Elliptic curves over the rationals with good reduction outside two odd primes

Abstract

We classify elliptic curves over Q with a rational point of order 2 or ≥4 and good reduction outside two odd primes. We also exhibit some families of elliptic curves with a rational point of order 3, collect some general existence/non-existence results, and present some information concerning upper bounds for the rank.

Full text

Elliptic curves over the rationals with good
reduction outside two odd primes
Andrzej D¡browski and Tomasz J¦drzejak
Abstract
. We classify elliptic curves over
Q
with a rational point of or-
der
2
or
≥4
and good reduction outside two odd primes. We also exhibit
some families of elliptic curves with a rational point of order
3
, collect some
general existence/non-existence results, and present some information con-
cerning upper bounds for the rank.
Key words: elliptic curve, rational point, diophantine equation, class
number
2010 Mathematics Subject Classication: 14G25, 14H52, 14G05
1 Introduction
It is well known (due to Shafarevich) that the number of isomorphism classes
of elliptic curves over a given number eld and having good reduction outside
a nite set of primes is nite. In particular, given a positive integer
N
,
there are only nitely many isomorphism classes of elliptic curves over
Q
of conductor
N
. Note that modularity of elliptic curves over
Q
gives, as a
trivial corollary, niteness of the set of isogeny classes of elliptic curves of a
given conductor.
The conductor
NE
of an elliptic curve
E
over
Q
is a positive integer that
encodes the primes of bad reduction. We always have
NE|∆E
(where
∆E
denotes the discriminant of
E
), and primes divisors of
NE
and
∆E
are the
same. If
p|NE
, then
p||NE
exactly if
E
has multiplicative reduction at
p
.
E
has additive reduction at
p
exactly if
p2+δp||NE
, where
δp= 0
for
p≥5
, and
0≤δ2≤6
,
0≤δ3≤3
can be calculated explicitly using Ogg-Saito formula
(see [20], Appendix C16).
1

The online tables by Cremona [5] exhibit all elliptic curves over
Q
of
conductors up to 400000, together with many additional information (tor-
sion subgroup, rank, etc.). Let us mention that the paper by Cremona and
Lingham [6] gives an explicit algorithm for nding all the elliptic curves
over a number eld with good reduction outside a given nite set of (nonar-
chimedean) primes.
Let us give an overview of known results concerning classication of el-
liptic curves over
Q
with good reduction outside at most two primes.
(i)
N=pk
, with
p
a prime. Elliptic curves of conductors
2k
(resp.
3k
)
were completely classied by Ogg [16] (resp. Hadano [9]). Setzer [19] proved,
that there is an elliptic curve over
Q
of conductor
p
with a rational point of
order
2
if and only if
p= 17
or
p=u2+ 64
for some integer
u
. Assuming
p≡ ±1(mod 8)
, and the class numbers of both
Q(√p)
and
Q(√−p)
are
not divisible by
3
, Setzer proved that in these cases each elliptic curve of
conductor
p
dened over
Q
has a rational point of order
2
. Edixhoven et
al. [8] proved that if
p≡5(mod 12)
, then every elliptic curve over
Q
of
conductor
p2
is a twist of one of conductor
p
.
(ii)
N= 2np
, with
p
odd prime. Elliptic curves of conductors
2k3
were
completely classied by Ogg [17]. Ivorra [12] has classied elliptic curves
over
Q
of conductor
2kp
with a rational point of order
2
. Let us mention the
following result by Hadano: assume
p≡1,7(mod 8)
, and the class numbers
of the following four elds
Q(√p)
,
Q(√−p)
,
Q(√2p)
and
Q(√−2p)
are not
divisible by
3
, then each elliptic curve of conductor
2kp
dened over
Q
has a
rational point of order
2
.
(iii)
N=pmqn
, with
p
,
q
dierent odd primes. Bennett, Vatsal and
Yazdani [3] classied all elliptic curves over
Q
with a rational
3
-torsion point
and good reduction outside the set
{3, p}
, for a xed prime
p
. It is an open
problem to classify elliptic curves over
Q
with a rational
3
-torsion point and
good reduction outside the set
{p, q}
, with
p
and
q
dierent primes
≥5
. In a
paper by Howe [10] it is proved that if there is an elliptic curve over
Q
of odd
conductor
pq
with a rational point of order
2
, then one of the diophantine
equation (from an explicit list) has a solution. In that paper, Howe also
stated some general existence and non-existence results. In a recent paper
by Sadek [18], the author nds all elliptic curves dened over
Q
with good
reduction outside two dinstinct primes and a rational point of xed order
N≥4
(actually, he only nds possible minimal discriminants of such curves).
It turns out that some of his claims are incomplete - for instance some curves
with a rational point of order
8
are missing (see subsection 6.1).
2

Our paper is a common extension (and clarication) of the work given by
Ogg, Hadano, Neumann, Setzer, Edixhoven-de Groot-Top, Ivorra, Bennett-
Vatsal-Yazdani, Howe, Sadek and others. In sections 3 and 5, we give explicit
description of elliptic curves over
Q
with good reduction outside two odd
primes and a rational point of order
2
or
≥4
, and exhibit some families of
elliptic curves with a rational point of order
3
. It turns out that an elliptic
curve with a rational point of order
2
belongs to one of
77
(conjecturally, in-
nite) families or to a nite "exceptional set". Elliptic curves with a rational
point of order
4
belong to one of
16
(conjecturally, innite) families or to a
nite "exceptional set". In section 6 we collect some general existence/non-
existence results. In section 7 we present some information concerning upper
bounds for ranks of elliptic curves of odd conductors
paqb
and with
Q
-rational
point of order
2
.
Acknowledgement.
We would like to thank the anonymous referee for
useful suggestions and comments which allow to improve the nal version of
the article.
2 Some Diophantine Equations
In this section we list some of the diophantine equations we use in the next
sections.
Consider a generalized Fermat equation
xp+yq=zr
, with
p
,
q
,
r
positive
integers satisfying
1/p+1/q +1/r < 1
, and where
x
,
y
,
z
are coprime integers.
If we x the triple
(p, q, r)
, then the number of solutions to such an equation
is nite [7]. It is expected (commonly known as Tijdeman-Zagier conjecture)
that the following ten solutions are the only solutions to the above equation:
1p+ 23= 32
,
25+ 72= 34
,
73+ 132= 29
,
27+ 173= 712
,
35+ 114= 1222
,
177+ 762713= 210639282
,
14143+ 22134592= 657
,
92623+ 153122832= 1137
,
438+ 962223= 300429072
,
338+ 15490342= 156133
.
Lemma 1
([18], Lemma 2.2) There are no integer solutions
(x, m, y, n)
to
the equation
16xm+ 1 = yn
, where
|x|
,
n
are primes,
m > 1
, and
y=lt
with
|l|
a prime and
t > 0
.
The diophantine equation (Lebesgue-Nagell equation)
x2+C=yn
(
x, y ≥
1
,
n≥3
) has a rich history. This equation has no solution for many values
3

of
C
(see, for instance, [4], where all solutions of this equation are given
when
1≤C≤100
). Barros [2] in his PhD thesis considered the range
−100 ≤C≤ −1
. Here are special cases for
C=±16
and
64
.
Lemma 2
There are no integer solutions
(x, y, n)
,
y
odd,
n≥3
, to the
equations
x2±16 = yn
and
x2+ 64 = yn
.
3 Elliptic curves over
Q
with good reduction
outside two odd primes and a rational point
of order two
Theorem 1
Let
p6=q
be odd primes. Any elliptic curve
E
dened over
Q
,
of conductor
N=pq
, and with
E(Q)[2] '(Z/2Z)2
is isomorphic (over
Q
)
to one of the following four curves of the form
Y2=X3+AX2+BX
:
A B ∆EE(Q)tors A B ∆EE(Q)tors
a
14 −15 32×52Z/2Z×Z/4Z
c
−2−63 34×72Z/2Z×Z/4Z
b
6−55 52×112(Z/2Z)2
d
−10 −39 32×132(Z/2Z)2
or belongs to one of the following two families:
A B
condition
∆E
e
±2p64 + pαqβpαqβpα−qβ=±16 p2αq2β
f
±2p64p2a+εqβεqβqβ−16pα=ε p2αq2β
where
=±1
and the sign before the quadratic square is chosen so that it is
congruent to
3
modulo
4
.
Remark.
(i) Primes of the type
p=qβ±16
(with
q
another odd prime
and
β≥1
) give an explicit (conjecturally, innite) family of elliptic curves
of conductor
pq
with three
Q
-rational points of order
2
. The diophantine
equations
p2=qβ±16
(
β≥2
) have solutions only for
β= 2
(use Lemma 2),
and produce the elliptic curve
Y2=X3−34X2+ 225X
of conductor
15
. If
we believe the (mentioned in Section 2) conjecture of Tijdeman and Zagier
then there are no other solutions (and, hence, there are no corresponding
elliptic curves).
4

