42 2. Elliptic curves

does not divide 13, it follows that E(Q)[5] must be trivial. Similarly,

one can show that E(Q)[7] is trivial, and we conclude that E(Q)torsion

is trivial.

However, notice that P = (1,2) ∈ E(Q) is a point on the curve.

Since we just proved that E does not have any points of finite order,

it follows that P must be a point of infinite order, and, hence, we have

shown that E has infinitely many rational points: ±P, ±2P, ±3P,....

In fact, E(Q)

∼

= Z and (1,2) is a generator of its Mordell-Weil group.

In the previous example, the Nagell-Lutz theorem (Theorem 2.5.5)

would have yielded the same result, i.e., the torsion is trivial, in an

easier way. Indeed, for the curve E :

y2

=

x3

+ 3, the quantity

4A3 +27B2

equals

35,

so the possibilities for y(P

)2,

where P is a tor-

sion point of order ≥ 3, are 1, 9 or 81 (it is easy to see that there are no

2-torsion points). Therefore, the possibilities for x(P

)3

= y(P

)2

− 3

are −2, 6 or 78, respectively. Since x(P ) is an integer, we reach a con-

tradiction. In the following example, the Nagell-Lutz theorem would

be a lengthier and much more tedious alternative, and Proposition

2.6.15 is much more effective.

Example 2.6.17. Let E/Q :

y2

=

x3

+ 4249388. In this case

4A3

+

27B2

=

24

·

33

·

112

·

132

·

172

·

192

·

232.

Therefore,

4A3

+

27B2

is divisible by 192 distinct positive squares,

which makes it very tedious to use the Nagell-Lutz theorem. The

(minimal) discriminant of E/Q is ΔE = −16(4A3 +27B2) and there-

fore E has good reduction at 5 and 7. Moreover, B = 4249388 ≡

3 mod 35 and therefore, by our calculations in Example 2.6.16, N5 = 6

and N7 = 13. Thus, Proposition 2.6.15 and the same argument we

used in Ex. 2.6.16 show that the torsion of E(Q) is trivial.

Incidentally, the curve E/Q : y2 = x3 + 4249388 has a rational

point P =

(

25502

169

,

6090670

2197

)

. Since the torsion of E(Q) is trivial, P

must be of infinite order. Another way to see this: since P has

rational coordinates that are not integral, the Nagell-Lutz theorem

implies that the order of P is infinite. In fact, E(Q) is isomorphic to

Z and it is generated by P .