Timeline for Congruence properties of $x_1^6+x_2^6+x_3^6+x_4^6+x_5^6 = z^6$?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 19, 2015 at 1:33 | vote | accept | Tito Piezas III | ||
| May 18, 2015 at 17:16 | comment | added | Will Sawin | 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 $p\neq 7, 31$. | |
| May 18, 2015 at 16:47 | comment | added | Max Alekseyev | @TitoPiezasIII: See update. | |
| May 18, 2015 at 16:47 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| May 17, 2015 at 1:30 | comment | added | Tito Piezas III | 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. | |
| May 16, 2015 at 20:08 | comment | added | Max Alekseyev | @TitoPiezasIII: OK, let me take a close look. Could you please give a reference to Ward's result? | |
| May 16, 2015 at 11:31 | comment | added | Tito Piezas III | 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. | |
| May 16, 2015 at 9:03 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| May 16, 2015 at 8:56 | history | answered | Max Alekseyev | CC BY-SA 3.0 |