(ii) Lemma 1 states that the diophantine equation
qβ−16pα=±1
has only
the obvious solutions
q= 16pα±1
, and we obtain an explicit (conjecturally,
innite) family of elliptic curves of conductor
pq
with three
Q
-rational points
of order
2
.
Theorem 2
Let
p6=q
be odd primes. Any elliptic curve
E
dened over
Q
,
of conductor
N=pq
, and with
E(Q)[2] 'Z/2Z
is isomorphic (over
Q
) to
one of the following ve curves of the form
Y2=X3+AX2+BX
:
A B ∆EE(Q)tors A B ∆EE(Q)tors
a
1 16 −32×7Z/4Z
c
−3 16 −5×11 Z/2Z
b
5 16 −3×13 Z/2Z
d
−7 16 −3×5Z/4Z
or belongs to one of the following families of elliptic curves (with
=±1
).
A B
conditions
∆E
ia
±2u qβu2=qβ−64p2α,±u≡3(mod 4) −p2αq2β
ib
±2u pαqβu2=pαqβ−64,±u≡3(mod 4) −p2αq2β
iia
±2u1u2= 64p2αq2β+1 + 1,±u≡3(mod 4) p2αq2β+1
iib
±2u pαu2= 64q2β+1 +pα,±u≡3(mod 4) p2αq2β+1
iic
±u16 u2=p2αq2β+1 + 64, ±u≡1(mod 4) p2αq2β+1
iid
±u16pαu2=q2β+1 + 64pα,±u≡1(mod 4) p2αq2β+1
iiia
±2u pαu2=pα−64q2β+1,±u≡3(mod 4) −p2αq2β+1
iiib
±u16pαu2= 64pα−q2β+1,±u≡1(mod 4) −p2αq2β+1
iva
±2u1u2= 64p2α+1q2β+1 + 1,±u≡3(mod 4) p2α+1q2β+1
ivb
±u16 u2=p2α+1q2β+1 + 64, ±u≡1(mod 4) p2α+1 q2β+1
Remark.
It is still a conjecture that there are innitely many elliptic
curves over
Q
with a rational point of order two and of odd conductor
pq
(
p
,
q
dierent odd primes). On the other hand, there are innitely many elliptic
curves over
Q
with rational point of order two, and of conductor
p
or
pq
. It
is enough to consider the Neumann-Setzer curves [19], and the (generalized
Neumann-Setzer) curves (ii.d) or (v.b) above, and apply the following result
of Iwaniec [13]: there are innitely many integers
n
such that
n2+ 64
is the
product of at most two primes.
Theorem 3
Let
p6=q
be odd primes. Any elliptic curve
E
dened over
Q
,
of conductor
N=paq
(
a≥2
), and with
E(Q)[2] '(Z/2Z)2
is isomorphic
5

(over
Q
) to one of the following curves of the form
Y2=X3+AX2+BX
(all with
E(Q)tors '(Z/2Z)2
):
A B ∆EA B ∆E
a
−75 −400 56×172
e
30 −351 38×132
b
45 −144 36×172
f
30 −1375 58×112
c
−51 144 38×52
g
14 −3087 34×78
d
85 400 32×58
h
−363 32912 116×172
or belongs to one of the following families (with
=±1
):
A B
conditions
∆E
i
±2up pm+2 u2= 64q2n+pm,±up ≡3(mod 4) p2m+6q2n
ii
±15p−16p2p6= 17,±p≡3(mod 4) 172p6
iii
±up 16p2m+2 u2=q2n+ 64p2m,±up ≡1(mod 4) p2m+6 q2n
iv
±2up p2qnu2= 64p2m+qn,±up ≡3(mod 4) p2m+6 q2n
v
±2up pm+2qnu2= 64 + pmqn,±up ≡3(mod 4) p2m+6 q2n
vi
±up 16p2qnu2=p2m−2+ 64qn,±up ≡1(mod 4) p2m+4q2n
vii
±up 16pm+2qnu2= 1 + 64pmqn,±up ≡1(mod 4 p2m+6q2n
Remark.
The conditions in (i), (iv) and (vii) lead to diophantine equa-
tions of the type
pa−16qb=±1
or
qb−16pa=±1
. The condition in (v)
leads to diophantine equation of the type
pa−qb=±16
. The conditions in
(iii) and (vi) lead to both types of diophantine equations.
Theorem 4
Let
p6=q
be odd primes. Any elliptic curve
E
dened over
Q
,
of conductor
N=paq
(
a≥2
), and with
E(Q)[2] 'Z/2Z
is isomorphic (over
Q
) to one of the following curves of the form
Y2=X3+AX2+BX
:
A B ∆EE(Q)tors A B ∆EE(Q)tors
a
−3 144 −38×7Z/2Z
k
41 656 −23 ×413Z/2Z
b
−7 784 −32×77Z/4Z
l
−47 752 −17 ×473Z/2Z
c
9 48 −33×37 Z/2Z
m
53 848 −11 ×533Z/2Z
d
−3 48 −33×61 Z/2Z
n
−59 944 −5×593Z/2Z
e
−15 80 −53×19 Z/2Z
o
61 976 −3×613Z/2Z
f
5 80 −53×59 Z/2Z
p
21 144 −37×5Z/2Z
g
−11 176 −113×53 Z/2Z
q
−15 144 −37×13 Z/4Z
h
17 272 −173×47 Z/2Z
r
−35 400 −3×57Z/2Z
i
−23 368 −233×41 Z/2Z
s
33 1936 −5×117Z/4Z
j
37 592 −33×373Z/2Z
t
65 2704 −3×137Z/4Z
6

