Timeline for A conjecture regarding prime numbers
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27, 2017 at 21:01 | vote | accept | BR Pahari | ||
| Nov 26, 2017 at 22:49 | answer | added | Gerhard Paseman | timeline score: 9 | |
| Nov 26, 2017 at 17:49 | comment | added | Yaakov Baruch | Well - Lucia did the work! | |
| Nov 26, 2017 at 17:46 | comment | added | Yaakov Baruch | Say, based on the highest quality $abc$ triple known to date, you have that $48407720661$ and $48407720661+15042$ have the same radical. Now even such a small number as $15042$ is so much larger than $\log(48407720661)$ that the chance of not there being primes in that interval is practically nil. So my expectation is that the OP conjecture is true, but this reasoning clearly needs work to be made into a clean euristic argument. | |
| Nov 26, 2017 at 17:40 | comment | added | Sylvain JULIEN | I'm sorry but I don't see how the radical of an integer could be anyhow greater than the integer itself, let alone this integer times another positive integer. But I may have misunderstood what you meant. | |
| Nov 26, 2017 at 17:38 | answer | added | Lucia | timeline score: 44 | |
| Nov 26, 2017 at 17:30 | comment | added | Yaakov Baruch | It feels like the $abc$ conjecture would have something to say on the how close numbers with the same radical can be. | |
| Nov 26, 2017 at 17:27 | comment | added | Yaakov Baruch | @SylvainJULIEN: the radical of $m$ would be an even larger lower bound. I didn't put well, but what I had in mind was some simple well behaved monotonic function (like, say, $\log(m)^2) that is a lower bound and gets "very close" to being optimal infinitely often. | |
| Nov 26, 2017 at 17:12 | comment | added | Sylvain JULIEN | @Yaakov Baruch: Wouldn't $(P_{-}(m)-1)m$ with $P_{-}(m)$ the smallest prime factor of $m$ be a rather obvious lower bound? | |
| Nov 26, 2017 at 17:07 | comment | added | Yaakov Baruch | It's a close miss between 48 and 54... | |
| Nov 26, 2017 at 16:52 | comment | added | Yaakov Baruch | Maybe then an easier and more interesting question would be to prove some bound function $b(m)$ on $n-m$ if $m<n$ have the same radical? | |
| Nov 26, 2017 at 16:28 | comment | added | fedja | If you just consider $n^2(n+1)$ and $n(n+1)^2$, then this would imply that there always is a prime between $n^3$ and $(n+1)^3$. That is known, but highly non-trivial. However, those two are just the simplest products with guaranteed same radical that jump to one's mind. Whether a more clever construction would allow one to deduce something that we certainly do not know today remains to be seen. | |
| Nov 26, 2017 at 15:51 | comment | added | Stanley Yao Xiao | This does not seem terribly easy... in fact, I don't have a good intuition as to whether it should be true or not. Let $S = \{p_1, \cdots, p_k\}$. The question can be thought of whether there exist two $S$-units $m, n$, say $N < m < n \leq 2N$, such that $|n - m| \ll \log N$ say. If the answer is yes, then one would expect that there are no primes between $m$ and $n$ typically, so the conjecture would be false. | |
| Nov 26, 2017 at 15:45 | comment | added | Konstantinos Gaitanas | This is equivalent to say that among two numbers having the same radical, there is at least one prime number. | |
| Nov 26, 2017 at 15:35 | review | First posts | |||
| Nov 26, 2017 at 16:01 | |||||
| Nov 26, 2017 at 15:32 | history | asked | BR Pahari | CC BY-SA 3.0 |