Consider the following elliptic curve


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.

  • $\begingroup$ It can't be that paper. There's no proof yet that $x_1^7+x_2^7+\dots+x_8^7 = 0$ has an infinite number of primitive solutions. $\endgroup$ – Tito Piezas III Dec 23 '14 at 5:12
  • $\begingroup$ You are right Tito, I mentioned the wrong article. The relevant paper is On Sums of Seventh Powers, Journal of Number Theory 81(2000), 266-269. My initial question was how did the author found the values of a, b, c, d at the bottom of page 267? I became familiar to Choudhry's work only very recently; I like his results quite a bit, but his style is usually very terse, and he skips a lot of details. Hope this clarifies things. Thanks for noticing. $\endgroup$ – Dan Ismailescu Dec 23 '14 at 12:08
  • $\begingroup$ I have read some of Choudhry's papers on equal sums of powers, which are all very interesting, but the terseness is certainly recognizable. In the mentioned paper how do the values of $a,b,c$ and $d$ relate to the elliptic curve above? If there is such a connection, can you clarify it? $\endgroup$ – Jesper Petersen Dec 23 '14 at 12:39
  • $\begingroup$ @DanIsmailescu: I realized that given your elliptic curve, $$(x + 1540)(x - 508)(x - 65024)=y^2$$ the transformation $x =4u+65024,\;y=8v$ transforms it to, $$u(u + 127^2)(u +129^2)=v^2$$ which is Rusin's version mentioned below. I should have seen it earlier. $\endgroup$ – Tito Piezas III Dec 27 '14 at 21:31

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,

$$(x + a)^7 + (x - a)^7 + (m x + b)^7 + (m x - b)^7 + (-x - c)^7 + (-x + c)^7 +\\ (-m x - d)^7 + (-m x + d)^7 + (-14(a^6 + m b^6 - c^6 - m d^6)N)^7 = 0\tag1$$


$$x =14^6(a^6 + m b^6 - c^6 - m d^6)^6N^7$$

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,


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,


The symmetry of $(4)$ allows that for non-zero $u_k$, then,


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,

$$u_9 = \small\Bigl(\frac{3399461793\sqrt{127}}{224312161}\Bigr)^2=-127^2+\Bigl(\frac{4236326896\sqrt{127}}{224312161}\Bigr)^2=-129^2+\Bigl(\frac{48010029072}{224312161}\Bigr)^2$$

$$u_{10} = \small\Bigl(\frac{125382401\sqrt{129}}{77684960}\Bigr)^2=-129^2+\Bigl(\frac{891195649\sqrt{129}}{77684960}\Bigr)^2=-127^2+\Bigl(\frac{9968236223}{77684960}\Bigr)^2$$

  • $\begingroup$ @JesperPetersen: In the link that I cited, Rusin mentions that the elliptic curve $$u\bigl(u+(n^7-1)^2\bigr)\bigl(u+(n^7+1)^2\bigr)=v^2$$ other than $n=2$, also has rational points for $n=5,6$. That was 5 years ago. Is it now feasible for Sage to search for those points, or not yet? $\endgroup$ – Tito Piezas III Dec 26 '14 at 22:07
  • $\begingroup$ For the case $n=2$ the generator of the free part is returned almost instantly, which is not the case for $n = 5,6$. As would be expected. How long it would take to find the generator I don't know how to estimate. I am wondering what sparks the interest in the more general case. Obviously, a point for $n > 2$ would unearth solutions to the sum-of-seventh-powers equation not mentioned in Choudhry's paper, but are there any other implications? $\endgroup$ – Jesper Petersen Dec 27 '14 at 9:59
  • $\begingroup$ @JesperPetersen: Other than knowing that it is doable? I guess it is the same curiosity why I wanted to see a third elliptic curve for $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ in this MO post. But thanks for the try. :) $\endgroup$ – Tito Piezas III Dec 27 '14 at 16:17
  • $\begingroup$ Doesn't the claim that $x_1^7+\cdots+ x_9^7=0$ has infinite many integral primitive solutions not contradict Falting's theorem ? $\endgroup$ – Peter Mar 8 '17 at 8:39
  • $\begingroup$ @Peter: It is not just a claim, but Choudhry showed that $x_1^7 + \dots + x_9^7 = 0$ has infinitely many primitive integer solutions, that is, the $x_i$ have $\gcd(x_i) = 1$. (The actual polynomial identity is given above.) I do not see how Faltings's theorem is applicable to this particular Diophantine equation. There are even $8$th deg and $10$ deg similar versions of this. $\endgroup$ – Tito Piezas III Mar 8 '17 at 12:50

EDIT: I had a typo in my first post. The conclusion is now different.

As suggested by Amzoti, you can use Sage freely at cloud.sagemath.com. With your curve you'll find out this:

sage: E = EllipticCurve([0,-63992,0,-67887088,50869575680])

sage: E.torsion_points()

[(-1540 : 0 : 1), (-508 : -262128 : 1), (-508 : 262128 : 1), (0 : 1 : 0), (508 : 0 : 1), (65024 : 0 : 1), (130556 : -33552384 : 1), (130556 : 33552384 : 1)]

sage: E.gens()

[(100132172429824908929/1508738252550400 : 143687064081412107244001809983/58603135399923860992000 : 1)]

So the points you have found is all of the torsion group, which is why applying the group law on these points will give you no further points. The rank is $1$ and so you can find further points by computing multiples of the generator of large height that Sage found.

  • $\begingroup$ Thank you Jesper, this is so cool! I will definitely give Sage a try. $\endgroup$ – Dan Ismailescu Dec 22 '14 at 21:00
  • $\begingroup$ If $P_i$ denotes the point on $E$ with $x$-coordinate $x_i$, then $P_2 = P_1 + T$ where $T$ is the torsion point with $x$-coordinate $508$. The curve $E$ and $V^2=U(U+1)(U+q^2)$, $q=\frac{-127}{129}$ are isomorphic -- they have the same minimal model according to Sage. I am not sure that I understand the significance of the numbers that repeat, but I guess that it has to do with the isomorphism between the curves. $\endgroup$ – Jesper Petersen Dec 26 '14 at 21:02
  • 1
    $\begingroup$ @JesperPetersen: I used your answer to find another "small" point in Rusin's elliptic curve. They have interesting relationships, it seems. $\endgroup$ – Tito Piezas III Dec 26 '14 at 21:08
  • $\begingroup$ I was not aware of Rusin's derivations. Here are my own computations. Hope this helps. Please check the link below. dropbox.com/s/ytc1hlcrl7mmjco/… $\endgroup$ – Dan Ismailescu Dec 26 '14 at 22:04
  • $\begingroup$ @DanIsmailescu: Thanks, Dan. By the way, I edited my answer to include both solutions of form $127p^2, 129q^2$ for balance. This MO post related to your elliptic curve might be of interest. $\endgroup$ – Tito Piezas III Dec 28 '14 at 3:27

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