# On Sixth Powers $x_1^6+x_2^6+x_3^6+\dots+x_6^6 = z^6$

Fourteen years ago, in 1999 (has it been that long?) Merignac started a search for,

$$x_1^6+x_2^6+x_3^6+\dots+x_7^6 = \color{red}z^6$$

with the hope that one $$x_i =0$$ using the five congruence classes,

\begin{aligned} &42^6(x_1^6+x_2^6+\dots+x_5^6)+(42x_6)^6+(1x_7)^6 = z^6\\ &42^6(x_1^6+x_2^6+\dots+x_5^6)+(21x_6)^6+(2x_7)^6 = z^6\\ &42^6(x_1^6+x_2^6+\dots+x_5^6)+(14x_6)^6+(3x_7)^6 = z^6\\ &42^6(x_1^6+x_2^6+\dots+x_5^6)+\,(7x_6)^6\,+\,(6x_7)^6 = z^6\\ &42^6(x_1^6+x_2^6+\dots+x_4^6)+(21x_5)^6+(14x_6)^6+(6x_7)^6 = z^6 \end{aligned}

So the first six $$x_i$$ are multiples of $$7$$, and primitive integer solutions are known for all five classes. The smallest known (found around 2000) by Meyrignac and Wannes de Smedt belongs to the 4th class,

$$42^6(195^6 + 260^6 + 440^6 + 506^6 + 580^6) + (7\times2747)^6 + (6\times5559)^6 = 34781^6$$

But it seems none has one $$x_i = 0$$ for $$\color{red}z<730000$$. (See Further work section of 2002 paper The Smallest Solutions to the Diophantine Equation $$a^6+b^6+c^6+d^6+e^6=x^6+y^6$$.)

To compare to 4th powers, one can primitively solve,

$$a^4(x_1^4+x_2^4+x_3^4)+x_4^4 = z^4$$

for $$a=10$$ or $$a=20$$ with the smallest being,

$$10^4(24^4+34^4+43^4)+599^4 = 651^4$$ $$20^4(19^4+83^4+94^4)+4907^4 = 4949^4$$

Question: Is $$b=42$$ overly restrictive? Can it be reduced to just $$b = 21$$? (As the fourth power example shows, $$a = 10$$ has a smaller solution. If so, maybe they overshot a solution in the $$z<730000$$ range?)

• I can only answer your first question: somehow, 1999 was, in fact, fourteen years ago. Commented May 15, 2013 at 21:13
• @Arkamis: Haha. It was supposed to be rhetorical. Time flies. Commented May 15, 2013 at 21:14

The restriction to $b = 42$ is correct because any solution of the Diophantine equation

$$x_1^6+x_2^6+\dots+x_6^6 = z^6$$

would have to meet constraints associated with each of the prime factors of 42:

a) If $a \not\equiv 0 \pmod{2}$ then $a^6 \equiv 1 \pmod{8}$;

b) If $a \not\equiv 0 \pmod{3}$ then $a^6 \equiv 1 \pmod{9}$;

c) If $a \not\equiv 0 \pmod{7}$ then $a^6 \equiv 1 \pmod{7}$.

(These can be found on this page of Meyrignac's website, albeit applied to a different 6th power equation.) To meet these constraints, a (primitive) solution must have $x_i \equiv 0 \pmod{2}$ for 5 of the 6 left hand terms, and similarly for mod 3 and mod 7. Hence a solution must have $x_i^6 \equiv 0 \pmod{42}$ for at least 3 of the 6 left hand terms.

There is also a restriction associated with 13 arising from the fact that:

d) If $a \not\equiv 0 \pmod{13}$ then $a^6 \equiv 1$ or $-1 \pmod{13}$.

This implies that a solution must have an odd number of terms (which could include $z$) congruent to 0 mod 13 and, among the remaining terms, an appropriate balance of sixth powers congruent to 1 and to -1 mod 13. To combine this restriction with those relating to 2, 3 and 7 would be quite tedious but it can be said that the five congruence classes are not restrictive enough to exclude all cases which are impossible for reasons of congruence.

• Thanks. I knew that for $x_1^k+x_2^k+\dots+x_k^k = y^k$ where $k+1=p$ is prime, then one must use a congruence involving $p$. I was trying to compare $k = 4,6$ but I guess the situation is different for $k=4$ since it has only one distinct prime factor. Commented May 29, 2013 at 18:13
• You may be interested in this question on $x_1^6+x_2^6+\dots+x_5^6 = y^6$. Commented May 10, 2015 at 3:17