Timeline for Show this number always is composite number
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
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| S Nov 1, 2018 at 17:00 | history | bounty ended | CommunityBot | ||
| S Nov 1, 2018 at 17:00 | history | notice removed | CommunityBot | ||
| S Oct 24, 2018 at 15:02 | history | bounty started | math110 | ||
| S Oct 24, 2018 at 15:02 | history | notice added | math110 | Authoritative reference needed | |
| May 15, 2018 at 3:00 | review | Close votes | |||
| May 16, 2018 at 3:06 | |||||
| S May 10, 2018 at 22:46 | history | suggested | user124222 | CC BY-SA 4.0 |
slight grammar improvement, removed redundant quantifier
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| May 10, 2018 at 21:26 | review | Suggested edits | |||
| S May 10, 2018 at 22:46 | |||||
| May 9, 2018 at 3:41 | comment | added | Dongryul Kim | If we use the heuristics that a number $n$ is prime with probability $1 / \log n$, the expected number of $m \le x$ with $f(m)$ prime is $\sim \log \log x$. I don't think checking for $m \le 1300$ is good enough support. | |
| May 9, 2018 at 1:21 | history | edited | math110 | CC BY-SA 4.0 |
added 30 characters in body
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| May 8, 2018 at 17:17 | review | Close votes | |||
| May 8, 2018 at 23:02 | |||||
| May 8, 2018 at 17:13 | comment | added | François Brunault | If $m \equiv 8$ mod $20$ then $f(m)$ is divisible by 5. Similarly if $m \equiv 4,16,22,30$ mod $42$ then $f(m)$ is divisible by 7. One can find other examples by noticing that if $p$ is prime then $f(m)$ mod $p$ depends only on $m$ mod $p(p-1)$ by Fermat's little theorem. | |
| May 8, 2018 at 17:00 | comment | added | Greg Martin | It's impossible for there to be a factorization $f(m)=g(m)h(m)$ where $g,h\in\Bbb Z[x]$, because $f(m)$ is not a polynomial. | |
| May 8, 2018 at 16:32 | comment | added | Sylvain JULIEN | Maybe it comes from the fact that the associated function of two variables $ f^{*}(a,b ; g)=(g(a))^{b} +a^{b}.b^{a}+b^{g(a)} $ is invariant under permutation of the variables $ g(a) $ and $ b $ on one hand outside the product and under permutation of $ a $ and $ b $ on the other hand in the product. When you think of a prime as a number of objects that can't be arranged as a rectangle, it seems to indicate that in some sense primes have no non-trivial symmetry. Ideas from Galois theory might help solve this intriguing question. | |
| May 8, 2018 at 14:43 | comment | added | Dietrich Burde | Crossposted at MSE. | |
| May 8, 2018 at 14:39 | history | edited | Joe Silverman |
Added top level nt.number-theory tag
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| May 8, 2018 at 13:04 | history | edited | Manuel Bärenz | CC BY-SA 4.0 |
Further clarification
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| May 8, 2018 at 12:58 | comment | added | Manuel Bärenz | @JoseArnaldoBebitaDris, I think it's a straightforward generalisation of the linked question. | |
| S May 8, 2018 at 12:57 | history | edited | Manuel Bärenz | CC BY-SA 4.0 |
Improved grammar and wording
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| May 8, 2018 at 12:30 | comment | added | math110 | I just find this form very interesting, and if this conjecture is correct, that is, the value of this formula just avoids the prime number situation, which is very convincing. | |
| May 8, 2018 at 12:30 | review | Suggested edits | |||
| S May 8, 2018 at 12:57 | |||||
| May 8, 2018 at 12:27 | comment | added | Jose Arnaldo Bebita | What is the motivation for this question? | |
| May 8, 2018 at 12:14 | history | edited | math110 | CC BY-SA 4.0 |
added 368 characters in body
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| May 8, 2018 at 12:09 | history | edited | math110 | CC BY-SA 4.0 |
added 368 characters in body
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| May 8, 2018 at 12:06 | history | undeleted | user90958 | ||
| May 8, 2018 at 11:31 | history | deleted | user90958 | via Vote | |
| May 8, 2018 at 11:15 | history | asked | math110 | CC BY-SA 4.0 |