should be congruent to 2 modulo 3. By substituting

x

, the equation becomes 31m

to

22

369 1kkmmm

2

ml . Since it has an integral

solution,

2

31kll l for an integer .

and

Now we show that there is no point of order 2 except

2

31,0l in

E. Assume that

2

31,0lE.

Then

2

31kll.

32

222432

633 62

311391812 2.

xkxkk

xlxlx lll

22432

139 18 122xlxlll

Qx be Let

3

93 11ll

with discriminant . If the solution of

Qx exists, then 311 0ll

0l. It gives us the

value

or 1. Hence and is singular. □ 0kE

Proposition 7. Let

23 2

:63362Ey xkxkk

k h

be the elliptic

curve with in . Assume that there is no integer

such that

2

433 1khhh or

2

41331hhh

. Then

E has no point of

order 3.

Proof. As we mentioned in the proof of the previous

lemma, the point

,Pxy is of order 3 if and only if

x

is the root of

32 2

31125 12123.

E

TX

XXX kXkk

E

SX be the polynomial Let

32 2

31

12512123 .

E

TX X

XXk Xkk

Since

2

1, 3k

E

is not in , it suffices to

check whether

x

is a root of or not.

0SX

E

Suppose that

0SX

E

x

has a root

k

in . Then

it is an integer. In other words, for an integer not the

form

2

433 1hh h

2

41331hhh

k or by

sorting again as , we can fine an integer such that

x

32 2

232

12 51212 3

1212153 0.

xxk xkk

kxkxxx

Copyright © 2013 SciRes. APM

Y. K. PARK 307

From, this must be a multiple of 12

and

32

53xx x

x

is one of 12 or 11 for a suitable inte-

ger .

3,5,9m

12 3m

m

When ,

x

E

S

32

96 16mm

2

1 13mm

2

21hh

23hh

12xm

k

22

2 5m

becomes

. Because it

has integral solutions as a quadratic equation with respect

to , its discriminant is a square.

That means that for an integer .

Through this we get or

.

2

4 144mkm

16 4

41m

43kh

31h

5,12 9m

12

k

41

12kk

2

3hh

12xm

1

11

If or then discri-

minant of the quadratic equations with respect to is

or

3125

31211m

1, 4mm

2

1m

2 23m

4

6

respectively. Neither case has a

perfect square discriminant and admit any integral root. □

Proof of Theorem 2. Use the Lemma 5 and Theorem 3

3), we can determine which finite abelian group has a

subgroup of

E for the case

1mod35

k

E

k

1mod7

, i.e.,

and. In fact, tors is a

subgroup of both

od51mk

9 and 6. It yields that it is

3 or trivial. Since our group has no point of order 3,

it is trivial.

Note that tors is a subgroup of order , if it is

a subgroup of order with

, then. So it

is resolved as trivial group in many cases.

EN

3, 1N

k

3rN

To observe easily, we can refer Table 2: In this table,

takes the value modulo 5 at the horizontal line and

modulo 7 at the vertical line respectively. The groups

n in the brackets at top line and at the very left

line are result from Lemma 5 .

Cn

Each entry implies that the type of group: “A”, “B” or

“C” implies one of subgroups of 4, 2 or trivial,

respectively. The same alphabet does not mean the same

group. And “D” means that both curves

55

and

77

are singular. In this table since “C” is trivial, it

remains that a few cases

or 34 .

E

E

mod 35

4,7,12,15, 20,k22, 25, 27, 29,32

For the cases that the subgroup is nontrivial Pro-

Table 2. Type of gr oup

tors

E

k

.

mod 7k

mod 5

0 1

position 6 makes us know which curve has the point of

order 2 or not. Hence, it is sufficient to check the value

having order 4 points.

k

Assume that

20,34mod 35k

l

31kll

and there exists an

integer such that . In fact

20mod 35k

34mod 35

5l (respectively, ) if and

only if

26mod 3519 (respectively, or

or

33mod 35).

2

31,0l is the unique point of order 2.

Using duplication formula for the elliptic curve, let

,Pxy

2

231,0Pl be the point satisfying . By

Substituting

,, 63xyk

2

362kk

,,

and for

x

yA B

and in (in the formulas for x2 and y2 in the

proof of Lemma 4), we get two equations affirming the

existence of point of order 4:

2

22432

22432

21 31818610

21 31818610

xlxlll

xlxlllFx

9

C 2

6

C3

3

C4

0 D C B C D

1

6

C

9

C

B C B C B

2 C C C C C

3

33

CC

6

C

C C C C C

4 B C B C B

5 D C B C D

6

12

C A C B C A

where

4234322

65432

87642

21 36918122

25416210854637

324 972 864270605.

F

xxlxl l lx

lllllx

lllll

To have an integral solution of

22432

21 31818610xlxlll

, its discrimi-

nant

3

36 3 2ll

have to be a square. Suppose that we

can find an integer such that and m

232mll

2

316

x

llm

2

316llm

ml

( or ). It is easy to check

that the integer satisfying the above condition exists

in each case determined by . Furthermore, by sub-

stituting

,

x

31kll

232mll and to the

right hand side of (1.4) we get a numerical formula

32

22

22 2

5432 652

963266 52

96652

llllm

lllllm

lml lm

0l

Since

makes the curve (1.4) singular,

2

66 52llm

23

:75506Ey xx

is a square of a suitable integer if and

only if there exists a point of order 4.

So we are done. □

3. Conclusions

By the help of Theorem 2, we explicitly calculate the

torsion part of Modell-Weil group.

Example 8. Let be the elli-

ptic curve. Then

2.

tors

E

12k

Given elliptic curve is the form in Theorem 2

and

2

123 212

. Therefore

2E

tors .

And

11,0 is the nontrivial torsion point on

E.

tors

E is able to be applied to The method to find

Copyright © 2013 SciRes. APM

Y. K. PARK

t © 2013 SciRes. APM

308

k

7p

Copyrigh

all elliptic curve without a condition for by choosing

another prime .

For example, in Theorem 2, there is a condition

359 4kk

for . This is one for nonsingular curve.

For the case that

k

35 9kk4

, choose the another

prime such that

7p

94kpk

. Calculate

p

p and eliminate the order 3 point and check the

condition for having order 2 point. Since

E

21p

p, the smaller gives simpler nece-

ssary condition. For example, if then the ellip-

tic curve is

Ep

16k

:93674x

23

x

7

Ey

with discriminant . Find

6

25

p

p

E

with

and,

11 17p

11 11 and

15E

17 1718E

.

Using Lemma 4, we observe that

E

23

:Eyxfkx gk

k

has no point of

order 3. So it is a trivial group.

and

for

maxdeg,deg2fkgk . We can

use the criterion for the quadratic equation to find a point

of order 2 or 3. Of course, it is indispensable to consider

some exceptional cases in the similar way to Proposition

7.

REFERENCES

[1] B. Mazur, “Modular Curves and the Eisenstein Ideal,”

Publications Mathématiques de l’Institut des Hautes Étu-

des Scientifiques, No. 47, 1977, pp. 33-168.

[2] A. Knapp, “Elliptic Curves,” Princeton University Press,

Princeton, 1992.

[3] D. Kim, J. K. Koo and Y. K. Park, “On the Elliptic

Curves Modulo p,” Journal of Number Theory, Vol. 128,

No. 4, 2008, pp. 945-953. doi:10.1016/j.jnt.2007.04.015

Remark 9. Generalize our elliptic curve