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

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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).

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    $\begingroup$ Thanks for the answer. However, perhaps I didn't make myself clear enough. Ward's constraint for $k=4$ that $z\pm x_j$ is divisible by $w_4 = 2^{10}$ $only$ works if one of the $x_i$ is zero. For $k=6$, I'm seeking a similar constraint $w_6 =\;?$ that $only$ works also if one of the $x_i$ is zero. $\endgroup$ – Tito Piezas III May 16 '15 at 11:31
  • $\begingroup$ @TitoPiezasIII: OK, let me take a close look. Could you please give a reference to Ward's result? $\endgroup$ – Max Alekseyev May 16 '15 at 20:08
  • $\begingroup$ I believe it is found in Morgan Ward, Euler’s problem on sums of three fourth powers, Duke Math. J. 15 (1948), 827–837. If not, maybe Euler’s three biquadrate problem, Proc. Nat. Acad. Sci. U. S. A. 31 (1945), 125–127. $\endgroup$ – Tito Piezas III May 17 '15 at 1:30
  • $\begingroup$ @TitoPiezasIII: See update. $\endgroup$ – Max Alekseyev May 18 '15 at 16:47
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    $\begingroup$ I can strengthen your "it is unlikely that" statement. Saying that $p$ divides $x^6+y^6+z^6+w^6$ for some coprime $x,z,t,w$ is equivalent to saying that the algebraic surface in $\mathbb P^3$ with equation $x^6+y^6+z^6+w^6$ has at least one $\mathbb F_p$ points. This is a smooth surface, so by the Weil conjectures its number of points is $p^2+a_p+1$ with $|a_p|< 106p$. (This can also be proved more directly using Gauss sums). Hence it has at least one point for $p>106$. Combined with your calculations, we see it has at least one point for all $p\neq 7, 31$. $\endgroup$ – Will Sawin May 18 '15 at 17:16

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