or belongs to one of the following families of elliptic curves (with
=±1
)
A B
conditions
∆E
ia
±2up pm+2 u2=pm−64q2n,±up ≡3(mod 4) −p2m+6q2n+6
ib
±2up p2qnu2+ 64 = qn,±up ≡3(mod 4) −p6q2n
ic
±2upm+1 p2qnu2p2m+ 64 = qn,±upm+1 ≡3(mod 4) −p6q2n
id
±2up p2qnu2+ 64p2m=qn,±up ≡3(mod 4) −p2m+6q2n
ie
±2up pm+2qnu2+ 64 = pmqn,±up ≡3(mod 4) −p2m+6q2n
iia
±66p p2p6= 17,±p≡3(mod 4) 17p6
iib
±2upm+1 p2u2p2m= 1 + 64q2n−1,±upm+1 ≡3(mod 4) p6q2n−1
iic
±2up p2u2= 1 + 64p2mq2n−1,±up ≡3(mod 4) p2m+6q2n−1
iid
±2up pm+2 u2= 64q2n−1+pm,±up ≡3(mod 4) p2m+6q2n−1
iie
±up 16p2u2=q2n−1+ 64, ±up ≡1(mod 4) p6q2n−1
iif
±upm+1 16p2u2p2m=q2n−1+ 64, ±upm+1 ≡1(mod 4) p6q2n−1
iig
±up 16p2u2=p2mq2n−1+ 64, ±up ≡1(mod 4) p2m+6q2n−1
iih
±up 16pm+2 u2=q2n−1+ 64pm,±up ≡1(mod 4) p2m+6q2n−1
iiia
±2up pm+2 u2=pm−64q2n−1,±up ≡3(mod 4) −p2m+6q2n−1
iva
±2upkp2k−2m+1 u2p2m−1= 64q2n+, ±upk≡3(mod 4) p6k−6m+3q2n
ivb
±2up p2u2= 1 + 64p2m−1q2n,±up ≡3(mod 4) p2m+5q2n
ivc
±upk16p2k−2m+1 u2p2m−1=q2n+ 64, ±upk≡1(mod 4) p6k−6m+3q2n
ivd
±up 16p2u2=p2m−1q2n+ 64, ±up ≡1(mod 4) p2m+5 q2n
ive
±2upkp2k−2m+1qnu2p2m−1= 64 + qn,±upk≡3(mod 4) p6k−6m+3q2n
ivf
±2up p2qnu2= 64p2m−1+qn,±up ≡3(mod 4) p2m+5q2n
ivg
±upk16p2k−2m+1qnu2p2m−1= 1 + 64qn,±upk≡1(mod 4) p6k−6m+3q2n
ivh
±up 16p2qnu2=p2m−1+ 64qn,±up ≡1(mod 4) p2m+5q2n
va
±2upkp2k−2m+1qnu2p2m−1=qn−64,±upk≡3(mod 4) −p6k−6m+3q2n
vb
±2up p2qnu2=qn−64p2m−1,±up ≡3(mod 4) −p2m+5q2n
vc
±upk16p2k−2m+1qnu2p2m−1= 64qn−1,±upk≡1(mod 4) −p6k−6m+3q2n
vd
±up 16p2qnu2= 64qn−p2m−1,±up ≡1(mod 4) −p2m+5q2n
via
±2upkp2k−2m+1 u2p2m−1= 64q2n−1+, ±upk≡3(mod 4) p6k−6m+3q2n−1
vib
±2up p2u2= 64p2m−1q2n−1+ 1,±up ≡3(mod 4) p2m+5q2n−1
vic
±upk16p2k−2m+1 u2p2m−1=q2n−1+ 64, ±upk≡1(mod 4) p6k−6m+3q2n−1
vid
±up 16p2u2=p2m−1q2n−1+ 64, ±up ≡1(mod 4) p2m+5 q2n−1
where
m, n
are positive integers, and
k=m, m + 1
.
Remark.
There are innitely many elliptic curves over
Q
with a rational
point of order two and of odd conductor
paq
(
p
,
q
dierent odd primes and
a≥2
). Take the family (ii) in Theorem 3 or the family (iia) in Theorem 4.
7

Theorem 5
Let
p6=q
be odd primes. Any elliptic curve
E
dened over
Q
,
of conductor
N=paqb
(
a, b ≥2
), and with
E(Q)[2] '(Z/2Z)2
is isomorphic
(over
Q
) to one of the following four curves of the form
Y2=X3+AX2+BX
(all with
E(Q)tors '(Z/2Z)2
):
A B ∆EA B ∆E
a
918 44217 36×178
c
−255 3600 38×58
b
−1275 −115600 56×178
d
−6171 9511568 116×178
or belongs to one of the following four families (with
=±1
):
A B
conditions
∆E
i
±2upq p2qn+2 u2= 64p2m+qn
,
±upq ≡3(mod 4) p2m+6 q2n+6
ii
±2upq pm+2qn+2 u2= 64 + pmqn
,
±upq ≡3(mod 4) p2m+6q2n+6
iii
±306p173p2p6= 17,±p≡3(mod 4) 178p6
iv
±upq 16pm+2qn+2 u2= 64pmqn+ 1
,
±upq ≡1(mod 4) p2m+6 q2n+6
Remark.
The conditions in (i) and (iv) lead to diophantine equations
of the type
pa−16qb=±1
or
qb−16pa=±1
. The condition in (ii) leads to
diophantine equations of the type
pa±qb= 16
.
Theorem 6
Let
p6=q
be odd primes. Any elliptic curve
E
dened over
Q
,
of conductor
N=paqb
(
a, b ≥2
), and with
E(Q)[2] 'Z/2Z
is isomorphic
(over
Q
) to one of the following curves of the form
Y2=X3+AX2+BX
:
A B ∆EE(Q)tors A B ∆EE(Q)tors
a
126 −63 36×73Z/2Z
l
−799 217328 −177×473Z/2Z
b
21 7056 −38×77Z/2Z
m
285 28880 −53×197Z/2Z
c
−63 1008 −36×73Z/2Z
n
−943 347024 −237×413Z/2Z
d
−130 65 53×133Z/2Z
o
333 65712 −33×377Z/2Z
e
65 1040 53×133Z/2Z
p
−943 618608 −233×417Z/2Z
f
165 48400 −57×117Z/2Z
q
−799 600848 −173×477Z/2Z
g
−195 24336 −37×137Z/2Z
r
−583 494384 −113×537Z/2Z
h
105 3600 −37×57Z/4Z
s
−295 278480 −53×597Z/2Z
i
−183 8784 −37×613Z/2Z
t
−183 178608 −33×617Z/2Z
j
−295 23600 −57×593Z/2Z
u
−111 5328 −39×373Z/2Z
k
−583 102608 −117×533Z/2Z
or belongs to one of the following families of elliptic curves (with
=±1
)
8

A B
conditions
∆E
ia
±2upm+1q p2qn+2 u2p2m+ 64 = qn,±upm+1q≡3(mod 4) −p6q2n+6
ib
±2upq pm+2q2u2=pm−64q2n,±upq ≡3(mod 4) −p2m+6 q2n+6
ic
±2upq pm+2qn+2 u2=pmqn−64,±upq ≡3(mod 4) −p2m+6q2n+6
iia
±2upm+1q p2q2u2p2m= 64q2n−1+ 1,±upm+1 q≡3(mod 4) p6q2n+5
iib
±2upq pm+2q2u2= 64q2n−1+pm,±upq ≡3(mod 4) p2m+6 q2n+5
iic
±2upq p2q2u2= 64p2mq2n−1+ 1,±upq ≡3(mod 4) p2m+6 q2n+5
iid
±2upqlpm+2 q2l+1−2nu2q2n−1= 64 + pm,±upql≡3(mod 4) p2m+6 q6l−6n+3
iie
±2upqlp2q2l+1−2nu2q2n−1= 64p2m+, ±upql≡3(mod 4) p2m+6q6l−6n+3
iif
±42p−7p2p6= 7,±p≡3(mod 4) 73p6
iig
±upm+1q16p2q2u2p2m=q2n−1+ 64, ±upm+1 q≡1(mod 4) p6q2n+5
iih
±upq 16pm+2q2u2=q2n−1+ 64pm,±upq ≡1(mod 4) p2m+6 q2n+5
iii
±upq 16p2q2u2=p2mq2n−1+ 64, ±upq ≡1(mod 4) p2m+6q2n+5
iij
±upql16p2q2l+1−2nu2q2n−1=p2m+ 64, ±upql≡1(mod 4) p2m+6q6l−6n+3
iik
±upql16p2m+2 q2l+1−2nu2q2n−1= 64p2m+ 1,±upql≡1(mod 4) p4m+6q6l−6n+3
iiia
±2upq pm+2q2u2=pm−64q2n−1,±upq ≡3(mod 4) −p2m+6q2n+5
iiib
±2upqlpm+2 q2l+1−2nu2q2n−1=pm−64,±upql≡3(mod 4) −p2m+6 q6l−6n+3
iiic
±upq 16pm+2q2u2= 64pm−q2n−1,±upq ≡1(mod 4) −p2m+6q2n+5
iiid
±upql16pm+2 q2l+1−2nu2q2n−1= 64pm−1,±upql≡1(mod 4) −p2m+6 q6l−6n+3
iiie
±21p112p2p6= 7,±p≡1(mod 4) −73p6
iva
±2upqlp2q2l+1−2nu2q2n−1= 64p2m−1+, ±upql≡3(mod 4) p2m+5q6l−6n+3
ivb
±2upq p2q2u2= 64p2m−1q2n−1+ 1,±upq ≡3(mod 4) p2m+5q2n+5
ivc
±upql16p2q2l+1−2nu2q2n−1=p2m−1+ 64, ±upql≡1(mod 4) p2m+5q6l−6n+3
ivd
±upq 16p2q2u2=p2m−1q2n−1+ 64, ±upq ≡1(mod 4) p2m+5q2n+5
where
m, n
are positive integers, and
l=n, n + 1
.
Remark.
There are innitely many elliptic curves over
Q
with a rational
point of order two and of odd conductor
paqb
(
p
,
q
dierent odd primes and
a, b ≥2
). Take the family (iii) in Theorem 5 or the families (iif), (iiie) in
Theorem 6.
4 Proofs of Theorems 1 - 6
4.1 Proofs of Theorems 1 and 2
Any semistable elliptic curve dened over
Q
, with
Q
-rational point of order
2
, has a unique Weierstrass model of the type
(1) Y2=X3+AX2+BX,
9

