑘′ + 3 with integers 𝑘 and 𝑘′ then, there are two relating equations for Γ as follows:

1)𝑁2 = 𝑀4 + (8𝑘 + 1)(16𝑘′ + 3)𝑒4 and

2)𝑁2 = (8𝑘 + 1)𝑀4 + (16𝑘′ + 3)𝑒4.

Equation 2) is 𝑁2 = (2𝑢4 − 4𝑢2𝑣2 + 𝑣4)𝑀4 + (2𝑢4 − 4𝑢2𝑣2 + 3𝑣4)𝑒4 . For searching the solution of this, it is necessary to consider two coefficients of 𝑀4 and 𝑒4.

The common numerical value 2𝑢4 − 4𝑢2𝑣2 exists in both coefficients of 𝑀4 and 𝑒4.

If 4𝑢4 − 8𝑢2𝑣2 is supposed as the part of resultant then, there should be shown 4𝑣4 after substituting the values into 𝑀 and 𝑒.

We can take it from the terms 𝑣4 and 3𝑣4. Moreover, there comes

2𝑢4 − 4𝑢2𝑣2 + 𝑣4 + 2𝑢4 − 4𝑢2𝑣2 + 3𝑣4

= 4𝑢4 − 8𝑢2𝑣2 + 4𝑣4

and so the integer 𝑁 is educed as 2𝑢2 − 2𝑣2. Therefore, the triple (1, 1, 2𝑢2 − 2𝑣2) is deduced as the solution of equation 2). For this reason, there is attained #α(Γ) = 4. Next, the curve 𝐸𝑝𝑞̅̅ ̅̅ ̅ is 𝑦

2 = 𝑥3 − 4(8𝑘 + 1)(16𝑘′ + 3)𝑥 from 𝐸𝑝𝑞 in the above.

And so we attain following relating equations for Γ̅:

1)𝑁2 = 𝑀4 − 4(8𝑘 + 1)(16𝑘′ + 3)𝑒4 2)𝑁2 = −𝑀4 + 4(8𝑘 + 1)(16𝑘′ + 3)𝑒4

3)𝑁2 = 2𝑀4 − 2(8𝑘 + 1)(16𝑘′ + 3)𝑒4 4)𝑁2 = −2𝑀4 + 2(8𝑘 + 1)(16𝑘′ + 3)𝑒4

5)𝑁2 = 4𝑀4 − (8𝑘 + 1)(16𝑘′ + 3)𝑒4 6)𝑁2 = −4𝑀4 + (8𝑘 + 1)(16𝑘′ + 3)𝑒4

7)𝑁2 = (8𝑘 + 1)𝑀4 − 4(16𝑘′ + 3)𝑒4

150 Shin-Wook Kim

8)𝑁2 = −(8𝑘 + 1)𝑀4 + 4(16𝑘′ + 3)𝑒4 9)𝑁2 = 2(8𝑘 + 1)𝑀4 − 2(16𝑘′ + 3)𝑒4

10)𝑁2 = −2(8𝑘 + 1)𝑀4 + 2(16𝑘′ + 3)𝑒4 11)𝑁2 = 4(8𝑘 + 1)𝑀4 − (16𝑘′ + 3)𝑒4

12)𝑁2 = −4(8𝑘 + 1)𝑀4 + (16𝑘′ + 3)𝑒4

If we treat a reduction modulo 𝑞 in relating equation 2) then, we obtain 𝑁2 ≡

−𝑀4(𝑚𝑜𝑑 𝑞) but we also attain that (−𝑀4

𝑞) = −1. And it is impossible that these

two facts coexist, thereby no solution exists in this equation.

After considering reduction modulo 16 in equation 3) then, we attain that 0, 4 ≡𝑁2 ≡ 2𝑀4 + 10𝑒4 ≡ 12(𝑚𝑜𝑑 16) and the sides 𝑅𝐻𝑆, 𝐿𝐻𝑆 are unmatched in this congruence, hence it cannot have a solution.

Reducing 6) by 8 implies that 1 ≡ 𝑁2 ≡ 4𝑀4 + 3 ≡ 7, 3(𝑚𝑜𝑑 8) and two sides do not match in this congruence. Accordingly, no solution exists in equation 6).

Cutting down on equation 8) by 4 gives that 1 ≡ 𝑁2 ≡ −𝑀4 ≡ 3(𝑚𝑜𝑑 4). Thus, a contradiction is gotten, hence this equation cannot have a solution.

