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 .
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