Congruence properties of $x_1^6+x_2^6+x_3^6+x_4^6+x_5^6 = z^6$?

Question

(This was posted previously in MSE without getting any answers.)

It is known that given primitive (co-prime) integer solutions to,

$$x_1^4+x_2^4+x_3^4+x_4^4 = z^4$$

then there is one $x_i$ such that $z^4-x_i^4$ is divisible by $d_4=5^4$. Additionally, Ward showed that if one of the $x_i$ is zero, then there is the further constraint that $z\pm x_j$ is divisible by $w_4=2^{10}$. For example,

$$673865^4+ 1390400^4+ 2767624^4 = 2813001^4$$

and,

$$z+x_3 = 2813001 + 2767624 = 5^4\cdot 8929$$

$$z-x_1 = 2813001 - 673865= 2^{10}\cdot 2089$$

Theorem: In general, for $k$ $k$th powers and co-prime terms,

$$x_1^k+x_2^k+x_3^k+\dots+x_k^k = z^k\tag1$$

if $k+1$ is prime, then there is one $x_i$ such that $z^k-x_i^k$ is divisible by $d_k = (k+1)^k$.

For $d_4 = 5^4 = 625$, this implies the smallest solution will have a term $>d_4/2 = \lfloor312\rfloor$. In fact,

$$30^4+120^4+315^4+\color{brown}{272}^4 = \color{brown}{353}^4$$

and $x_3+z =272+353=5^4.^{\color{brown}{Note}}$

For $d_6 = 7^6 = 117649$, the smallest solution (not yet found as of 2015) will have a term $>d_6/2 = \lfloor58824\rfloor$ hence will be relatively large and somehow "explains" why $k=8$ was found first,

$$90^8 + 223^8 + 478^8 + 524^8 + 748^8 + 1088^8 + 1190^8 + 1324^8 = 1409^8$$

since $8+1$ is composite and doesn't have the divisibility constraint $d_k$.

Question: Given $(1)$ where one of the $x_i$ is zero, is there an analogue to Ward's $w_k$ for $k=6$? If there is, what is it for general $k$?

$\color{brown}{Note}$: This has an analogue for $6$th powers when using seven addends,

$$1344^6+ 23268^6+ 25263^6+ 39088^6+ 48090^6+ 54138^6+ \color{brown}{54018}^6 = \color{brown}{63631}^6$$

and $x_7+z =54018+63631=7^6$.

Answer 1

To answer this question, one needs to understand why Theorem holds. In fact, it is a corollary of a stronger statement: if $$x_1^k+x_2^k+x_3^k+\dots+x_k^k = z^k\qquad (1)$$ and $k+1$ is prime, then all $x_j$, except possibly one, are multiples of $k+1$. (Letting $x_i$ be this exceptional value, we trivially get that $z^k-x_i^k$ is divisible by $(k+1)^k$ as claimed by Theorem.)

The proof of this statement is based on the Fermat little theorem, implying that for any integer $y$, we have $y^k\equiv 0$ or $1\pmod{k+1}$. In particular, this holds for all $x_j$ and $z$ and this (1) modulo $k+1$ represents the sum of $k$ 0-or-1 values equal 0 or 1. Clearly, there is at most one summand 1 in this sum (and such summand exists iff $(k+1)\nmid z$). All other $x_j$ must be 0 modulo $k+1$. QED

So, for $k=6$, all summands in (1), except possibly one, are divisible by 7. (If we talk about a minimal solution, then such exceptional summand must exist as otherwise we would obtain a smaller solution by divide all its terms by 7.)

For $k=6$, similar arguments apply to some prime divisors other than $k+1=7$:

• for any integer $y$, we have $y^6\equiv 0$ or $1\pmod{2^3}$, implying (together with $k<2^3$) that in the minimal solution (1) all $x_j$, except one, are divisible by 2;

• for any integer $y$, we have $y^6\equiv 0$ or $1\pmod{3^2}$, implying (together with $k<3^2$) that in the minimal solution (1) all $x_j$, except one, are divisible by 3.

For general $k$, candidates for such prime divisors are the prime divisors of $k$.

UPDATE. The Ward constraint for the equation $$x_1^4 + x_2^4 + x_3^4 = z^4$$ is essentially equivalent to $z^4 - x_j^4=(z^2+x_j^2)(z+x_j)(z-x_j)$ being divisible by $2^{12}$. To get this constraint, Ward uses quadratic reciprocity, which restricts possible prime factors of numbers of the form $u^4+v^4 = (u^2+v^2)^2 - 2(uv)^2$ with co-prime $u,v$ to primes $\equiv 1\pmod{8}$.

It is unlikely that such arguments can be generalized for sixth powers as it would require to have a modular-type restriction on possible prime factors of $p^6 + q^6 + r^6 + s^6$ for co-prime $p,q,r,s$, which apparently does not exist (e.g., among the primes below $10^4$ only primes $7$ and $31$ cannot divides such sum of sixth powers).