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
- If $x_1=0,$ is it a congruence requirement that one term be a multiple of $11$?
- As a long shot, can we use the symmetric structure of Solution 1 to find similar solutions?
Best Answer
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:
(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 $\equiv 0 \pmod{11}$.