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?

  • $\begingroup$ and what is the $y$ solution? $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '18 at 19:29
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    $\begingroup$ Once you have one rational point that’s not a torsion point, you have infinitely more. $\endgroup$ – Lubin Jan 6 '18 at 19:31
  • $\begingroup$ @Dr.SonnhardGraubner $y=\pm\frac{87808249505851020222496918348032}{11286478484123596330229281}$ $\endgroup$ – Carl Schildkraut Jan 6 '18 at 19:32
  • $\begingroup$ oh man how Can we check this? $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '18 at 19:32
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    $\begingroup$ @Dr.SonnhardGraubner: Verifying it is easily done in Mathematica. Finding $x$ in the first place is harder. $\endgroup$ – Tito Piezas III Jan 6 '18 at 19:33

The point on the curve


has $u$-coordinate \begin{equation} u=\frac{-2^2*43^2*229^2*449*19531^2*36965783^2*24591784649^2}{3^2*61^2* 518245943877809163227^2} \end{equation}

I found this using my home-grown Pari-software and an old version of Cremona's mwrank. The height of the point is roughly 51.9/103.8 (depending on which height normalisation you like).

The points of order $4$ have a simpler formula, namely $u=\pm (1-a^2)$.


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

has rational point,

$$x = \Bigl(\frac{1115500181050003597405480\sqrt{2245}}{94839007729639076870541}\Bigr)^2$$

  • 1
    $\begingroup$ Beautiful! I've been searching for this rational point since 2009. Rusin suggested it might be feasible within 10 years. How long did it take your computer? $\endgroup$ – Tito Piezas III Jan 7 '18 at 16:45
  • $\begingroup$ About 4 hours, but a lot of that was just fiddling about. $\endgroup$ – Allan MacLeod Jan 7 '18 at 16:50
  • $\begingroup$ I added an explanatory note below. $\endgroup$ – Tito Piezas III Jan 7 '18 at 18:14
  • $\begingroup$ By the way, did you see this post about an elliptic curve and $n=809$? Fisher remarks that more than 10 years ago, this prime was a bit troublesome in your's and Rathbun's list for $n<1000$. $\endgroup$ – Tito Piezas III Jan 12 '18 at 16:05

(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.


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