Infinitely rational points in $y^2 = x^3 - 4$? If the $x$-coordinate of a rational point $P$ of $y^2 = x^3 - 4$ is given by $m/n$, the $x$-coordinate of $2P$ is given by$${{(m^3 + 32n^3)m}\over{4(m^3 - 4n^3)n}}.$$Using this fact, how do I show that$$144 \cdot H(x\text{-coordinate of }2P) \ge H(x\text{-coordinate of }P)^4.$$Using this fact, can we conclude that there exist infinitely many rational points in $y^2 = x^3 - 4$?
EDIT: Here, let $a$ be a rational number and write $a = m/n$ as a fraction in lowest terms. Define the height $H(a)$ to be $\max(|m|, |n|)$.
 A: Let $m$ and $n$ be relatively prime integers and let$$A = |(m^3 + 32n^3)m|,\text{ }B = |4(m^3 - 4n^3)n|.$$Denote by $D$ the greatest common divisor of $A$ and $B$. In order to show the inequality in question, it suffices to show that $D$ is a divisor of $144$. For, if that is the case and if the $x$-coordinate of $P$ is given by $m/n$ ($n \neq 0$) in lowest terms, then we have$$H(x\text{-coordinate of }2P) = H\left({A\over{B}}\right) = {1\over{D}}\max(A, B)$$$$\ge {1\over{D}}\max(m, n)^4 = {1\over{D}}H(x\text{-coordinate of }P)^4.$$Let $p$ be a prime number. We have$$\text{ord}_p(D) = \min(\text{ord}_p(A), \text{ord}_p(B)),$$since $\text{ord}_p$ indicates how many times $p$ divides the number. If $p$ is a prime factor of $D$, then $p$ does not divide $n$, since if it does, $p$ does not divide $m$, and thus $p$ does not divide $m^3 + 32n^3$ and $A$. If $p$ is a prime factor of $D$ and $p \neq 2$, then $p$ does not divide $m$, since if $p \neq 2$ and $p$ divides $m$, then $p$ does not divide $B$. Thus, if $p$ is a prime factor of $D$ and $p \neq 2$, then we have$$\text{ord}_p(D) = \min(\text{ord}_p(m^3  32n^3),\text{ord}_p(m^3 - 4n^3))$$$$\le \text{ord}_p((m^3 + 32n^3) - (m^3 - 4n^3))$$$$=\text{ord}_p(36n^3) = \text{ord}_p(36).$$Hence, we have $p = 3$ and $\text{ord}_3(D) \le 2$.
Next, we consider $\text{ord}_2(D)$. If $m$ is odd, then $\text{ord}_2(A) = 0$. If $m$ is even, then $\text{ord}_2(m^3 - 4n^3) = 2$ since $n$ is even. Hence, we have $\text{ord}_2(B) = 4$. Therefore, $D$ is a divisor of $2^4 \times 3^2 = 144$, and thus the inequality in question is proved.
If $r \ge 6$, then we have the inequality $r^4/144 > r$. If a rational point $P$ on the elliptic curve in equation satisfies$$H(x\text{-coordinate of }P) \ge 6,$$then we have$$H(x\text{-coordinate of }P) > H(x\text{-coordinate of }2P).$$Then the height of the $x$-coordinate of$$P,\text{ }2P,\text{ }4P,\text{ }8P,\text{ }16P, \dots$$are all different. Thus, these points are all distinct. This means there are infinitely many rational points on this elliptic curve.
To be more precise, we can show the following. If $m$ and $n$ satisfy $m \not\equiv 0 \text{ (mod }3\text{)}$ or $n \not\equiv 0 \text{ (mod }3\text{)}$, then$$m^3 - 4n^3 \not\equiv 0 \text{ (mod }9\text{)}.$$This can be done by checking all the possibilities of $0 \le m \le 8$, $0 \le n \le 8$. Thus we can see that $D$ is a divisor of $2^4 \times 3 = 48$ and that$$48 \cdot H(x\text{-coordinate of }2P) \ge H(x\text{-coordinate of }P)^4.$$If $r \ge 4$, then $r^4/48 > r$. Thus, if $P = (5, 11)$, then the $x$-coordinates of$$P,\text{ }2P,\text{ }4P,\text{ }8P, \dots$$all have different heights. This implies that we see the existence of infinitely many rational points as soon as we find one rational point $(5, 11)$.