If there exists a solution in 9) then, derived congruences are 𝑁2 ≡ −2(16𝑘′ +3)𝑒4(𝑚𝑜𝑑 𝑝) and 𝑁2 ≡ 2(8𝑘 + 1)𝑀4(𝑚𝑜𝑑 𝑞) from cutting down on it by 𝑝 and 𝑞, thus there must be shown that

1 = (−2(16𝑘′+3)𝑒4

𝑝) = (

𝑞

𝑝) and

1 = (2(8𝑘+1)𝑀4

𝑞) = − (

𝑝

𝑞).

𝐿𝐷𝑉 is deduced but there educed 𝐿𝑆𝑉 between 𝑝 and 𝑞 . Thus, we take a contradiction and so 9) cannot have a solution. The triple (1, 1, 2𝑣2) is educed as the solution of equation 10). There is no solution in 12) because reducing it by 8 shows that 1 ≡ 𝑁2 ≡

4𝑀4 + 3 ≡ 7, 3(𝑚𝑜𝑑 8) and the sides 𝐿𝐻𝑆 and 𝑅𝐻𝑆 do not match. Whence, it cannot take a solution.

On that account, there is educed #�̅�(Γ̅) ≥ 4. Thus, we reach the consequence 𝑟𝑎𝑛𝑘(𝐸(2𝑢4−4𝑢2𝑣2+𝑣4)(2𝑢4−4𝑢2𝑣2+3𝑣4)(𝑄)) ≥ 2

on account of 𝑟≥4.4. Now due to proposition 2.1(1) and lemma 3.1(1) we complete the proof of (1).

(2). Define 𝐸−𝑝𝑞 as an elliptic curve 𝑦2 = 𝑥3 − 𝑝𝑞𝑥 with different odd primes

𝑝 and 𝑞 as 𝑝 = 4𝑢4 + 20𝑢2𝑣2 + 27𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1and 𝑝 ≡ 3(𝑚𝑜𝑑 16) and 𝑞 =4𝑢4 + 20𝑢2𝑣2 + 23𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑞 ≡ 15(𝑚𝑜𝑑 16) . From [10], there are remained two relating equations 3)𝑁2 = 𝑝𝑀4 − 𝑞𝑒4 for Γ and 5)𝑁2 = 2𝑝𝑀4 +2𝑞𝑒4 for Γ̅ that is needed to inspect the solvability. First equation in 3) replacing 1 into both 𝑀 and 𝑒 derives that 27𝑣4 − 23𝑣4 = 4𝑣4 and thus we attain a solution of it as (1, 1, 2𝑣2) . Thereby, deduced conclusion is #α(Γ) = 4. Second

Enumeration in ranks of various elliptic curves 151

equation in 5), there exists common arithmetical value 8𝑢4 + 40𝑢2𝑣2 in both coefficients of 𝑀4 and 𝑒4, hence we can acquire 16𝑢4. Henceforth, a probability for being shown the square of polynomial whose components are two variables 𝑢 and 𝑣 exists. If we take a supposition that 80𝑢2𝑣2 is another component of resultant then, there ought to be induced the term 100𝑣4. From the terms 54𝑣4 and 46𝑣4 in coefficients of 𝑀4 and 𝑒4 respectively we can obtain this. Furthermore, owing to the numeration 8𝑢4 + 40𝑢2𝑣2 + 54𝑣4 + 8𝑢4 + 40𝑢2𝑣2 +46𝑣4 the integer 𝑁 is given as 4𝑢2 + 10𝑣2. Thus, the triple (1, 1, 4𝑢2 +10𝑣2) is educed as the solution of relating equation 5). On that account, there derived that #�̅�(Γ̅) = 4 . For this reason, 𝑟𝑎𝑛𝑘(𝐸−(4𝑢4+20𝑢2𝑣2+27𝑣4)(4𝑢4+20𝑢2𝑣2+23𝑣4)(𝑄)) =

2 is given due to 𝑟4.4. Now owing to proposition 2.1(2) and lemma 3.1(2) we accomplish the proof of (2).

(3). Take 𝐸−𝑝𝑞 as an elliptic curve 𝑦2 = 𝑥3 − 𝑝𝑞𝑥 that satisfies different odd

primes 𝑝 and 𝑞 are the forms 𝑝 = 36𝑢4 + 36𝑢2𝑣2 + 11𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡11(𝑚𝑜𝑑 16) , 𝑞 = 36𝑢4 + 36𝑢2𝑣2 + 7𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑞 ≡ 7(𝑚𝑜𝑑 16) . Put 𝑝 and 𝑞 as 𝑝 = 16𝑘 + 11 and 𝑞 = 16𝑘′ + 7 with integers 𝑘 and 𝑘′ then, educed relating equations for Γ are the followings:

1)𝑁2 = 𝑀4 − (16𝑘 + 11)(16𝑘′ + 7)𝑒4 and

2)𝑁2 = −𝑀4 + (16𝑘 + 11)(16𝑘′ + 7)𝑒4 and

3)𝑁2 = (16𝑘 + 11)𝑀4 − (16𝑘′ + 7)𝑒4 and

4)𝑁2 = −(16𝑘 + 11)𝑀4 + (16𝑘′ + 7)𝑒4.

