/div>
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 lInstitut 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