where
A, B ∈Z
and
gcd(A, B)=1
([14], Lemme 1). This model is minimal
outside 2, with the discriminant
∆ = 24B2(A2−4B)
and
c4= 24(A2−3B)
.
The minimal discriminant
∆E= 2−8B2(A2−4B)
(note that
c4/16
is odd).
Lemma 3
We can choose
A, B
such that
(2) A≡6(mod 8),B≡1(mod 8)
or
(3) A≡1(mod 4),B≡0(mod 16).
Proof
. See [19], p. 374 or [14], Remarque on p. 176.
Proof of Theorem 1
. Our assumptions lead to
∆ = 212psqt
, with
2|s
and
2|t
(note that
∆
must be a square of a nonzero integer, since the equation
X2+AX +B= 0
has two rational solutions). Hence
B2(A2−4B)=28p2αq2β
with the assumption
gcd(A, B, pq) = 1
(since
E
has multiplicative reduction
at
p
and
q
).
From (2) it follows that
A= 2A0
,
A0≡3(
mod
4)
, hence
B2(A2
0−B) =
26p2αq2β
and
26|A2
0−B
,
(B, A2
0−B)=1
. Therefore we have the following
possibilities:
(4) A2
0−B= 64p2αq2β, B = 1
,
(5) A2
0−B= 64p2α, B2=q2β(A2
0−B= 64q2β, B2=p2α)
,
(6) A2
0−B= 64, B2=p2αq2β
.
Consider the case (4). Then we have
A2
0−1 = (A0−1)(A0+1) = 64p2αq2β
,
A0−1≡2(
mod
4)
,
32 |A0+ 1
and
gcd(A0−1, A0+ 1) = 2
. Elementary
calculations show that there exist no elliptic curve satisfying these conditions.
Consider the case (5). We have
A2
0−64p2α=±qβ
. Since
(A0−8pα) +
16pα= (A0+8pα)
, hence
gcd(A0−8pα, A0+8pα, q) = 1
. Moreover,
A0−8pα≡
A0+ 8pα≡3(
mod
4)
. Taking into account the above conditions, we are
arriving at the following two possibilities:
A0−8pα=−qβ
and
A0+ 8pα=−1
,
A0−8pα=−1
and
A0+ 8pα=qβ
.
The corresponding families of elliptic curves are given by the following
models:
10

Y2=X3±2p26p2α+qβX2+qβX,
if
qβ−16pα=,
where
=±1
, and the sign before the quadratic square is chosen so that it
is congruent to
3
modulo
4
.
Let us consider the remaining case (6). Here we have
A2
0−26=±pαqβ
,
gcd(A0−8, A0+8, pq)=1
and
A0−8≡A0+8 ≡3(
mod
4)
. The possibilities
(7) A0−8 = −1
and
A0+ 8 = 15 = pαqβ
,
(8) A0−8 = −pα
and
A0+ 8 = qβ
,
(9) A0−8 = −qβ
and
A0+ 8 = pα
,
produce the four elliptic curves (a), (b), (c) and (d). The remaining cases
lead to the equations
pαqβ=±17
(which have no solution) or
pα−qβ=±16
.
The last equations produce the family
Y2=X3±2p64 + pαqβX2+pαqβX,
if
pα−qβ=±16,
where the sign before the quadratic square is chosen so that it is congruent
to
3
modulo
4
.
We will show that the case (3) does not produce any new curve. In this
case
B= 16B0
,
B2
0(A2−64B0) = p2αq2β
,
2-B0
, and
gcd(B0, A2−64B0) = 1
.
Therefore we have the following possibilities:
(10) A2−64B0=p2αq2β, B2
0= 1
,
(11) A2−64B0=p2α, B2
0=q2β(A2−64B0=q2β, B2
0=p2α)
,
(12) A2−64B0= 1, B2
0=p2αq2β
.
Note that (10) implies
A2−p2αq2β=±64
. The equation
A2−p2αq2β=
−64
leads to conditions
pαqβ=±17
, hence does not produce any elliptic
curve. On the other hand, the equation
A2−p2αq2β= 64
leads to an elliptic
curve
Y2=X3+ 17X2+ 16X
(isomorphic to (a)).
The case (11) implies
A2−64qβ=±p2α
. From
(A−pα)+2pα=A+pα
it follows that one of the factors
A−pα, A +pα
is divisible by
25
, and the
other by 2. Moreover, of course
gcd(A−pα, A +pα, q)=1
. Checking all
the possibilities, we obtain the curves (isomorphic to) (b), (c) and (d) if
pα+qβ= 16
, and the following families:
Y2=X3±pp2α+ 64qβX2+ 16qβX,
if
16qβ−pα= 1
or
pα−qβ=−16,
Y2=X3±pp2α−64qβX2−16qβX,
if
16qβ−pα=−1
or
pα−qβ= 16,
11

where the sign before the quadratic square is chosen so that it is congruent
to
1
modulo
4
. If
pα−qβ=±1
(respectively,
16pα−qβ=±1
), then the
change of variables
x7→ x±qβ
,
y7→ y
(respectively,
x7→ x−1
,
y7→ y
) leads
to elliptic curves from the family (e) (respectively (f)).
Let us consider the remaining case (12). The conditions
(A−1)(A+ 1) =
64pαqβ
,
A+ 1 ≡2(
mod
4)
imply
2|A+ 1
,
32 |A−1
. Here we produce the
elliptic curve
Y2=X3−31X2+ 240X
(isomorphic to (a)), and the family
Y2=X3±p64pαqβ+ 1X2+16pαqβX,
if
16pα−qβ=±1
or
16qβ−pα=±1,
where the sign before the quadratic square is chosen so that it is congruent
to
1
modulo
4
. The change of variables
x7→ x+qβ
,
y7→ y
, leads to elliptic
curves from the family (f).
Proof of Theorem 2
. We start in the same way as in the proof of Theorem
1. Note that now
∆
is not a square of a nonzero integer. Also notice that
still
A0=A/2
in the case (2), and
B0=B/16
in the case (3) (c.f. Lemma
3).
∆<0
,
s= 2α
,
t= 2β
. In this case we have
B2(A2−4B) = −28p2αq2β
,
hence
B > 0
.
Assume, that
A, B
satisfy the conditions (2). We have the following cases:
(13) A2
0−B=−64p2αq2β, B = 1
,
(14) A2
0−B=−64p2α, B =qβ(A2
0−B=−64q2β, B =pα)
,
(15) A2
0−B=−64, B =pαqβ
.
The case
A2
0−B=−1
does not hold, since otherwise
B= 8pαqβ
, which
contradicts the condition for
B
in (2). Let us consider the case (13). We have
A2
0−B=A2
0−1 = (A0−1)(A0+ 1) = −64p2αq2β
, where
A0−1≡2(
mod
4)
,
A0+ 1 ≡0(
mod
32)
and
gcd(A0−1, A0+ 1) = 2
. Hence it is easy to see that
(13) does not produce any elliptic curve.
Let us consider the case (14). We have
A2
0+ 64p2α=qβ
. Consequently,
we obtain the following models of elliptic curves:
Y2=X3±2pqβ−64p2αX2+qβX,
where the sign before the quadratic square is chosen so that it is congruent
to
3
modulo
4
.
Let us consider the case (15). Then we have
A2
0−pαqβ=−64
, and we
obtain the following models of elliptic curves:
Y2=X3±2ppαqβ−64X2+pαqβX,
12