Taking a reduction modulo 16 in equation 2) then, we attain the congruence 0, 1, 4,9 ≡ 𝑁2 ≡ 15𝑀4 + 13𝑒4 ≡ 12, 15, 13(𝑚𝑜𝑑 16) and two sides 𝑅𝐻𝑆, 𝐿𝐻𝑆 do not match, therefore this equation has no solution.

Cutting down on equation 4) by 16 then, induced relation is 0, 1, 4, 9 ≡ 𝑁2 ≡5𝑀4 + 7𝑒4 ≡ 12, 7, 5(𝑚𝑜𝑑 16). It is unmatched in both sides and so we cannot expect an appearance of solution in 4).

Equation 3) possess a solution (1, 1, 2𝑣2) from the enumeration 11𝑣4 − 7𝑣4 =4𝑣4.

As a result, we acquire that #α(Γ) = 4. Now, the curve 𝐸−𝑝𝑞̅̅ ̅̅ ̅̅ is 𝑦

2 = 𝑥3 + 4(16𝑘 + 11)(16𝑘′ + 7)𝑥 from 𝐸−𝑝𝑞.

Thereby, we obtain relating equations for Γ̅ as follows:

1)𝑁2 = 𝑀4 + 4(16𝑘 + 11)(16𝑘′ + 7)𝑒4 and

2)𝑁2 = 2𝑀4 + 2(16𝑘 + 11)(16𝑘′ + 7)𝑒4 and

3)𝑁2 = 4𝑀4 + (16𝑘 + 11)(16𝑘′ + 7)𝑒4 and

152 Shin-Wook Kim

4)𝑁2 = (16𝑘 + 11)𝑀4 + 4(16𝑘′ + 7)𝑒4 and

5)𝑁2 = 2(16𝑘 + 11)𝑀4 + 2(16𝑘′ + 7)𝑒4 and

6)𝑁2 = 4(16𝑘 + 11)𝑀4 + (16𝑘′ + 7)𝑒4.

After cutting down on equation 2) by 𝑝 there derived that 𝑁2 ≡ 2𝑀4(𝑚𝑜𝑑 𝑝)

but we also get (2𝑀4

𝑝) = −1. Therefore, a contradiction is gotten.

The equations 4) and 6) cannot possess a solution since 4)1 ≡ 𝑁2 ≡ 11𝑀4 ≡3(𝑚𝑜𝑑 4) and 6)1 ≡ 𝑁2 ≡ 7𝑒4 ≡ 3(𝑚𝑜𝑑 4) are educed after reducing it by 4.

Relating equation 5) is rewritten as 𝑁2 = 2(36𝑢4 + 36𝑢2𝑣2 + 11𝑣4)𝑀4 +2(36𝑢4 + 36𝑢2𝑣2 + 7𝑣4)𝑒4. For finding the solution of this equation, treating the coefficients is indispensable. On account of existence the term 72𝑢4 in two coefficients of 𝑀4 and 𝑒4 we can expect that there can be deduced the polynomial’s square that is consisted of the variables 𝑢 and 𝑣 after choosing the values 𝑀 and 𝑒. If we suppose that the term 144𝑢2𝑣2 comprises another part of resultant then, there should be emerged the square 36𝑣4 . The bigness of the solution is meaningless in finding the solution of relating equation. Only one

triple is found which satisfies 5) then, it is enough to became the solution of equation 5) . We appoint that 𝑀 = 𝑒 = 1 then, there derived 2 ∙ 36𝑢4 + 2 ∙36𝑢2𝑣2 + 22𝑣4 + 2 ∙ 36𝑢4 + 2 ∙ 36𝑢2𝑣2 + 14𝑣4. For that reason, the integer 𝑁 is gotten as 12𝑢2 + 6𝑣2 . On this account, the triple ( 1 , 1 , 12𝑢2 + 6𝑣2 ) is produced as the solution of relating equation 5).

Eventually, we obtain #�̅�(Γ̅) = 4. Therefore, there educed 𝑟𝑎𝑛𝑘(𝐸−(36𝑢4+36𝑢2𝑣2+11𝑣4)(36𝑢4+36𝑢2𝑣2+7𝑣4)(𝑄)) = 2

owing to 𝑟4.4. In the next step, due to proposition 2.1(3) and lemma 3.1(3) the proof of (3) is

done.

