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This is a follow-up to $6$th powers of form $(6,1,7)$ in this post. For $8$th powers of form $(8,1,9)$, we already have a solution with $x_1=0$,

$$190^8 + 284^8 + 348^8 + 366^8 + 558^8 + 560^8 + 1040^8 + 1094^8 + 271^8 = 1167^8$$ $$\quad \color{red}{0^8} + 90^8 + 478^8 + 524^8 + 748^8 + 1088^8 + 1190^8 + 1324^8 + 223^8 = 1409^8$$

with the first by Nuutti Kuosa, and the second by Scott Chase found in 2000. (Update: My thanks to John Ca for bringing to my attention the first which was not in Meyrignac's database http://euler.free.fr/database.txt but was in Mathworld.)

Note that $1167 \equiv 0\, \text{(mod 3)},$ while $1409 \not\equiv 0\, \text{(mod 3)}$ and I assume it is the $2$nd type that we can find a zero term, a result semi-analogous to $(6,1,7)$. Re-arranging terms, we get the more informative, $$4^8(71^8 + 87^8 + 140^8 + 260^8) + 2^8(95^8 + 183^8 + 279^8 + 547^8) + 271^8 = 1167^8$$ $$4^8(131^8 + 187^8 + 272^8 + 331^8) + 2^8( \color{red}{0^8}+ 45^8 + 239^8 + 595^8) + 223^8 = 1409^8$$

with the unexpected (?),

$$1167-271 \equiv \text{0 mod 32}\quad$$ $$1409+223 \equiv \text{0 mod 32}\quad$$

and the expected,

$$z^8-x_9^8 \equiv \text{0 mod 256}\quad$$

Assume primitive solutions. Compare to the more hefty congruence restrictions for $6$th powers of form $(6,1,6)$,

$$42^6(x_1^6+x_2^6+x_3^6+x_4^6)+(7x_5)^6+x_6^6 = z^6$$

with the huge constraint,

$$\qquad z^6-x_6^6 \,\equiv\, \text{0 mod 117649}$$

where $7^6=117649.\,$ This huge mod $(k+1)^k$ constraint for $(k,1,k)$ exists when $k+1 = p$ is prime which partly explains why $k=6$ hasn't been found yet. For $k=4$, the smallest is,

$$15^4(2^4 + 8^4 + 21^4) + 272^4 = 353^4$$

where $272+353 = 625 = 5^4.$ This special case has a $(6,1,7)$ analogue,

$$42^6(32^6 + 554^6 + 1145^6 + 1289^6) + 7^6(3609^6 + 5584^6) + 54018^6 = 63631^6$$

where $54018+63631 = 117649 = 7^6.$

Questions:

Assume a primitive solution to $(8,1,8)$.

  1. Did S. Chase found this only solution with the slow computers of more than 20 years using brute force and/or some congruence $x_8^n\pm z^n \equiv \text{0 (mod w)}$ to speed up the search?
  2. If so, with more computing power nowadays, surely we can find another one for both (8,1,8) and (8,1,9)?
Tito Piezas III
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2 Answers2

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I don't know what S. Chase did, but a congruence relation that might have helped, assuming only $z$ and one left-hand side term $x$ are odd, is:

Either $z+x\equiv 0\pmod{32}$ or $z-x\equiv 0\pmod{32}$

Why $32$? We have that $2^8 | z^8-x^8$ implies:

$$2^8|(z^4+x^4)(z^2+x^2)(z+x)(z-x)$$

Since odd squares (including odd fourth powers) are always $\equiv1 \pmod8$, the expressions $z^4+x^4$ and $z^2+x^2$ are each $\equiv2 \pmod8$, implying:

$$2^6|(z+x)(z-x)$$

It is not possible for a higher power of $2$ than $2$ itself to be a factor of both $z+x$ and $z-x$ since then we would have:

$$2^2| (z+x)+(z-x)$$

$$2^2|2z$$

implying $z$ is even, a contradiction. So one of $z+x$ and $z-x$ must have $2^{6-1} = 32$ as a factor.

Adam Bailey
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  • I'm glad I brought this to your attention. After your congruence answers on $5$th and $6$th powers, I knew you'd be able to shed light on $8$th powers also. And if primitive solutions to $(8,1,9)$ have only one odd term on either side, then your analysis covers that as well. I'll see if I can come up with something for $7$th power, just to complete the sequence. :) – Tito Piezas III Aug 19 '23 at 05:56
  • I answered your question on $a^6+b^6+c^6 = d^6$ in this post. As usual, an elliptic curve is involved, but only two terms on the LHS were made $6$th powers. – Tito Piezas III Aug 20 '23 at 08:00
2

One obvious simplification for $(8,1,8)$ involves the parity of the terms. All odd $8$th powers are $\equiv1\bmod32$ and all even $8$th powers are $\equiv0\bmod32$. So to match residues you must have all terms even on both sides (which is nonprimitive) or an odd term for the total and just one odd term in the eight-term sum.

Oscar Lanzi
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  • I slightly revised the post. I checked $z^8-x_8^8\equiv 0,\text{mod m}$ using Chase's result and the only thing that popped out was the obvious $m=2^8=256$. No higher powers of $3$ or some other prime at all. – Tito Piezas III Aug 17 '23 at 13:08
  • The two odd terms of Chase's solution also satisfies $z+x_8\equiv 0,\text{mod 32}$, but I don't know if this is just a peculiarity of his result, or is a congruence for all $(8,1,8)$. Hard to tell with only one known solution. – Tito Piezas III Aug 17 '23 at 13:33
  • There is a further simplification arising from the fact that if $n$ is not divisible by $5$ then $n^8 \equiv 1 \pmod 5$. This implies that if the right hand term in $(8,1,8)$ or $(8,1,9)$ is not divisible by $5$, then the left hand side must have exactly one or (as in the two known solutions above) exactly six terms not divisible by $5$. Moreover the 'exactly one' case requires $5^8 | z^8-x^8$ which looks possible only for very large $z$. – Adam Bailey Aug 19 '23 at 21:34
  • So putting the considerations mod $2$ and mod $5$ together, the main chance for $(8,1,8)$ solutions appears to be the pattern: $$10^8(x_1^8+x_2^8)+2^8(x_3^8+x_4^8+\cdots+x_7^8)+x_8^8=z^8$$ where $x_8$ and $z$ are divisible by neither $2$ nor $5$. – Adam Bailey Aug 19 '23 at 21:45