Timeline for Is the $n$-th prime $p_n$ expressible as the difference of coprime $A, B$ such that the set of prime divisors of $AB$ is $\{p_1, \dots, p_{n-1}\}$?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Dec 1, 2015 at 14:33 | comment | added | Denis Serre | As noted by François, I asked a few years ago a closely related question. Moret-Bailly wrote a useful comment. | |
| Dec 1, 2015 at 13:23 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Made the title of the question more informative.
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| Mar 9, 2013 at 22:07 | answer | added | Aaron Meyerowitz | timeline score: 6 | |
| Mar 9, 2013 at 22:05 | comment | added | François Brunault | Related question : mathoverflow.net/questions/38794/constructing-prime-numbers | |
| Mar 9, 2013 at 21:53 | comment | added | François Brunault | It seems that there are two questions here, the one in the title and the one in the body, which are not equivalent. | |
| Mar 9, 2013 at 15:33 | answer | added | Stefan Kohl♦ | timeline score: 7 | |
| Mar 9, 2013 at 15:21 | comment | added | Gerhard Paseman | Likely the answer is yes by using applications of Euclid's GCD algorithm, or by methods involving Stormer's theorem. I hope to say more later today. Gerhard "Is Starting An Exam Soon" Paseman, 2013.03.09 | |
| Mar 9, 2013 at 14:43 | comment | added | Barry Cipra | It's clear that if $p_1$ to $p_{n-1}$ are the first $n-1$ primes, then $p_n$ cannot be smaller than the next prime. It's also easy to get $5=3^2-2^2$, $7=2\cdot5-3$, and $11=3\cdot7-2\cdot5$. Can you supply the next several examples? How far have you computed things? | |
| Mar 9, 2013 at 13:23 | history | edited | André Henriques | CC BY-SA 3.0 |
added 76 characters in body
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| Mar 9, 2013 at 13:10 | comment | added | Asterios Gkantzounis | By $p_i | (A \mathrm{or} B )$ I meant that $p_i | A\cdot B \forall 1 \leq i \leq n-1$ I am sorry if that was not clear | |
| Mar 9, 2013 at 13:05 | comment | added | Duchamp Gérard H. E. | $p_4=7=2^3-1$, is it legal in your definition ? do you admit the zeroth power ? | |
| Mar 9, 2013 at 12:56 | comment | added | user9072 | Sorry, it is still not clear to me what you are trying to say. What I now assume is that $A$ and $B$ are both products of $p_1, \dots, p_{n-1}$ (allowing repetead factors). But then how should this ever work? Assuming $p_3$ will be indeed $5$ given by $(4,9)$. But then how is the condition to be read that in the next step $(3,5)$ is excluded for a minimum of $2$? | |
| Mar 9, 2013 at 12:56 | history | edited | Asterios Gkantzounis | CC BY-SA 3.0 |
added 40 characters in body
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| Mar 9, 2013 at 12:30 | comment | added | Asterios Gkantzounis | @quid yes i wanted to say that only $p_i$ divide A and B,but i think that it doesnt affect the definition of minimum,thus $p_n$ is the same. | |
| Mar 9, 2013 at 12:22 | history | edited | Asterios Gkantzounis | CC BY-SA 3.0 |
added 4 characters in body
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| Mar 9, 2013 at 12:18 | comment | added | user9072 | Could you please explain more verbally (or by a correct formula) how $F_{n-1}$ is defined. Am I correct to assume that both $A$ and $B$ are divisible by (at least) one of the $p_i$ but not nececessarily the same one. And that the second use of $(A,B)$ means the gcd, while the first means the couple. | |
| Mar 9, 2013 at 12:11 | history | asked | Asterios Gkantzounis | CC BY-SA 3.0 |