(4). Denote 𝐸−𝑝𝑞 as an elliptic curve 𝑦2 = 𝑥3 − 𝑝𝑞𝑥 where two distinct odd

primes 𝑝 and 𝑞 are the forms 𝑝 ≡ 7(𝑚𝑜𝑑 16) and 𝑞 ≡ 7(𝑚𝑜𝑑 16) as −𝑝 + 𝑞 =𝑡2 with integer 𝑡 and 2𝑢4 + 2𝑝𝑞 = 𝑣2 with integers 𝑢 and 𝑣. Set 𝑝 and 𝑞 as 𝑝 =16𝑘 + 7 and 𝑞 = 16𝑘′ + 7 with integers 𝑘 and 𝑘′ then, we have relating equations for Γ as follows:

1)𝑁2 = 𝑀4 − (16𝑘 + 7)(16𝑘′ + 7)𝑒4 and

2)𝑁2 = −𝑀4 + (16𝑘 + 7)(16𝑘′ + 7)𝑒4 and

3)𝑁2 = (16𝑘 + 7)𝑀4 − (16𝑘′ + 7)𝑒4 and

4)𝑁2 = −(16𝑘 + 7)𝑀4 + (16𝑘′ + 7)𝑒4.

Enumeration in ranks of various elliptic curves 153

Equation 2) has no solution because cutting down on this by 𝑞 educes that 𝑁2 ≡

−𝑀4(𝑚𝑜𝑑 𝑞) but there is also gotten that (−𝑀4

𝑞) = −1 and these two results

cannot coexist.

From hypothesis the triple (1, 1, 𝑡) satisfies the solution of equation 4). Suppose that equation 3) possess a solution then, we confront to the structure

(16𝑘 + 7) ∙ (−(16𝑘 + 7)) ≡ −1 ∈ α(Γ)(𝑚𝑜𝑑 𝑄×2) but no solution exists in

equation 2)𝑁2 = −𝑀4 + (16𝑘 + 7)(16𝑘′ + 7)𝑒4 in the above, thus we obtain a contradiction.

Henceforth, we get #α(Γ) = 4. Next, the curve 𝐸−𝑝𝑞̅̅ ̅̅ ̅̅ is 𝑦

2 = 𝑥3 + 4(16𝑘 + 7)(16𝑘′ + 7)𝑥 due to 𝐸−𝑝𝑞.

Thus, educed relating equations for Γ̅ are the followings:

1)𝑁2 = 𝑀4 + 4(16𝑘 + 7)(16𝑘′ + 7)𝑒4 and

2)𝑁2 = 2𝑀4 + 2(16𝑘 + 7)(16𝑘′ + 7)𝑒4 and

3)𝑁2 = 4𝑀4 + (16𝑘 + 7)(16𝑘′ + 7)𝑒4 and

4)𝑁2 = (16𝑘 + 7)𝑀4 + 4(16𝑘′ + 7)𝑒4 and

5)𝑁2 = 2(16𝑘 + 7)𝑀4 + 2(16𝑘′ + 7)𝑒4 and

6)𝑁2 = 4(16𝑘 + 7)𝑀4 + (16𝑘′ + 7)𝑒4.

Equation 2) has a solution (𝑢, 1, 𝑣) from hypothesis. Cutting down on 4) and 6) by 4 shows that 4)1 ≡ 𝑁2 ≡ 7𝑀4 ≡ 3(𝑚𝑜𝑑 4) and

6)1 ≡ 𝑁2 ≡ 3(𝑚𝑜𝑑 4) and two sides are unmatched in both cases, thus neither equation takes a solution.

Reducing 5) by 32 gives that 0 , 4 , 16 ≡ 𝑁2 ≡ 14𝑀4 + 14𝑒4 ≡ 28(𝑚𝑜𝑑 32) and hence we attain a contradiction.

With regard to equations from 4) to 6) insolvability can also be shown by the method of 𝐿𝐷𝑉 and 𝐿𝑆𝑉. We appoint that there is a solution in equation 4) then, there are given 𝑁2 ≡

(16𝑘 + 7)𝑀4(𝑚𝑜𝑑 𝑞) , 𝑁2 ≡ 4(16𝑘′ + 7)𝑒4(𝑚𝑜𝑑 𝑝) from reducing a modulo 𝑞 and 𝑝 in it. Henceforth, there ought to be educed that

1 = ((16𝑘+7)𝑀4

𝑞) = (

𝑝

𝑞) and

1 = (4(16𝑘′+7)𝑒4

𝑝) = (

𝑞

𝑝).

There comes 𝐿𝑆𝑉 in the above but it is given 𝐿𝐷𝑉 between 𝑝 and 𝑞. Whence, we have a contradiction and it makes it impossible to find a solution in equation 4).