where the sign before the quadratic square is chosen so that it is congruent
to
3
modulo
4
.
Assume that
A, B
satisfy the conditions (3). We have the following cases:
(16) A2−64B0=−p2αq2β, B0= 1
,
(17) A2−64B0=−p2α, B0=qβ(A2−64B0=−q2β, B0=pα)
,
(18) A2−64B0=−1, B0=pαqβ
.
Let us consider the case (16). The corresponding equation
A2+p2αq2β=
64
has no solution (reduce modulo
8
).
The case (17) (resp. the case (18)) leads to
A2+p2α= 64qβ
(resp. to
A2+ 1 = 64pαqβ
), with no solution.
∆>0
,
s= 2α
,
t= 2β+ 1
.
Assuming (2), we obtain
B2(A2
0−B) = 64p2αq2β+1
, and, in consequence:
(19) A2
0−B= 64p2αq2β+1, B = 1
,
(20) A2
0−B= 64q2β+1, B =p2α
.
The conditions (19) lead to the following pairs of equations:
A0−1 = ±2p2α
and
A0+ 1 = ±32q2β+1
,
A0−1 = ±2q2β+1
and
A0+ 1 = ±32p2α
,
A0−1 = ±2p2αq2β+1
and
A0+ 1 = ±32
.
Consequently, if
16q2β+1 −p2α=±1
or
16p2α−q2β+1 =±1
, we obtain
the following models of elliptic curves:
Y2=X3±2p64p2αq2β+1 + 1X2+X,
where the sign before the quadratic square is chosen so that it is congruent
to
3
modulo
4
.
The conditions (20) lead to the following models of elliptic curves:
Y2=X3±2p64q2β+1 ±pαX2±pαX,
where the sign before the quadratic square is chosen so that it is congruent
to
3
modulo
4
.
Assume that
A, B
satisfy the conditions (3). Then we obtain the following
classes of elliptic curves:
Y2=X3±pp2αq2β+1 + 64X2+ 16X
, if
p2α−q2β+1 =±16
,
Y2=X3±pp2αq2β+1 −64X2−16X
,
Y2=X3±pq2β+1 + 64pαX2+ 16pαX
,
Y2=X3±pq2β+1 −64pαX2−16pαX
,
13

where the sign before the quadratic square is chosen so that it is congruent
to
1
modulo
4
.
∆<0
,
s= 2α
,
t= 2β+ 1
. If we assume the conditions (2), then
B= 1
or
B=pα
. The case
B= 1
does not produce any elliptic curve. On the
other hand, the case
B=pα
(and conditions (2)) lead to:
Y2=X3±2ppα−64q2β+1X2+pαX,
where the sign before the quadratic square is chosen so that it is congruent
to
1
modulo
4
.
The conditions (3) lead to the following models of elliptic curves:
Y2=X3−X2+ 16X
,
Y2=X3±p64pα−q2β+1X2+ 16pαX
,
where the sign before the quadratic square is chosen so that it is congruent
to
1
modulo
4
.
∆<0
,
s= 2α+ 1
,
t= 2β+ 1
. In this case
B2(A2−4B) = −28p2α+1q2β+1
.
Of course
B > 0
. The conditions (2) lead to
A2
0−B=−64p2α+1q2β+1 , B = 1,
hence
(A0−1)(A0+ 1) = −64p2α+1q2β+1
, where
A0−1≡2(
mod
4)
and
A0+ 1 ≡0(
mod
32)
. This case does not produce any elliptic curve.
Assume that
A, B
satisfy the conditions (3). Then we obtain
A2−64B0=−p2α+1q2β+1 , B0= 1,
hence
A2+p2α+1q2β+1 = 64
, where
A≡1(
mod
4)
. Consequently, we obtain
the following models of elliptic curves:
Y2=X3+ 5X2+ 16X
,
Y2=X3−3X2+ 16X
,
Y2=X3−7X2+ 16X
.
∆>0
,
s= 2α+ 1
,
t= 2β+ 1
. Assuming (2) we obtain
B2(A2
0−B) =
64p2α+1q2β+1
, hence
A2
0−B= 26p2α+1q2β+1 , B2= 1.
B=−1
leads to
A2
0+ 1 = 64p2α+1q2β+1
with no solution.
14

If
B= 1
, then we have the following possibilities:
A0−1 = ±2
and
A0+ 1 = ±32p2α+1 q2β+1
,
A0−1 = ±2p2α+1
and
A0+ 1 = ±32q2β+1
,
A0−1 = ±2q2β+1
and
A0+ 1 = ±32p2α+1
,
A0−1 = ±2p2α+1q2β+1
and
A0+ 1 = ±32
.
In this case we obtain the family of elliptic curves
Y2=X3±2p64p2α+1q2β+1 + 1X2+X,
if
p2α+1 −16q2β+1 =±1
,
where the sign before the quadratic square is chosen so that it is congruent
to
3
modulo
4
.
Assuming (3) we obtain
B2
0(A2−64B0) = p2α+1q2β+1
, and hence
A2−26B0=p2α+1q2β+1 , B2
0= 1.
If
B0=−1
, then we obtain the family of elliptic curves
Y2=X3±pp2α+1q2β+1 −64X2−16X,
where the sign before the quadratic square is chosen so that it is congruent
to
3
modulo
4
.
If
B0= 1
, then
(A−8)(A+8) = p2α+1q2β+1
, when
gcd(A−8, A+8, pq) = 1
,
A−8≡A+8 ≡1(
mod
4)
. In this case we obtain the family of elliptic curves
Y2=X3±pp2α+1q2β+1 + 64X2+ 16X,
if
p2α+1 −q2β+1 =±16,
where the sign before the quadratic square is chosen so that it is congruent
to
1
modulo
4
.
4.2 Proofs of Theorems 3 - 6
Proof of Theorem 3
. Now
E
has additive reduction at
p
(so
p|∆E
and
p|c4
).
Hence, in this case,
E
has a unique Weierstrass model of the type
Y2=
X3+AX2+BX
, where
A, B ∈Z
with
A=apk
,
B=bpl
,
gcd(p, ab) = 1
,
and
gcd(a, b)=1
. Our assumptions lead to
∆ = 212p2αq2β
. Hence
b2(a2p2k−4bpl) = 28p2α−2kq2β
with the assumptions
gcd(ab, p)=1
, and
gcd(a, b)=1
. We have four cases
to consider: (i)
b2= 1
, (ii)
b2= 28
, (iii)
b2=q2β
, and (iv)
b2= 28q2β
.
15

The case (i) leads to four subcases (with
a0=a/2
):
a2
0−64q2β=±1
(no solution),
a2
0p2m−64q2β=±1
(no solution),
a2
0−64p2mq2β=±1
(no solution),
a2
0−64q2β=pm
(produces the family (i)).
The case (ii) leads to four subcases:
a2±64 = q2β
(produces the family (ii)),
a2p2m±64 = q2β
(produces examples (a), (b)),
a2±64 = p2mq2β
(produces examples (c), (d)),
a2−64p2m=q2β
(produces the family (iii),
The case (iii) leads to four subcases (with
a0=a/2
):
a2
0−64 = ±qβ
(produces the family (ii)),
a2
0p2m−64 = ±qβ
(produces an example (b),
a2
0−64p2m=qβ
(produces the family (iv)),
a2
0−64 = ±pmqβ
(produces the family (v) and examples (e), (f), (g)).
The case (iv) leads to four subcases:
a2−1 = ±64qβ
(produces the family (ii)),
a2p2m−1 = ±64qβ
(produces examples (b), (h)),
a2−p2m= 64qβ
(produces the family (vi)),
a2−1 = ±64pmqβ
(produces the family (vii)).
The equations in (i) - (vii) have discriminants
212p2αq2β
. These equations
are minimal (the conductor is even) or non-minimal at
2
(the conductor is
odd). To decide this we apply Tate's algorithm at
2
, obtaining the conditions
for
±up
,
±2up
, etc.
Proof of Theorem 4
. Follows the same lines as above.
Proof of Theorem 5
. Now
E
has additive reduction at
p
and at
q
(so
pq|∆E
and
pq|c4
). Hence, in this case,
E
has a unique Weierstrass model of
the type
Y2=X3+AX2+BX
, where
A, B ∈Z
with
A=apkql
,
B=bpsqt
,
gcd(pq, ab) = 1
, and
gcd(a, b) = 1
. Our assumptions lead to
∆=212p2αq2β
.
Hence
b2(a2p2kq2l−4bpsqt)=28p2α−2sq2β−2t
16

