TITLE: An elliptic curve $x(x - 78126^2)(x - 4\times5^7) = y^2$ for $x_1^7+x_2^7+\dots +x_8^7= x_9^7$
QUESTION [7 upvotes]: Ajai Choudhry found an infinite number of primitive solutions to,
$$x_1^7+x_2^7+\dots +x_8^7= x_9^7$$
by using an elliptic curve with a parameter $m=2$. There are variants, the one I used is,
$$x(x - 129^2) (x - 4\times2^7) = y^2\tag1$$
I tried the online Magma and, after some time, it found one rational point,
$$x  =\Bigl(\frac{48010029072}{224312161}\Bigr)^2$$

Dave Rusin years ago suggested $m=5$,

$$x(x - 78126^2) (x - 4\times5^7) = y^2\tag2$$

The online Magma is limited only to 120 seconds, and discontinued its calculation. Note that the change $x = u+78126^2$ transforms $(2)$ to the more symmetric,
$$u(u+78124^2)(u+78126^2) = y^2$$
In general, curves of the form,
$$u\big(u+(a-1)^2\big)\big(u+(a+1)^2\big) = y^2$$
have at least five torsion points,
$$u = 0,\quad -(a-1)^2,\quad  -(a+1)^2,\quad  \tfrac{(a - 1)^2 + (a + 1)^2}{2}-2,\quad -\,\tfrac{(a - 1)^2 + (a + 1)^2}{2}+2$$

Q: Does $(2)$ really have another rational solution, other than its five  integer torsion points?

REPLY [5 votes]: (This is an addendum to MacLeod's answer and where I got that elliptic curve.)
To see how to use MacLeod's solution to find infinitely many eight $7$th powers equal to a $7$th power, expand the expression,
$$F(x) = (x+a)^7+(x-a)^7+(mx+b)^7+(mx-b)^7\\-(x+c)^7-(x-c)^7-(mx+d)^7-(mx-d)^7$$
and collecting powers of $x$,
$$F(x) = 42(a^2+m^5b^2-c^2-m^5d^2)x^5 + 70(a^4+m^3b^4-c^4-m^3d^4)x^3 + 14(a^6+mb^6-c^6-md^6)x$$   
We get rid of the $x^5$ and $x^3$ terms by finding $a,b,c,d,m$ such that, 
$$a^2+m^5b^2=c^2+m^5d^2\tag1$$
$$a^4+m^3b^4=c^4+m^3d^4\tag2$$
and only the linear term is left.  Since $x$ is arbitrary, let $x = 14^6(a^6+mb^6-c^6-md^6)^6 N$. Thus, what remains is,
$$F(x) = 14^7(a^6+mb^6-c^6-md^6)^7 N$$ 
for arbitrary $N=t^7$.

Let $\color{blue}{m=5}$. Using Brahmagupta's identity,
$$(p r + m q s)^2 +m(-q r + p s)^2 = (p r - m q s)^2 +m(q r + p s)^2$$
eq. $(1)$ is easily solved as,
$$a,\;b,\;c,\;d = p r + 3125 q s,\; -q r + p s,\; p r - 3125 q s,\; q r + p s$$
and eq. $(2)$ becomes,
$$(25 p^2 - q^2) r^2 = (p^2 - 5^{12} q^2) s^2\tag3$$
Without loss of generality, let $q=1$. To solve $(3)$, it must be,
$$(25 p^2 - 1)(p^2 - 5^{12} ) = w^2$$
Assume it to be the square,
$$(25 p^2 - 1)(p^2 - 5^{12} ) = \big(5p^2 + (-x/10 + 5^6)\big)^2\tag4 $$
Expand to get,
$$100 p^2 (x - 78126^2) = x(x - 4\times5^7)$$ 
Therefore the elliptic curve,
$$x(x - 78126^2)(x - 4\times5^7)=y^2\tag5$$
with MacLeod's solution,
$$x = \Bigl(\frac{1115500181050003597405480\sqrt{2245}}{94839007729639076870541}\Bigr)^2$$ 
Retracing the steps and scaling, one can then get integer $p,q,r,s$, 
$$p =26495849279233847921447334543677665845247100\\ 
q =848947621992638248200474987563356720299231909\\
r =626004927447821946628577048253429765183543983525\\
s =39573486494385206930358492946164451904255363$$
then $a,b,c,d$ as $93$-digit integers to solve $(1),(2)$ for $m=5$. This implies $F(x)$ are integers with at least $93\times6\times7 \approx 3900$ digits.