Deje $m$ $n$ ser relativamente primer enteros y dejar$$A = |(m^3 + 32n^3)m|,\text{ }B = |4(m^3 - 4n^3)n|.$$Denote by $D$ the greatest common divisor of $$ and $B$. In order to show the inequality in question, it suffices to show that $D$ is a divisor of $144$. For, if that is the case and if the $x$-coordinate of $P$ is given by $m/n$ ($n \neq 0$) in lowest terms, then we have$$H(x\text{-coordinate of }2P) = H\left({A\over{B}}\right) = {1\over{D}}\max(A, B)$$$$\ge {1\over{D}}\max(m, n)^4 = {1\over{D}}H(x\text {coordenada de }P)^4.$$Let $p$ be a prime number. We have$$\text{ord}_p(D) = \min(\text{ord}_p(A), \text{ord}_p(B)),$$since $\texto{ord}_p$ 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 $$. If $p$ is a prime factor of $D$ and $p \neq 2$, then $p$ does not divide $m$, since if $p \neq 2$ and $p$ divides $m$, then $p$ does not divide $B$. Thus, if $p$ is a prime factor of $D$ and $p \neq 2$, then we have$$\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 de$2.
A continuación, se consideran $\text{ord}_2(D)$. Si $m$ es impar, entonces $\text{ord}_2(A) = 0$. Si $m$ es incluso, a continuación, $\text{ord}_2(m^3 - 4n^3) = 2$ desde $n$ es incluso. Por lo tanto, tenemos $\text{ord}_2(B) = 4$. Por lo tanto, $D$ es un divisor de a $2^4 \times 3^2 = 144$, y por lo tanto la desigualdad en cuestión está probado.
Si $r \ge 6$, luego tenemos a la desigualdad de $r^4/144 > r$. Si un punto racional $P$ en la curva elíptica en la ecuación satisface$$H(x\text{-coordinate of }P) \ge 6,$$then we have$$H(x\text{-coordinate of }P) > H(x\text{-coordinate of }2P).$$Then the height of the $x$-coordinate of$$P,\text{ }2P,\text{ }4P,\text{ }8P,\text{ }16P, \dots$$son todos diferentes. Por lo tanto, estos puntos son todas diferentes. Esto significa que hay un número infinito de puntos racionales en esta curva elíptica.
Para ser más precisos, se puede mostrar lo siguiente. Si $m$ $n$ satisfacer $m \not\equiv 0 \text{ (mod }3\text{)}$ o $n \not\equiv 0 \text{ (mod }3\text{)}$, luego$$m^3 - 4n^3 \not\equiv 0 \text{ (mod }9\text{)}.$$This can be done by checking all the possibilities of $0 \le m \le el 8$, $0 \le n \le el 8$. 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 $(5, 11)$.