with the assumptions
gcd(ab, pq) = 1
, and
gcd(a, b) = 1
. We have two cases
to consider: (i)
b2= 1
and (ii)
b2= 28
. The case (i) produces an example
(a), and the families (i), (ii), (iii), and the case (ii) produces examples (a),
(b), (c), (d), and the family (iv). To decide the signs of
±p
and
±upq
, we
use Tate's algorithm at
2
.
Proof of Theorem 6
. Theorem 5 treats the case
∆E=p2αq2β
. Theorem
6 treats the remaining cases, and the proof follows under the same (case by
case) method.
5 Elliptic curves over
Q
with good reduction
outside two odd primes and a rational point
of order
≥3
5.1 Elliptic curves over
Q
with odd conductor
paqb
and
a rational point of order
3
Bennett, Vatsal and Yazdani [3] completely characterized elliptic curves over
Q
that possess both a rational
3
-torsion point and conductor
3aqb
for nonneg-
ative integers
a
and
b
, and
q
a prime
>3
. They proved that such an elliptic
curve
E
is isogenous over
Q
to a curve of the form
Y2+a1XY +a3Y=X3
,
with coecients given by explicit list of values ([3], Prop. 6.1). To do this,
they rst develop machinery to solve ternary Diophantine equations of the
shape
Axn+Byn=Cz3
for various choices of coecients
(A, B, C )
. Next,
they solve (among others) Diophantine equations of the type
xn+3αyn=Cz3
for specic choices of
C
([3], Theorem 1.5). When
q6 |a1
, then there exists a
nonzero integer
x
and a nonnegative integer
m
such that one of the following
occurs: (i)
x3=qn±3m
, (ii)
3mx3=qn±1
, (iii)
x3= 3mqn±1
. They use
Theorem 1.5 to conclude that the largest prime factor of
n
is at most
3
. Such
a reduction is crucial for the proof of Proposition 6.1.
The next task is to characterize elliptic curves over
Q
that possess both
a rational
3
-torsion point and a conductor
paqb
for any dierent odd primes
p, q ≥5
, and
1≤a, b ≤2
. In general, it is a dicult problem. Variants
of the equations (i) - (iii) above lead to new cases of Diophantine equations
Axn+Byn=Cz3
, which are very dicult to solve.
Below we exhibit several families of semistable elliptic curves over
Q
that
17

possess a rational point of order
3
of conductor
pq
, where
p
and
q
are dierent
primes
>3
. First let us observe the following easy result.
Lemma 4
Assume that an elliptic curve (over
Q
)
E:Y2+a1XY +a3Y=
X3
has conductor
pα1
1. . . pαk
k
, with
p1, . . . , pk
dierent primes
≥5
. If
vpi(a1) =
0
for all
i= 1, . . . , k
, then
E
is semistable.
Proof
. We have
c4=a1(a3
1−27a3)
and
∆ = a3
3(a3
1−27a3) = ±pβ1
1. . . pβk
k
.
It is easy to observe that
vpi(a1) = 0
implies
vpi(c4) = 0
, and the assertion
follows.
Theorem 7
The following families give elliptic curves over
Q
of conductor
pq
and a rational point of order
3
.
(i)
Y2+(pk+ 3)XY +Y=X3
, where
p
and
q=p2k+9pk+ 27
are primes
≥5
,
(ii)
Y2+aXY +pY =X3
, where
p
and
q= 27p−a3
are primes
≥5
,
Proof
. The minimal discriminants are
pq
in case (i) and
−p3q
in case (ii),
respectively. In both cases, the point
(0,0)
has order
3
.
Remark.
The families in Theorem 7 have the torsion subgroups isomor-
phic to
Z/3Z
(use the results of Sadek [18] that there are no elliptic curves
over
Q
of conductor odd
pq
and a rational point of order
6
or
9
).
5.2 Elliptic curves over
Q
with odd conductor
paqb
and
a rational point of order
4
Theorem 8
Any elliptic curve
E
over
Q
with odd
NE=pq
and
E(Q)[4] 6=
{0}
is isomorphic (over
Q
) to one of the following curves of the form
Y2=
X3+AX2+BX
:
A B ∆EE(Q)tors A B ∆EE(Q)tors
14 −15 32×52Z/2Z×Z/4Z54 25 54×11 Z/4Z
−2−63 34×72Z/2Z×Z/4Z1 16 −32×7Z/4Z
−34 225 34×54Z/2Z×Z/4Z6 841 −13 ×294Z/4Z
14 625 −32×58Z/8Z62 1 3 ×5Z/4Z
38 169 3 ×134Z/4Z−7 16 −3×5Z/4Z
46 81 38×7Z/8Z
18

or belongs to one of the following families
A B
conditions
∆E
−2(p2m−16) (p2m+ 16)2q=p2m+ 16 −p2mq4
2(16p2m−1) (16p2m+ 1)2q= 16p2m+ 1 −p2mq4
2pm+ 64 p2mpm+ 16 = q2n−1, q ≡3(mod 4) p4mq2n−1
p2m+ 8 16 q=p2m+ 16 p2mq
1+8pm16p2mq2n−1= 16pm+ 1 p4mq2n−1
−2(pm−32) p2mpm=q2n−1+ 16, q ≡1(mod 4) −p4mq2n−1
1−8pm16p2mq2n−1= 16pm−1−p4mq2n−1
where
m, n
are positive integers.
Theorem 9
Any elliptic curve
E
over
Q
with odd
NE=paq
(
a≥2
) and
E(Q)[4] 6={0}
is isomorphic (over
Q
) to one of the following ve curves of
the form
Y2=X3+AX2+BX
with
E(Q)tors 'Z/4Z
:
A B ∆EA B ∆E
−7 784 −32×7765 2704 −3×137
−15 144 −37×13 582 9 37×72
33 1936 −5×117
or belongs to one of the following families of elliptic curves
A B
conditions
∆E
p(p−8) 16p2p= 16 + q2np7q2n
2p(1 + 16p2m−1)p2q2nqn= 16p2m−1−1, p ≡3 (mod 4) p2m+5q4n
p(1 + 8qn) 16p2q2n16qn=p2m−1−1, p ≡1 (mod 4) p2m+5q4n
2p(16p2m−1−1) p2q2nqn= 16p2m−1+ 1, p ≡1 (mod 4) −p2m+5q4n
p(8qn−1) 16p2q2n16qn=p2m−1+ 1, p ≡3 (mod 4) −p2m+5q4n
2p(32p2m−1+ 1) p2q2n−1= 16p2m−1+ 1, p ≡3 (mod 4) p2m+5q2n−1
2p(32p2m−1−1) p2q2n−1= 16p2m−1−1, p ≡1 (mod 4) p2m+5q2n−1
p(p2m−1−8) 16p2p2m−1=q2n−1+ 16 p2m+5q2n−1
p(p2m−1+ 8) 16p2p2m−1=q2n−1−16 p2m+5q2n−1
where
m, n
are positive integers.
Theorem 10
If
E
is an elliptic curve over
Q
with odd
NE=paqb
(
a, b ≥2
)
and
E(Q)[4] 6={0}
, then
E
is isomorphic (over
Q
) to the curve
Y2=X3+
105X2+ 3600X
of the (minimal) discriminant
∆ = −37×57
.
19