154 Shin-Wook Kim

If there is a solution in equation 5) then, we encounter to 𝑁2 ≡ 2(16𝑘 +7)𝑀4(𝑚𝑜𝑑 𝑞) and 𝑁2 ≡ 2(16𝑘′ + 7)𝑒4(𝑚𝑜𝑑 𝑝) by cutting down on it modulo 𝑞 and 𝑝. Thereby, there must be shown that

1 = (2(16𝑘+7)𝑀4

𝑞) = (

𝑝

𝑞) and

1 = (2(16𝑘′+7)𝑒4

𝑝) = (

𝑞

𝑝).

𝐿𝑆𝑉 is produced in the above but we acquire 𝐿𝐷𝑉 between the primes 𝑝 and 𝑞. Therefore, a contradiction is emerged and so no solution exists in 5). If a solution exists in equation 6) then, we attain the congruences 𝑁2 ≡

4(16𝑘 + 7)𝑀4(𝑚𝑜𝑑 𝑞) and 𝑁2 ≡ (16𝑘′ + 7)𝑒4(𝑚𝑜𝑑 𝑝) from reducing it by 𝑞 and 𝑝 and so we should gain

1= (4(16𝑘+7)𝑀4

𝑞) = (

𝑝

𝑞) and

1= ((16𝑘′+7)𝑒4

𝑝) = (

𝑞

𝑝)

It is given 𝐿𝑆𝑉 in the above but 𝐿𝐷𝑉 is appeared between primes 𝑝 and 𝑞 . Accordingly, we obtain a contradiction and thus we cannot look for the solution in

equation 6). Resultantly, there comes #�̅�(Γ̅) = 4. For that reason, the consequence 𝑟𝑎𝑛𝑘(𝐸−(−𝑝+𝑞,2𝑢4+2𝑝𝑞)(𝑄)) = 2 is derived on

account of 𝑟4.4. Moreover, from proposition 2.1(4) and lemma 3.1(4) the proof of (4) is

completed. □

Remark 4.2. In above result in (1), the rank is not decided. There is a probability

that it became 3. This is up to whether the equations 4) and 7) and 11) for Γ̅ possess a solution or not. The values �̅�(𝑃) in 7) and 11) are the same in 𝑄×/𝑄×2 and so only one of it is necessary to examine the solvability. Due to algebraic structure if only one equation of above three equations takes a solution

then, the rank became 3.

Remark 4.3. The forms of primes 𝑝 and 𝑞 in each case from (1) to (4) are all different. Except (1) the rank of other three cases are all determined as 2.

Particularly, in forms of primes 𝑝 and 𝑞 in (2) and (3) in curve 𝐸−𝑝𝑞 there are found often the value of rank 2 compared with other conditions.

Remark 4.4. The case of (4) the rank is also 2. In primes of the forms 𝑝 ≡7(𝑚𝑜𝑑 16) and 𝑞 ≡ 7(𝑚𝑜𝑑 16) in curve 𝐸−𝑝𝑞 searching the rank 2(systematized)

Enumeration in ranks of various elliptic curves 155

is more difficult than other forms.

Remark 4.5. It is just one value in difference but finding the generalized elliptic

curve 𝐸−𝑝𝑞 with rank 3 is more difficult than the rank 2(systematized). In theoretically, it is trivial but this is matter of calculation. In practical treatment,

considering the solvability of relating equation is core thing.

Remark 4.6. In above equations 3)𝑁2 = (16𝑘 + 7)𝑀4 − (16𝑘′ + 7)𝑒4 for Γ in (4) the method of 𝐿𝐷𝑉 and 𝐿𝑆𝑉 is of no use because after reducing this by 𝑞 and 𝑝 there derived the congruences as 𝑁2 ≡ (16𝑘 + 7)𝑀4(𝑚𝑜𝑑 𝑞) and 𝑁2 ≡

−(16𝑘′ + 7)𝑒4(𝑚𝑜𝑑 𝑝) and there also given that 1 = ((16𝑘+7)𝑀4

𝑞) = (

𝑝

𝑞) and 1=

(−(16𝑘′+7)𝑒4

𝑝) = − (

𝑞

𝑝). Thus, 𝐿𝐷𝑉 is educed and this is matched to the relation of

primes 𝑝 and 𝑞.

Remark 4.7. In elliptic curve, average rank is significant notation which is related

to distribution of ranks([15]). It is defined as lim𝑋→∞

∑ 𝑟𝑎𝑛𝑘(𝐸𝑑)𝑑∈𝑆(𝑋)

#(𝑆(𝑋)) ([15]). This rank

is another rank(different from Mordell-Weil rank) which is necessary to treat

meaning the rank(Mordell-Weil rank) in elliptic curve.

