2.7. The rank and the free part of E(Q) 45 (2) For all P ∈ E(Q) and m ∈ Z, h(mP ) = m2·h(P ). (Note: in particular, the height of mP is much larger than the height of P , for any m = 0,1.) (3) Let P ∈ E(Q). Then h(P ) ≥ 0, and h(P ) = 0 if and only if P is a torsion point. For the proofs of these properties, see [Sil86], Ch. VIII, Thm. 9.3, or [Mil06], Ch. IV, Prop. 4.5 and Thm. 4.7. As we mentioned at the beginning of this section, we can calculate upper bounds on the rank of a given elliptic curve (see [Sil86], p. 235, exercises 8.1, 8.2). Here is an example: Theorem 2.7.4 ([Loz08], Prop. 1.1). Let E/Q be an elliptic curve given by a Weierstrass equation of the form E : y2 = x3 + Ax2 + Bx, with A, B ∈ Z. Let RE be the rank of E(Q). For an integer N ≥ 1, let ν(N) be the number of distinct positive prime divisors of N . Then RE ≤ ν(A2 − 4B) + ν(B) − 1. More generally, let E/Q be any elliptic curve with a non-trivial point of 2-torsion and let a (resp. m) be the number of primes of additive (resp. multiplicative) bad reduction of E/Q. Then RE ≤ m + 2a − 1. Example 2.7.5. Pierre de Fermat proved that n = 1 is not a con- gruent number (see Example 1.1.2) by showing that x4 + y4 = z2 has no rational solutions. As an application of the previous theorem, let us show that the curve E1 : y2 = x3 − x = x(x − 1)(x + 1) only has the trivial solutions (0,0), (±1,0), which are torsion points of order 2. Indeed, the minimal discriminant of E1 is ΔE 1 = 64. Therefore p = 2 is the unique prime of bad reduction. Moreover, the reader can check that the reduction at p = 2 is multiplicative. Now thanks to Theorem 2.7.4 we conclude that RE 1 = 0 and E1 only has torsion points. Finally, using Proposition 2.6.15 or Theorem 2.5.5, we can show that the only torsion points are the three trivial points named above.

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