Proof
. For the proofs of Theorems 8, 9 and 10, we consider all elliptic
curves listen in Theorems 1 - 6, and check whether exists a point
P∈E(Q)
such that
2P
has order two or not. Let
E:Y2=X3+AX2+BX
. Existence
of a point
P∈E(Q)
such that
2P= (0,0)
leads to the conditions (and is
equivalent to) (i)
B
is a square of an integer, (ii) at least one of the numbers
A±2√B
is a square of an integer. If
E
has three points of order two (
(0,0)
,
P1
,
and
P2
, say), then we have additionally to consider the equations
2P=Pi
,
i= 1,2
.
5.3 Elliptic curves over
Q
with odd conductor
paqb
and
a rational point of order
5
Theorem 11
If
E
is an elliptic curve over
Q
with (possibly even)
NE=paqb
and
E(Q)[5] 6={0}
, then
E
is isomorphic to one of the following curves of
the form
Y2+a1XY +a3Y=X3+a2X2
with
E(Q)tors 'Z/5Z
:
a1a2a3NEa1a2a3NE
a
3 2 4 38
g
8 7 49 203
b
5 4 16 58
h
1−2−4 50
c
9 8 64 50
i
6−7−49 175
d
4 3 9 75
j
14 13 169 325
e
10 9 81 57
k
38 37 1369 1147
f
6 5 25 155
or belongs to one of the following two families of elliptic curves with odd
conductor
pq
(
p6= 5
)
a1a2a3
conditions
∆E
l
qv−1−qv−q2vp2k+1 =q2v+ 11qv−1
,
k, v ≥0−p2k+1q5v
m
qv+ 1 qvq2vp2k+1 =q2v−11qv−1
,
k, v ≥0p2k+1q5v
We will use the following elementary result.
Lemma 5
The equation
1 + 11qv−q2v=pu
has exactly seven solutions in
positive integers
u, v
and primes
p, q
:
q v p u q v p u
2 1 19 1 3 2 19 1
2 2 29 1 5 1 31 1
2 3 5 2 7 1 29 1
3 1 5 2
20

Proof
. Easy calculations.
Proof of Theorem 11
. Less explicit version of this theorem is also given
in [18]; below we give independent and short proof of this result.
Any elliptic curve over
Q
, with a rational
5
-torsion point is isomorphic to
a curve
E:Y2+(1 −c)XY −cY =X3−cX2
, where
c
is any nonzero rational
number [11]. Writting
c=a/b
, (
a, b ∈Z
,
b > 0
,
gcd(a, b)=1
) we obtain
E:Y2+(b−a)XY −ab2Y=X3−abX2
. We have
∆E=a5b5(a2−11ab−b2)
.
Assume
∆E=±pαqβ
. The case
a2−11ab −b2=±1
does not produce
any elliptic curve of conductor
pnqm
. The remaining case we divide into two
subcases.
(i)
a= 1
,
b=qv
,
1−11qv−q2v=±pu
. Of course,
1−11qv−q2v<0
.
Assuming
p= 5
, we deduce
u= 2
,
q= 2
,
v= 1
or
u= 3
,
q= 7
,
v= 1
,
and obtain the elliptic curves
(h)
and
(i)
.
Assume
p6= 5
. It turns out that the equation
q2v+ 11qv−pu−1 = 0
has
no solution in primes
p6= 5
and
q
, if
u= 2k
. The case
u= 2k+ 1
leads to
the family
(l)
.
(ii)
a=−1
,
b=qv
,
1+11qv−q2v=±pu
. The equation
1+11qv−q2v=pu
has exactly seven solutions with
u, v ≥1
and
p, q
primes (cf. Lemma 5). The
corresponding elliptic curves are given by
(a)
,...,
(g)
.
Consider the equation
1 + 11qv−q2v=−pu
. Assuming
p= 5
, we deduce
u= 2
,
q= 13
,
v= 1
; the corresponding elliptic curve is
(j)
. Now assume
p6= 5
. It turns out that the equation
q2v−11qv−1−pu= 0
has no solution
in primes
p, q
,
p6= 5
, if
u= 2k+ 2
. The case
u= 2
leads to the elliptic curve
(k)
:
p= 31
,
q= 37
,
u= 2
,
v= 1
. The case
u= 2k+ 1
leads to the family
(m)
.
We have
c4=q4v+ 12q3v+ 14q2v−12qv+ 1 = p4k+2 −10qvp2k+1 + 5q2v
in
case
(l)
, and
c4=q4v+−2q3v+ 14q2v+ 12qv+ 1 = p4k+2 + 10qvp2k+1 + 5q2v
in case
(m)
. Therefore, if
p6= 5
, then the elliptic curves from families
(l)
and
(m)
have multiplicative reductions at
p
and
q
.
Remark.
We expect that the family
(l)
consists of the curves with
k= 0
only. Similarly for the family
(m)
. It leads to the following question, which
may be of independent interest.
Question.
Let
p
be an odd prime, dierent from
5
. Does the equation
t2−125 = 4p2k+1
has any solution in positive integers
t, k
?
21

5.4 Elliptic curves over
Q
with odd conductor
paqb
and
a rational point of order
≥6
From the results of Sadek [18] it follows that there are no elliptic curves over
Q
of odd conductor
paqb
and a rational point of order
≥6
. On the other hand,
it is easy to check that, for instance, the curves
A:Y2=X3+ 14X2+ 625X
(with
∆ = −32×58
) and
B:Y2=X3+ 46X2+ 81X
(with
∆ = 38×7
)
both have a rational point of order
8
. We have checked, using Theorems 8 -
10, that these are the only elliptic curves over
Q
of odd conductor
paqb
and
a rational point of order
8
.
Proof of Theorem 3.7 in [18] is erroneous, since the Weierstrass equations
considered there are, in general, not minimal. For instance, the curve
A
has
non-minimal model
y2−14xy −120y=x3−20x2
(i.e.
s= 1
,
t= 6
), and in
this case the proof in [18] doesn't work. The same for the second curve
B
.
6 Existence and non-existence of elliptic curves
over
Q
with certain odd conductors
paqb
The following results by Howe [10] concern the problem of existence/non-
existence of elliptic curves over
Q
with certain odd conductors
pq
, and gener-
alize Theorems 1 and 3 from [19]. Proofs of these results use the methods of
Ogg [16] [17] and Setzer [19], applying in particular basic class eld theory.
Theorem 12
([10], Theorem 4.8) Let
N=pq
, with
p6=q
, and
p, q ≡
±1(mod 8)
. Assume that the class numbers of the quadratic elds
Q(ñp)
,
Q(ñq)
, and
Q(ñpq)
are not divisible by
3
. Then any elliptic curve over
Q
of conductor
N
has a rational point of order
2
.
Theorem 13
([10], Theorem 6.1) Assume that
p
,
q
are dinstinct primes
satisfying
p≡7(mod 16)
,
q≡15(mod 16)
. Assume furthermore that the
class numbers of the quadratic elds
Q(ñp)
,
Q(ñq)
, and
Q(ñpq)
are
not divisible by
3
. Then there are no elliptic curve over
Q
of conductor
pq
.
Edixhoven, de Groot and Top in [8] considered elliptic curves over
Q
with bad reduction at only one prime
p
and proved
inter alia
that for
p≡
5 (mod 12)
any such a curve with conductor
p2
is a twist of one with conductor
p
. Here we prove analogous result for elliptic curves with conductor
p2q2
.
22