Remark 4.8. In [4], the author managed the curve whose rank is 4 in curve 𝐸𝑝𝑞. The Parity Conjecture was used in calculating the rank. Getting systematized

rank(bigger than 2) in this curve is not simple work. If two primes 𝑝 and 𝑞 are the forms 𝑝, 𝑞 ≡ 1(𝑚𝑜𝑑 8) then, there is a possibility that rank can become 4.

Remark 4.9. In curve 𝐸−𝑝𝑞𝑠 (𝑝 and 𝑞 and 𝑠 are different odd primes) if three

primes 𝑝 and 𝑞 and 𝑠 are the forms 𝑝, 𝑞, 𝑠 ≡ 1(𝑚𝑜𝑑 8) then, the maximal value of #α(Γ) and #�̅�(Γ̅) are both 16. Thus, the maximal rank in 𝐸−𝑝𝑞𝑠 is derived as 6.

Generalizing this value in 𝐸−𝑝𝑞𝑠 is not simple dispute. Finding the forms of three

primes 𝑝 and 𝑞 and 𝑠 that induces the rank 6 is complex matter without any conjecture or some condition.

5 Form 𝑬−𝒑

In fifth section, we shall treat the rank of 𝐸−𝑝: 𝑦

2 = 𝑥3 − 𝑝𝑥.

Corollary 5.1. (1). Assume that prime 𝑝 is 8101𝑢4 + 720𝑢2𝑣2 + 16𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 5(𝑚𝑜𝑑 16) in 𝐸−𝑝 then, we take that

𝑟𝑎𝑛𝑘(𝐸(2𝑢4−4𝑢2𝑣2+𝑣4)(2𝑢4−4𝑢2𝑣2+3𝑣4)(𝑄)) >

𝑟𝑎𝑛𝑘(𝐸−(8101𝑢4+720𝑢2𝑣2+16𝑣4)(𝑄)).

156 Shin-Wook Kim

(2). If prime 𝑝 is 5𝑢4 + 256𝑢2𝑣2 + 4096𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1and 𝑝 ≡ 5(𝑚𝑜𝑑 16) in curve 𝐸−𝑝 then, we attain that

𝑟𝑎𝑛𝑘(𝐸−(4𝑢4+20𝑢2𝑣2+27𝑣4)(4𝑢4+20𝑢2𝑣2+23𝑣4)(𝑄)) >

𝑟𝑎𝑛𝑘(𝐸−(5𝑢4+256𝑢2𝑣2+4096𝑣4)(𝑄)).

(3). Define prime 𝑝 is 5𝑢4 + 112𝑢2𝑣2 + 784𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡ 5(𝑚𝑜𝑑 16) in curve 𝐸−𝑝 then, we get that

𝑟𝑎𝑛𝑘(𝐸−(36𝑢4+36𝑢2𝑣2+11𝑣4)(36𝑢4+36𝑢2𝑣2+7𝑣4)(𝑄)) >

𝑟𝑎𝑛𝑘(𝐸−(5𝑢4+112𝑢2𝑣2+784𝑣4)(𝑄)).

(4). Take that prime 𝑝 is 117𝑢4 + 240𝑢2𝑣2 + 400𝑣4 𝑤. 𝑖. 𝑢. 𝑣. 1 and 𝑝 ≡5(𝑚𝑜𝑑 16) in 𝐸−𝑝 then, there comes

𝑟𝑎𝑛𝑘(𝐸−(−𝑝+𝑞,2𝑢4+2𝑝𝑞)(𝑄)) >

𝑟𝑎𝑛𝑘(𝐸−(117𝑢4+240𝑢2𝑣2+400𝑣4)(𝑄)).

Proo𝑓. (1). The only equation that requires to research the solvability is

2)𝑁2 = −𝑀4 + (8101𝑢4 + 720𝑢2𝑣2 + 16𝑣4)𝑒4 for Γ

on account of [12].

Set 𝑀 and 𝑒 as 𝑢 and 1 then, there induced that

−𝑢4 + (8101𝑢4 + 720𝑢2𝑣2 + 16𝑣4)

= 8100𝑢4 + 720𝑢2𝑣2 + 16𝑣4.

Resultantly, the value 𝑁 is gotten as 90𝑢2 + 4𝑣2. Wherefore, the triple (𝑢, 1, 90𝑢2 + 4𝑣2) is derived as the solution of 2). Thereby, it is obtained that #α(Γ) = 4. And so we encounter to 𝑟4.2. On that account, we attain that 𝑟𝑎𝑛𝑘(𝐸−(8101𝑢4+720𝑢2𝑣2+16𝑣4)(𝑄)) = 1.

Now from theorem 4.1(1) the proof of (1) is completed.

(2). The remanent equation which is necessary to check the solvability is

2)𝑁2 = −𝑀4 + (5𝑢4 + 256𝑢2𝑣2 + 4096𝑣4)𝑒4 for Γ

Enumeration in ranks of various elliptic curves 157

owing to [12].

