Timeline for Fifth powers modulo a prime
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 18, 2011 at 3:24 | comment | added | Noam D. Elkies | @Greg: just saw your comment ("...I conjecture that this disjointness property is impossible unless either $p$ is small or $a=−1$..."); you're right, see my answer. | |
| Oct 18, 2011 at 2:44 | answer | added | Noam D. Elkies | timeline score: 15 | |
| Oct 18, 2011 at 0:05 | vote | accept | Andrés E. Caicedo | ||
| Oct 17, 2011 at 19:19 | comment | added | GH from MO | I think I have solved the general problem, using a result of Inkeri's. See below. | |
| Oct 17, 2011 at 19:13 | answer | added | GH from MO | timeline score: 14 | |
| Oct 17, 2011 at 18:12 | comment | added | Greg Kuperberg | Note also that (in the general question) if $a$ is any $k$th root of unity mod $p$, then $1,2,\ldots,n$ is disjoint from $a,2a,\ldots,an$ mod $p$, assuming that $p$ is a counterexample to Andres' question. I conjecture that this disjointness property is impossible unless either $p$ is small or $a = -1$, even without other restrictions on $a$. | |
| Oct 17, 2011 at 17:55 | comment | added | Greg Kuperberg | @darij Experimental fact (i.e., SAGE-supported conjecture): Let $f(x,k)$ be the polynomial interpolation of $1^k + 2^k + \cdots + x^k$. Then $k^{k+1}f(-1/k,k)$ is 1 mod $k$ when $k$ is odd and $(k-2)/2$ mod $k$ when $k$ is even. | |
| Oct 17, 2011 at 16:52 | comment | added | darij grinberg | @Greg (and everybody else wondering): I may have answered the question, but I haven't answered the Question, namely the case of $k$ general. I feel that at least the non-divisibility of the polynomial $1^k+2^k+...+n^k$ by $kn+1$ is a doable (i. e., not entirely out of reach), yet nontrivial problem, and whoever solves it should be considered the actual answerer of this question. But if Andres feels he is not interested in that, he is free to self-answer the question with my proof. I have enough undeserved rep here already since the alg geo question I posted some weeks ago (if not earlier). | |
| Oct 17, 2011 at 16:43 | comment | added | Greg Kuperberg | @darij Could you please just post your nice solution? By the way, you can refine the method by taking $jk$th powers for various $j$. | |
| Oct 16, 2011 at 17:42 | comment | added | darij grinberg | Oh, of course, it is clear that for every $k$, it will work for all but finitely many values of $n$, unless the polynomial $kn+1$ divides the polynomial $1^k+2^k+...+n^k$. The question is when this happens, and what are the finitely many values of $n$ otherwise. | |
| Oct 16, 2011 at 17:36 | comment | added | darij grinberg | Andres: The general case seeems much more interesting; I wouldn't say that it is dealt with by my argument. Probably it will use some congruences with Stirling and Bernoulli numbers. If you have something useful in the way of solving it, please post it! | |
| Oct 16, 2011 at 17:31 | comment | added | darij grinberg | Yes, exactly... | |
| Oct 16, 2011 at 17:31 | comment | added | Andrés E. Caicedo | @darij: Ah, ok, you are right, the same observation should deal with the general case. Sorry, this turned out to be too easy, I wasn't thinking about this appropriately. Would you like to write this as an answer? Otherwise, I'll wait a bit and write something myself. | |
| Oct 16, 2011 at 17:15 | comment | added | j.p. | @Darij: Your point being that for $p = 5n+1$ prime, it has to divide one of the factors? | |
| Oct 16, 2011 at 17:01 | comment | added | darij grinberg | ... polynomial division shows that each of $n$, $n+1$ and $2n^2+2n-1$ is coprime to $5n+1=p$ unless $p=11$. | |
| Oct 16, 2011 at 17:00 | comment | added | j.p. | As $-1$ has to be a $5$-th power, you need the prime $p$ to be a fifth power plus $1$. | |
| Oct 16, 2011 at 16:58 | comment | added | darij grinberg | This is probably wrong or otherwise too easy: If the powers $1^5$, $2^5$, ..., $n^5$ would be distinct modulo $p$, then they would be a list of all $5$-th powers modulo $p$ (since $p=5n+1$, so there are only $n$ $5$-th powers modulo $p$), so their sum would be the sum of all distinct $5$-th powers modulo $p$, and that latter sum is known to be $\equiv 0\mod p$ (see artofproblemsolving.com/Forum/viewtopic.php?t=40171 ). In other words, we would have $1^5+2^5+...+n^5\equiv 0\mod p$. But $1^5+2^5+...+n^5=\frac{1}{12}n^2\left(n+1\right)^2\left(2n^2+2n-1\right)$, and some ... | |
| Oct 16, 2011 at 16:02 | history | asked | Andrés E. Caicedo | CC BY-SA 3.0 |