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This post discussed the more symmetric equation,

$$x_1^5+x_2^5+x_3^5 = y_1^5+y_2^5+y_3^5$$

where we assume all terms $\in \mathbb{Z},$ solutions as primitive, and $(x_1, x_2, x_3) = (y_1, y_2, y_3)$ as trivial.

Jyrki Lahtonen pointed out that for such equations, if one term happens to be a multiple of $11$, then another term must also be a multiple of $11$. For the three known solutions of the special case $x_1 = 0$ (a counterexample to Euler's Sum of Powers conjecture), it seems to be unmentioned that all three have terms with $11m$.

I. Solution 1

While not the first discovered, this has a lot of structure,

$$11^5\,(\color{red}{0^5} + 20^5) + 14132^5 = 11^5(457^5 + 567^5) + 14068^5$$ $$14132-14068 = 2^6\quad$$ $$\; 457+567 = 2^{10}$$

Having only two odd terms, it must obey a general congruence discused in this post.

II. Solution 2

Likewise, this also has just two odd terms, and follows the aforementioned congruence.

$$11^5\,(\color{red}{0^5} + 10^5) + 27^5 + 84^5 + 133^5 = 144^5\quad$$ $$27+133\equiv 0 \text{ mod }2^5$$

This was the first discovered (by accident) by Lander & Parkin in 1967. Incidentally,

$$-144+133\equiv 0 \text{ mod }11\;\;$$

a property shared by the other solution below.

III. Solution 3

Last one to be discovered (1996). This has four odd terms, so,

$$11^5\,(\color{red}{0^5} + 5^5) + 3183^5 + 28969^5 + 85282^5 = 85359^5$$

$$55+28969\equiv 0 \text{ mod }2^5$$ $$-3183+85359\equiv 0 \text{ mod }2^5\;$$

though it is uncertain if both pairs of odd terms will always obey the congruence. And,

$$-85282+85359\equiv 0 \text{ mod }11\;\;$$

IV. Questions

  1. If $x_1=0,$ is it a congruence requirement that one term be a multiple of $11$?
  2. As a long shot, can we use the symmetric structure of Solution 1 to find similar solutions?
Tito Piezas III
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1 Answers1

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The answer to question 1 is YES. This is a consequence of the fact that, for all $n$, $n^5 \equiv 0 \pmod{11}$ iff $n \equiv 0 \pmod{11}$, otherwise $n^5 \equiv \pm1 \pmod{11}$.

A somewhat more general consequence can be stated as follows:

Let $x_i \in \mathbb{Z}.\,$ If $\,\sum_{i=1}^k x_i^5=0$ for some $k<11$, then the sum must contain an even number of terms not divisible by $11$.

(This formulation avoids the possible ambiguity over whether a zero "term" should be considered a term.)

For example, Mathworld lists several (5,1,6) solutions, the first being:

$$4^5+5^5+6^5+7^5+9^5+11^5=12^5$$

All, like this one, have 6 of their 7 terms not divisible by 11.

The qualification $k<11$ is needed because if 11 terms were either all $\equiv 1$ or all $\equiv -1 \pmod{11}$ then their sum would be $\equiv 0 \pmod{11}$.

Adam Bailey
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    Going higher, I assume the same rule holds for $\large{\sum_{i=1}^k x_i^n=0}$, if $2n+1 = \text{prime }p,$ doesn't it? For example, I checked $n^9 \equiv \pm1 \pmod{19}$. But it seems things get fuzzy with $n=7$ since $2n+1 = 15$ is not prime. Hmm, I wonder how this affects finding a $(7,1,6)$ equation, analogous to the $(5,1,4)$ of this post? Easier, harder, or impossible? – Tito Piezas III Aug 31 '23 at 14:41
  • @TitoPiezasIII I think the rule generalises to any exponent $n$ for which $2n+1$ is prime, subject to the qualification $k<2n+1$ (which I overlooked earlier), but haven't been able to prove it. You might like to ask a new question on $7$th power equations. – Adam Bailey Sep 01 '23 at 12:57
  • For this post, I focused more on odd powers $n$ but I remember you mentioned in an answer that there is an additional constraint for $n=6$ involving $\text{mod } 13$. I guess that may make the $(6,1,6)$ even harder to find. P.S. Regarding a new question, I was thinking of jumping to $n=9$ since it was easier to find patterns because $2n+1 = 19$ is prime. It should be done in a day or two. – Tito Piezas III Sep 01 '23 at 14:38
  • I totally forgot to click on the button. Mea culpa. – Tito Piezas III Dec 08 '23 at 15:54