/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
 
 

  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 Y. K. PARK
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
 B. Mazur, “Modular Curves and the Eisenstein Ideal,”
Publications Mathématiques de lInstitut des Hautes Étu-
des Scientifiques, No. 47, 1977, pp. 33-168.
 A. Knapp, “Elliptic Curves,” Princeton University Press,
Princeton, 1992.
 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