Let
p
and
q
be distinct prime numbers greater than 3. Given an elliptic
curve over the rationals with additive reduction at
p
and
q
, and good reduc-
tion at all other primes (i.e. with the conductor
p2q2
), one can twist it over
quadratic extension of
Q
unramied outside
{p, q}
. The discriminant of the
resulting elliptic curve is a product of powers of
p
and
q
as well. Moreover,
we have the following
Lemma 6
Let
p, q ≥5
be distinct primes and
K
be the quadratic extension
of
Q
unramied outside
{p, q}
. Then up to quadratic twist over
K
all elliptic
curves over
Q
with good reduction away from
p
and
q
are
(i) the ones with multiplicative reduction at
p
and
q
,
(ii) the ones with multiplicative reduction at
p
and additive reduction at
q
or vice versa,
(iii) the ones with discriminant
±piqj
, where
i, j ∈ {2,3,4}
.
Proof
. We proceed similarly as in the proof of Lemma 1 in [8]. We omit
the details.
Now we present an auxiliary diophantine result.
Lemma 7
If
p
and
q
are distinct primes congruent to
5
modulo
12
and
p
q= 1
, then the diophantine equation
(21) pqx3−y2=±1728psqt
, where
s, t ∈ {0,1,2}
has no integer solution
(x, y)
with
y6= 0
.
Proof
. Since
p≡5 (mod 12)
, by quadratic reciprocity laws we get
−1 =
(2
3)=(p
3)=(3
p)
, and the same holds for
q
. Consequently,
±1728
p=
±1728q
p=±1728
q=±1728p
q=−1
. If (21) has a solution for
s= 0
,
then
−y2≡ ±1728qt(mod p)
, so
±1728qt
must be a square modulo
p
, but
it is not true. If
t= 0
, then we obtain a contradiction by the same way.
Assume now that (21) has a solution
(x, y)
for
s= 2
. Then
p
divides
y
, and
so
p
divides
x
too. Putting
x0=x/p
,
y0=y/p
, we get
p2qx03−y02=±1728qt
.
Hence again
±1728qt
must be a square modulo
p
which is impossible. The
similar argument holds for
t= 2
. Now assume that
pqx3−y2=±1728pq
has a solution
(x, y)
. Then
y=pqy0
for some integer
y0
. Consequently
23

pqy02=x3∓1728
, and we obtain the rational ane point on the quadratic
twist of the elliptic curve
Y2=X3+ 1
by
∓3pq
. Nagell [15] showed that if
an integer
d
is not divisible by primes congruent to
±1
or
7
modulo
12
then
the only rational solution of
dy2=x3+ 1
is the one with
y= 0
. Therefore
the assertion follows.
The above two lemmata imply:
Theorem 14
If
p
and
q
are distinct primes congruent to
5
modulo
12
and
p
q= 1
, then every elliptic curve over
Q
with conductor
p2q2
is a twist of
one with conductor
pq
,
p2q
or
pq2
.
Proof
. Suppose that such elliptic curve do not come from the ones with
multiplicative reduction at
p
or at
q
. Then by Lemma 6, it has discriminant
∆ = ±piqj
, where
i, j ∈ {2,3,4}
. Moreover, the invariants
c4
and
c6
satisfy
the relation
c3
4−c2
6= 1728∆
, and by assumption
p
and
q
divide
c4
. Conse-
quently also
p
and
q
divide
c6
, and we obtain equation (21). But by Lemma
7, this equation has no nontrivial integral solution, and we are done.
7 Upper bounds for the ranks
Here we present some information about upper bounds for ranks of elliptic
curves
E
dened over
Q
of odd conductor
NE=paqb
(
a, b ≥1
) and with a
Q
-rational point of order
2
. If
E
has three
Q
-rational points of order
2
and
NE=pq
, we also compute the coecients
ap(E)
and
aq(E)
explicitly (where
al(E) := l+ 1 −#E(Fl)
for any prime
l
, and any elliptic curve
E
over
Q
),
and consequently we (conjecturally) determine the ranks. We start with the
following
Lemma 8
Let
E
be an elliptic curves over
Q
with a
Q
-rational point of order
2
. Let
m
and
n
denote the number of primes of multiplicative and additive
reduction of
E
respectively. Then
rank (E(Q)) ≤m+ 2n−1
.
Proof
. It follows by descent via
2
-isogeny. See also ([1], Prop. 1.1).
We immediately obtain the following corollaries.
Corollary 1
Any elliptic curve
E
over
Q
of conductor
pq
and with a
Q
-
rational point of order
2
has rank
≤1
.
24

Corollary 2
Any elliptic curve
E
over
Q
of conductor
paq
(
a > 1
) and with
a
Q
-rational point of order
2
has rank
≤2
.
Corollary 3
Any elliptic curve
E
over
Q
of conductor
paqb
(
a, b > 1
) and
with a
Q
-rational point of order
2
has rank
≤3
.
Now we restrict to the case when
NE=pq
and
E
has three
Q
-rational
points of order
2
(i.e., to the curves from Theorem 1). It is easy to check
using Magma (or nd in Cremona online tables [5]) that the elliptic curves
a)-d) from Theorem 1 all have rank 0. For the elliptic curves given by e) and
f) we have the following results which allow us to compute their ranks under
the parity conjecture.
Proposition 1
Let
E1:Y2=X3±2p64 + pαqβX2+pαqβX
, where
pα−
qβ=±16
, and
E2:Y2=X3±2p64p2α+qβX2+qβX
, where
qβ−16pα=
,
be the curves given by e) and f) in Theorem 1. Then
ap(E1) = 1
if
p≡1 (mod 4)
or
α
is even
−1
if
p≡3 (mod 4)
and
α
is odd,
aq(E1) = 1
if
q≡1 (mod 4)
or
β
is odd
−1
if
q≡3 (mod 4)
and
β
is even,
ap(E2)=1,
aq(E2) = 1
if
q≡1 (mod 4)
−1
if
q≡3 (mod 4) .
Proof
. First consider the family
E1:Y2=X3+AX2+BX
, where
A= 2ηp64 + pαqβ
,
B=pαqβ
,
η=±1
and
ηp64 + pαqβ≡3 (mod 4)
.
Without loss of generality
pα−qβ= 16
(in the case
pα−qβ=−16
we
switch
p
and
q
). Note that
A= 2η(pα−8) = 2η(qβ+ 8)
, in particular
ηpα≡ηqβ≡3 (mod 4)
. Hence the curve
E1
after reduction modulo
p
has
the form
Y2=X2(X−16η)
. Then
#E1(Fp) = 1 +
p−1
X
x=0 1 + x2(x−16η)
p
25

= 1 + p+
p−1
X
x=1 x−16η
p= 1 + p−−η
p
,
hence
ap(E1) = −η
p
. Similarly,
E1
modulo
q
is
Y2=X2(X+ 16η)
, so
aq(E1) = η
q
. Since
ap(E1) = 1 ⇔η=−1
or
η= 1
and
p≡1 (mod 4)
,
we obtain the above formula for
ap(E1)
. In similar way, we get the formula
for
aq(E1)
. Now consider the curve
E2
i.e.,
A= 2ηp64p2α+εqβ
and
B=
εqβ
, where
qβ−16pα=ε=±1
. Then
A= 2η(8pα+ε) = ηqβ+ε
, in
particular
ηε ≡3 (mod 4)
hence
ε=−η
. Then reducing modulo
p
we have
Y2=X(X−1)2
, so
ap(E2)=1
, and modulo
q
we obtain
Y2=X2(X−1)
,
hence
aq(E2) = −1
q
, and the assertion follows.
Corollary 4
Let
w(Ei) = ±1
denote the global root number of
Ei
(
i= 1,2
).
Then
w(E1) = −ap(E1)aq(E1) = −1
if and only if
q≡1 (mod 4)
, and
w(E2) = −aq(E2) = −1
if and only if
q≡1 (mod 4)
. Therefore under the
parity conjecture
rank (E1(Q)) = 0
if
q≡3 (mod 4)
,
1
if
q≡1 (mod 4)
.
rank (E2(Q)) = 0
if
q≡3 (mod 4)
,
1
if
q≡1 (mod 4)
.
In particular, the curve
E2
with
ε=−1
has always rank zero.
References
[1] J. Aguirre, Á. Lozano-Robledo, J.C. Peral,
Elliptic curves of maximal
rank
, Revista Matemática Iberoamericana, Proceedings of the confer-
ence Segundas Jornadas de Teoria de Números, (2010), 1-28
[2] C. F. Barros,
On the Lebesgue-Nagell equation and related subjects
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Andrzej D¡browski, Institute of Mathematics, University of Szczecin, 70-
451 Szczecin, Poland,
e-mail: dabrowskiandrzej7@gmail.com and andrzej.dabrowski@usz.edu.pl
Tomasz J¦drzejak, Institute of Mathematics, University of Szczecin, 70-
451 Szczecin, Poland,
e-mail: tjedrzejak@gmail.com
28