Choose 𝑀 and 𝑒 as 𝑢 and 1 then, there comes the enumeration

−𝑢4 + (5𝑢4 + 256𝑢2𝑣2 + 4096𝑣4)

= 4𝑢4 + 256𝑢2𝑣2 + 4096𝑣4.

Whence, there induced 𝑁 = 2𝑢2 + 64𝑣2. As a result, the triple (𝑢, 1, 2𝑢2 + 64𝑣2) is attained as the solution of equation

2). Consequentially, we get the conclusion #α(Γ) = 4. Now there comes 𝑟4.2 and thus 𝑟𝑎𝑛𝑘(𝐸−(5𝑢4+256𝑢2𝑣2+4096𝑣4)(𝑄)) = 1 is gotten.

Moreover, by theorem 4.1(2) we accomplish the proof of (2).

(3). It is sufficient that we only examine the solvability of equation

2)𝑁2 = −𝑀4 + (5𝑢4 + 112𝑢2𝑣2 + 784𝑣4)𝑒4 for Γ

because of [12].

If 𝑀 and 𝑒 are selected as 𝑢 and 1 then, we are confronted with −𝑢4 + (5𝑢4 +112𝑢2𝑣2 + 784𝑣4) = 4𝑢4 + 112𝑢2𝑣2 + 784𝑣4 . Hence, we have 𝑁 as 2𝑢2 +28𝑣2 . Thereby, the triple (𝑢, 1, 2𝑢2 + 28𝑣2) satisfies the solution of relating equation 2). In conclusion, there is obtained #α(Γ) = 4. Next, from 𝑟4.2 we gain 𝑟𝑎𝑛𝑘(𝐸−(5𝑢4+112𝑢2𝑣2+784𝑣4)(𝑄)) = 1. In addition, on account of theorem 4.1(3)

the proof of (3) is finished.

(4). We only have to investigate the solvability of following equation

2)𝑁2 = −𝑀4 + (117𝑢4 + 240𝑢2𝑣2 + 400𝑣4)𝑒4 for Γ

owing to [12].

Suppose that 𝑀 and 𝑒 as 3𝑢 and 1 then, there is gotten the calculation

−(3𝑢)4 + (117𝑢4 + 240𝑢2𝑣2 + 400𝑣4)

= 36𝑢4 + 240𝑢2𝑣2 + 400𝑣4.

For this reason, it is produced that 𝑁 = 6𝑢2 + 20𝑣2. Therefore, we gain the triple (3𝑢, 1, 6𝑢2 + 20𝑣2 ) as the solution of 2). At last there is gotten #α(Γ) = 4. Next, because of 𝑟4.2 we conclude that 𝑟𝑎𝑛𝑘(𝐸−(117𝑢4+240𝑢2𝑣2+400𝑣4)(𝑄)) =

1.

Furthermore, by theorem 4.1(4) we complete the proof of (4). □

158 Shin-Wook Kim

Remark 5.2. In above, the prime 𝑝 is the case 𝑝 ≡ 5(𝑚𝑜𝑑 16) but we also attain the systematized rank 1 in 𝐸−𝑝 where prime 𝑝 is the form 𝑝 ≡ 7, 15(𝑚𝑜𝑑 16).

Not so much as the case of 𝑝 ≡ 5(𝑚𝑜𝑑 16) it were found generalized rank 1 in these forms.

Remark 5.3. In curve of 𝐸−𝑝 the maximal rank is 2 and for this the prime 𝑝 must

be 𝑝 ≡ 1(𝑚𝑜𝑑 8). But in addition to this condition 𝑝 also has to be satisfied some other form that induces the rank 2. Searching this particular form is difficult. It is

different from making the generalized rank 1.

6 Examples

In sixth section, the examples of previous results will be submitted. The prime

numbers can be examined in [1].

There are appeared examples of proposition 2.1(1):

The taken examples of proposition 2.1(2) are the followings:

There shown the examples of proposition 2.1(3):

Deduced examples of proposition 2.1(4) are the followings:

Induced examples of lemma 3.1(1) are given as:

(𝑝, 𝑞, 𝑢, 𝑣) (739, 757, 3, 1)

(5939, 6037, 7, 1)

(40739, 41077, 13, 1)

(𝑝, 𝑞, 𝑢, 𝑣) (3, 5, 1, 1)

(59107, 72229, 5, 9)

(𝑝, 𝑞, 𝑢, 𝑣) (3, 5, 1, 1)

(20707, 33829, 7, 9)

(𝑝, 𝑞, 𝑢, 𝑣) (211, 373, 5, 3)

(31891, 45013, 5, 9)

