Timeline for Growth of $n=n(k)$ for which there's a non-trivial solution to $x_1^k+\cdots+x_n^k=y^k$?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2012 at 0:15 | comment | added | Will Jagy | @Gerry, found it, you are right. | |
| Mar 29, 2012 at 23:09 | comment | added | Gerry Myerson | @Will, isn't D1 more to the point than D7? We have (thanks to Noam Elkies) 3 4th powers that sum to a 4th power, and (thanks to Lander and Parkin) 4 5th powers that sum to a 5th power. Guy says that for $k\ge6$ there is no known sum of $k$ $k$th powers giving a $k$th power. | |
| Mar 29, 2012 at 22:11 | comment | added | zeb | euler.free.fr has a lot of numerical examples. | |
| Mar 20, 2012 at 7:52 | comment | added | Kevin Buzzard | @Will: he did, but perhaps here is not the place to discuss it -- email me Will and I'll see if I can find a pdf. | |
| Mar 19, 2012 at 18:40 | comment | added | Will Jagy | Did your student ever turn in the paper on principal binary quadratic form plus monic cubic one-variable polynomial, the sum representing all numbers? | |
| Mar 19, 2012 at 15:44 | comment | added | Daniel Loughran | Isn't Trevor Wooley on mathoverflow? | |
| Mar 19, 2012 at 10:27 | history | edited | Kevin Buzzard | CC BY-SA 3.0 |
fixed a mathematical slip pointed out by Will Jagy.
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| Mar 19, 2012 at 10:26 | comment | added | Kevin Buzzard | Will -- you're right. I guess you can use not $N^k$ but $N^k-1$ in the $G(k)$ result, giving $n(k)\leq G(k)+1$ which is still good enough for what follows. Thanks. I'll fix it. | |
| Mar 18, 2012 at 22:32 | comment | added | Will Jagy | $ 19^5 + 43^5 + 46^5 + 47^5 + 67^5 = 72^5 $ | |
| Mar 18, 2012 at 21:02 | comment | added | Will Jagy | $3^3 + 4^3 + 5^3 = 6^3, \; \; \; 1^3 + 6^3 + 8^3 = 9^3, \; \; 3^3 + 10^3 + 18^3 = 19^3, \; \; 7^3 + 14^3 + 17^3 = 20^3$ | |
| Mar 18, 2012 at 20:27 | comment | added | Will Jagy | Oh, well. In Unsolved Problems in Number Theory by Richard K. Guy, the closest thing is section D7, page 153 in my edition, very specific, your $x_j = j$ and $y = n+1,$ Rufus Bowen conjectured no solutions. Does anyone know an all positive solution to $x^3 + y^3 + z^3 = t^3?$ I don't think $10^3 + 9^3 = 12^3 + 1^3$ counts. | |
| Mar 18, 2012 at 20:08 | comment | added | Will Jagy | I don't see why $G(k)$ is an upper bound for $n(k).$ Since every large $k$-th power is already a $k$-th power, I think they are irrelevant to the value of $G(k).$ | |
| Mar 16, 2012 at 16:45 | comment | added | Boris Bukh | I asked about lower bounds at mathoverflow.net/questions/64649/… | |
| Mar 16, 2012 at 15:36 | history | asked | Kevin Buzzard | CC BY-SA 3.0 |