On the elliptic curve $u(u+2186^2)(u+2188^2)=v^2$ and 7th powers?

Question

I. Elliptic curve

Given some constant integer $m$, a solution to the elliptic curve,

$$E:=u\big(u+(m^7−1)^2\big)\big(u+(m^7+1)^2\big)=v^2$$

which is not a torsion point implies a polynomial identity to,

$$(a+x)^7+(-a+x)^7+(c-x)^7+(-c-x)^7\\+(b+mx)^7+(-b+mx)^7+(d-mx)^7+(-d-mx)^7=z^7$$

for arbitrary $x$. Note that the integers $(a,b,c,d,z)$ depend on $u$ and are trivial if $u$ is one of the 5 torsion points of $E$. But non-torsion points have been found for $\color{blue}{m=2,5}.$

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II. Multi-grade

For the moment, there are about 30 known primitive integer solutions to,

$$(a_1+y)^k+(-a_1+y)^k+(a_2-y)^k+(-a_2-y)^k\\+(a_3+my)^k+(-a_3+my)^k+(a_4-my)^k+(-a_4-my)^k=\color{blue}0$$

simultaneously true for $k=1,3,7$, a third of which have either $\color{blue}{m=2,3,5}.$ Hmm.

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III. Connections

Other than being equal sums of 7th powers, the obvious connections between the two Diophantine equations are the LHS obeys,

$$x_1+x_2+x_3+x_4 =0 \\ x_5+x_6+x_7+x_8 = 0$$

thus,

$$\frac{x_1+x_2}{x_5+x_6} = \frac{x_3+x_4}{x_7+x_8} = \frac1m$$

The $x_i$ are large but a third of the multi-grade solutions have either $\color{blue}{m=2,3,5}$ (a small integer) so it is a bit curious.

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IV. Question

Does it follow then that $E$ for $m=3$,

$$E:=u\big(u+(3^7−1)^2\big)\big(u+(3^7+1)^2\big)=v^2\\ E:=u\big(u+2186^2\big)\big(u+2188^2\big)=v^2$$

also has a non-torsion point?

Answer 1

According to the online magma calculator, the rank of $E$ is $0$ as follows.

      E := EllipticCurve( [0, 9565940, 0, 22876782889024, 0]);
      SetClassGroupBounds("GRH");
      Rank(E);

Result: $0$ true