Y. K. PARK 307
From, this must be a multiple of 12
is one of 12 or 11 for a suitable inte-
. 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
If or then discri-
minant of the quadratic equations with respect to is
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
E for the case
and. In fact, tors is a
subgroup of both
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.
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 .
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
are singular. In this table since “C” is trivial, it
remains that a few cases
or 34 .
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
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.
and there exists an
integer such that . In fact
5l (respectively, ) if and
26mod 3519 (respectively, or
31,0l is the unique point of order 2.
Using duplication formula for the elliptic curve, let
231,0Pl be the point satisfying . By
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:
0 D C B C D
B C B C B
2 C C C C C
C C C C C
4 B C B C B
5 D C B C D
C A C B C A
324 972 864270605.
xxlxl l lx
To have an integral solution of
, its discrimi-
36 3 2ll
have to be a square. Suppose that we
can find an integer such that and m
( or ). It is easy to check
that the integer satisfying the above condition exists
in each case determined by . Furthermore, by sub-
232mll and to the
right hand side of (1.4) we get a numerical formula
makes the curve (1.4) singular,
is a square of a suitable integer if and
only if there exists a point of order 4.
So we are done. □
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
Given elliptic curve is the form in Theorem 2
11,0 is the nontrivial torsion point on
E is able to be applied to The method to find
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