Enumeration in ranks of various elliptic curves 159

Examples of lemma 3.1(2) are emerged as follows:

(𝑝, 𝑢, 𝑣) (1013, 1, 1)

(1326821, 1, 10) (58807013, 1, 26)

There derived examples of lemma 3.1(3):

We have examples of lemma 3.1(4) as follows:

Some examples of theorem 4.1(1) are derived as follows:

There deduced examples of theorem 4.1(2):

(𝑝, 𝑢, 𝑣) (20347, 1, 1) (238747, 1, 4) (515803, 1, 5)

(𝑝, 𝑢, 𝑣) (653, 1, 1)

(34781, 3, 1) (2674733, 9, 1)

(5955773, 11, 1)

(𝑝, 𝑢, 𝑣) (421, 1, 1)

(24421, 1, 7) (26261, 3, 1) (60901, 1, 9)

(427813, 1, 15) (756421, 7, 1)

(𝑝, 𝑞, 𝑢, 𝑣) (17, 19, 2, 1)

(17, 179, 4, 3)

(121937, 122099, 16, 3)

(𝑝, 𝑞, 𝑢, 𝑣) (9187, 8863, 5, 3)

(20611, 20287, 7, 3) (82531, 82207, 11, 3)

160 Shin-Wook Kim

Choose the pair (𝑢 , 𝑣 ) as (13, 3), (17, 3) then, we are confronted with the primality as (𝑋, 𝑂)(𝑋 indicates that the primality doesn’t hold and 𝑂 denotes it is satisfied the primality.)

There induced examples of theorem 4.1(3):

We confront to examples of theorem 4.1(4):

There comes the examples of corollary 5.1(1) :

(𝑝, 𝑢, 𝑣) (8837, 1, 1)

(15877, 1, 3) (19485797, 7, 1)

Examples of corollary 5.1(2) are derived:

We are confronted with examples of corollary 5.1(3):

We get examples of corollary 5.1(4):

(𝑝, 𝑞, 𝑢, 𝑣) (85931, 83431, 6, 5)

(239963, 181399, 4, 11)

(𝑝, 𝑞, 𝑢, 𝑣) (7, 23, 1, 18)

(7, 71, 3, 34)

(7, 151, 1, 46)

(7, 263, 3, 62)

(103,167, 7, 198)

(𝑝, 𝑢, 𝑣) (4357, 1, 1)

(94261, 5, 2)

(𝑝, 𝑢, 𝑣) (6709, 5, 1)

(87541, 11, 1) (279109, 15, 1)

(𝑝, 𝑢, 𝑣) (757, 1, 1)

(12037, 3, 1) (787477, 9, 1)

Enumeration in ranks of various elliptic curves 161

References

[1] C. Caldwell, http://primes.utm.edu/curios/includes/primetest.php

[2] A. Dabrowski and M. Wieczorek, On the equation 𝑦2 = 𝑥(𝑥 − 2𝑚)(𝑥 + 𝑞 −2𝑚), J. of Number Theory, 124 (2007), 364-379.

[3] History of elliptic curves rank records,

https://web.math.pmf.unizg.hr>~duje>tors>rankhist.

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[5] D. Husemöller, Elliptic curves, Springer, 2004.

[6] J. A. Johnstone and B. K. Spearman, Congruent number elliptic curves with

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[9] S. W. Kim, Crucial function of prime’s form, Int. J. of Algebra, 10 (2016),

283 - 290. https://doi.org/10.12988/ija.2016.6428

[10] S. W. Kim, Different odd primes in curve 𝑦2 = 𝑥3 − 𝑝𝑞𝑥, Far East J. Math. Sci. (FJMS), 107 (2018), 155 - 165.

[11] S. W. Kim, Comparison of ranks in some elliptic curves, JP J. Algebra,

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[12] S. W. Kim, Searching the ranks of elliptic curves 𝑦2 = 𝑥3 − 𝑝𝑥 , Int. J. of Algebra, 12 (2018), 311 - 318. https://doi.org/10.12988/ija.2018.8934.

[13] S. W. Kim, Various forms in components of primes, Int. J. of Algebra, 13

(2019), 59-72. https://doi.org/10.12988/ija.2019.913

[14] F. Lemmermeyer and R. Mollin, On Tate-Shafarevich groups of 𝑦2 =𝑥(𝑥2 − 𝑘2), Acta Math. Univ. Comenianae, LXXII (2003), 73 - 80.

[15] K. Rubin, Ranks, Ross program, July (2003), 1 - 25.

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162 Shin-Wook Kim

[16] J. H. Silverman and J. Tate, Rational points on elliptic curves, Springer,

New York, 1992. https://doi.org/10.1007/978-1-4757-4252-7

Received: January 27, 2020; Published: February 24, 2020