Timeline for existence of equal sum, equal sum of square, equal sum of cube, etc
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 1, 2011 at 15:34 | comment | added | Igor Rivin | @Gerry: this is purely academic, given @Mark's pointer, but counting variables is not entirely satisfying, since this is a nonlinear system (over the reals...) | |
| Feb 1, 2011 at 3:47 | comment | added | Aaron Meyerowitz | With equal number of terms a solution stays a solution under linear transformations. In particular, it is no problem to shift everything up to have all terms positive. | |
| Feb 1, 2011 at 3:25 | comment | added | Gerry Myerson | @Igor, if positivity (and integrality) is (are) not required, then my guess is it's a simple matter of counting up conditions and variables; if there are more of the latter than of the former, I would expect there to be solutions. @Mark, of course in Prouhet's solution the number of variables is exponential in the number of conditions. The big question (I'm sure you know this) is whether, for all $n$, there is a (non-trivial, integral) solution up to $n$th powers with just $n+1$ terms on each side. I concede that this may be of limited interest to WAB but still recommend searching multigrade. | |
| Feb 1, 2011 at 3:22 | comment | added | user6976 | Also I think Prouhet in his paper mentioned that the problem goes back to Euler. | |
| Feb 1, 2011 at 3:16 | comment | added | user6976 | The problem with equal number of terms in each side has a nice solution found by Prouhet in around 1860. Consider the word $w_n$ in $a,b$ defined as $\phi^n(a)$ where $\phi$ is the substitution $a\to ab, b\to ba$. It has exactly $2^n$ $a$'s and $2^n$ $b$'s. Let $x_1,\ldots,x_{2^n}$ be the numbers of places in $w_n$ where letter $a$ occurs, and $y_1,\ldots, y_{2^n}$ the places where $b$ occurs. Then these $x$'s and $y$'s satisfy the first $n$ Prouhet's conditions. Note that later the words $w_n$ were rediscovered by Thue, Morse, Arshon and many others. These are called Thue words now. | |
| Feb 1, 2011 at 1:15 | comment | added | Igor Rivin | Not only that, the OP explicitly wants to know about not-necessarily-integer (but positive) solutions as well. Even if positivity were not required, and there were the same number of terms on each side, the question becomes: can you have $M^{-1} N \mathbf{1} = \mathbf{1},$ where $M, N$ are Vandermonde matrices, and does not seem to be entirely trivial. | |
| Jan 31, 2011 at 22:24 | comment | added | user6976 | @Gerry: The Prouhet-Tarry-Escott problem is different as you yourself mention, it has the same number of terms on each side. | |
| Jan 31, 2011 at 22:20 | history | answered | Gerry Myerson | CC BY-SA 2.5 |