Consider the following elliptic curve
$y^2=(x+1540)(x-508)(x-65024)$.
It is trivial that the points $P_1(-1540,0)$, $P_2(508,0)$ and $P_3(65024,0)$ lie on this curve. It is also quite easy to find four other integer points $P_4(-508, 262128)$, $P_5(-508, -262128)$, $P_6(130556, 33552384)$ and $P_7(130556, -33552384)$.
I want to find some other rational points on this curve. If one uses the usual group law we obtain that for every $1\le i \le j \le 7$ we have $P_i+P_j=P_k$ for some $1\le k\le 7$. Thus we fail to obtain any new points.
Background info: I am reading a paper by Ajai Choudhry, Equal sums of seventh powers. Rocky Mountain J. Math. 30 (2000), no. 3, 849–852. The entire proof relies on finding a rational point on the curve above. Choudhry finds one such point $(x,y)$ with the denominator of $x$ being $12$ digits long. He does not explain how he achieved this discovery. Are there any other simpler rational points? Any suggestion would be appreciated.
As a background to the question, Choudhry showed that the eqn,
$$x_1^7 + x_2^7 + \dots + x_n^7 = 0$$
has an infinite number of primitive integer solutions for $n=9$. He used the polynomial identity,
$$(x + a)^7 + (x - a)^7 + (m x + b)^7 + (m x - b)^7 + (-x - c)^7 + (-x + c)^7 +\\ (-m x - d)^7 + (-m x + d)^7 + (-14(a^6 + m b^6 - c^6 - m d^6)N)^7 = 0\tag1$$
where,
$$x =14^6(a^6 + m b^6 - c^6 - m d^6)^6N^7$$
for arbitrary $N$ though one must solve the simultaneous equations,
$$\begin{aligned} a^2+m^5b^2 &= c^2+m^5d^2\\ a^4+m^3b^4 &= c^4+m^3d^4 \end{aligned}\tag2$$
In appropriate cases, this can be reduced to an elliptic curve. For $m=2$, Choudhry found the smallest integer solutions $a,b,c,d$ as the 33-digit numbers,
$$292565171139318137956759657471297,\\ 863420822620431936290192229011966,\\ 534407060429869176086407612538177,\\ 859793943610761912321826231621886$$
implying the coefficients of $(1)$ are as high as $x \approx (28d^6)^6 \approx 10^{1180}$.
Dave Rusin reduced the system $(2)$ to an elliptic curve. Let $q=(1-m^7)/(1+m^7)$, then,
$$U(U+1)(U+q^2)=V^2\tag3$$
With $(2)$ being the case $m=2$, so $q=-127/129$. For simplicity, let $U=u/129^2,\;V=v/129^3$ and we get,
$$u(u+127^2)(u+129^2)=v^2\tag4$$
The symmetry of $(4)$ allows that for non-zero $u_k$, then,
$$u_{k+1}=\frac{(127\cdot129)^2}{u_k}\tag5$$
is also a solution. Excluding the torsion points, there are eight "small" solutions (with $u_5$ courtesy of Jeremy Rouse in this MO post),
$$u_1 =\small \Bigl(\frac{\color{blue}{77684960}\sqrt{129}}{79\cdot12497}\Bigr)^2=-129^2+\Bigl(\frac{\color{blue}{78490049}\sqrt{129}}{79\cdot12497}\Bigr)^2$$
$$u_3 =\small -\Bigl(\frac{891195649\sqrt{129}}{78490049}\Bigr)^2=-129^2+\Bigl(\frac{15796208\sqrt{258}}{78490049}\Bigr)^2$$
$$u_5 =\small \Bigl(\frac{\color{green}{224312161}\sqrt{127}}{1079\cdot24423}\Bigr)^2=-127^2+\Bigl(\frac{\color{green}{372170768}\sqrt{127}}{1079\cdot24423}\Bigr)^2$$
$$u_7 =\small -\Bigl(\frac{264770431\sqrt{127}}{23260673}\Bigr)^2=-127^2-\Bigl(\frac{26352417\sqrt{254}}{23260673}\Bigr)^2$$
and distinct $u_2,\,u_4\,u_6,\,u_8$ derived by using $(5)$. Note that,
$$\color{blue}{77684960}+\color{blue}{78490049} = 12497^2$$
$$\color{green}{224312161}+\color{green}{372170768} = 24423^2$$
P.S. The first four $u_i$, after removing common factors, yield the same $a,b,c,d$ as above. Why are the $u_i$ seem so "structured"?
Edit: (Jan 2018) There are other "smallish" solutions,
$$u_9 = \small\Bigl(\frac{3399461793\sqrt{127}}{224312161}\Bigr)^2=-127^2+\Bigl(\frac{4236326896\sqrt{127}}{224312161}\Bigr)^2=-129^2+\Bigl(\frac{48010029072}{224312161}\Bigr)^2$$
$$u_{10} = \small\Bigl(\frac{125382401\sqrt{129}}{77684960}\Bigr)^2=-129^2+\Bigl(\frac{891195649\sqrt{129}}{77684960}\Bigr)^2=-127^2+\Bigl(\frac{9968236223}{77684960}\Bigr)^2$$