Timeline for A prime number determined by its congruence relation?
Current License: CC BY-SA 3.0
27 events
| when toggle format | what | by | license | comment | |
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| Jul 15, 2017 at 1:31 | comment | added | user56983 | Fair enough. I answered too quick, without you clarifying. $\mathbb{F}$ for effort. Terribly sorry, then. | |
| Jul 14, 2017 at 23:40 | comment | added | mesel | @user7212389: Dear idt, your answer is not true since you do not understand the question even if I ıty to say what i mean step by step you reject to understand. | |
| Jul 12, 2017 at 23:53 | history | edited | user56983 | CC BY-SA 3.0 |
added 35 characters in body
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| Jul 12, 2017 at 23:44 | comment | added | mesel | I am begging you, can you please delete? | |
| Jul 12, 2017 at 23:37 | comment | added | user56983 | No. I won't delete my answer, because its not wrong. | |
| Jul 12, 2017 at 23:34 | comment | added | mesel | @user7212389: Can you please delete your answer ? | |
| Jul 12, 2017 at 23:21 | comment | added | user56983 | {433} = (1,1,3,6,4,4, ...) and none of the rest of the $x_i$ are zero until $p_i = 433$. | |
| Jul 12, 2017 at 23:17 | comment | added | user56983 | Counterexample: (1,1,3,6,0) = {13}. However (1,1,3,6) can also be formed by 433, see: primes.utm.edu/lists/small/1000.txt | |
| Jul 12, 2017 at 23:13 | comment | added | user56983 | Any time you intersect $\bigcap_{i<n}\{x_i \pmod{p_i} \}$ you end up with a prime residue class and there are infinitely many primes with those $x_i$ for each $p_i$. You end up with the lowest possible value for $p_n$, and you can not determine it uniquely. | |
| Jul 12, 2017 at 23:03 | comment | added | mesel | In my second comment, try to explain it can be said that $$p_n\leq 2^{n-k(n)}p_{[k(n)]}$$, which is nothing but a generaliazition of above idea. | |
| Jul 12, 2017 at 22:59 | comment | added | mesel | Let us assume that $k(n)=n-1$ for example. Then we know all prime numbers $p_1,p_2\ldots , p_{n-1}$. It is known that there exists a prime number between $m$ and $2m$ for $m\geq 2$. Thus, there exists a prime between $p_{n-1}$ and $2p_{n-1}$, which means $p_n\leq 2p_{n-1}$. That is, $p_n$ can not be arbitrarily large. By using thi idea, in my first comment I showed that "good $k(n)\leq n-1$". | |
| Jul 12, 2017 at 22:55 | history | edited | user56983 | CC BY-SA 3.0 |
added 90 characters in body just in case I missed anything.
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| Jul 12, 2017 at 22:40 | history | edited | user56983 | CC BY-SA 3.0 |
added 90 characters in body
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| Jul 12, 2017 at 22:18 | comment | added | user56983 | How is $p_n$ bounded by $k(n)$? | |
| Jul 12, 2017 at 22:17 | comment | added | user56983 | "You are claiming there are infinitely many prime number satisfying the congruence relation, which is true. But, $p_n$ is bounded by $k(n)$." And nowhere do you specify in the question (not the comment) that $p_n$ is bounded by $k(n)$. | |
| Jul 12, 2017 at 21:19 | comment | added | mesel | @user7212389: I did not understand most of your comments, even if I would like to understand. I edited the question. By the way I did not mean that $p_i<k(n)$. I mean that $i\leq k(n)$. That is why I do not understand your sentence "You failed to specify $p_i<k(n)$ in the question." | |
| Jul 12, 2017 at 19:12 | comment | added | user56983 | see comment. then give your own answer to a different problem down below by clicking "Add Another Answer" and accept that one instead. | |
| Jul 12, 2017 at 19:07 | comment | added | Gerhard Paseman | Except you are also given n and thus an upper bound of 2nlog n on p_n. Thus with n, k(n) can be O(log n). Gerhard "Sometimes Helps Having All Information" Paseman, 2017.07.12. | |
| Jul 12, 2017 at 19:04 | comment | added | user56983 | That's why the solution is precisely $k(n) = n$ There are infinitely many other primes in the residue class formed by the intersection because it is in a multiplicative group mod a primorial number. That implies the residue class is prime and therefore there is no singleton solution ${p}$. $p_{k(n)}$ still has to be prime, so it cannot be a negative number or zero, since otherwise the answer would be rational. You failed to specify $p_i < k(n)$ in the question. Your wording implied $\i < k(n)$ and $p_i < p_k(n)$ rather than $p_i < k(n)$ | |
| Jul 12, 2017 at 17:01 | comment | added | user56983 | It's okay. The only time you find 0 residue is $\pmod{ p_n }$ the rest are in the multiplicative groups $\pmod{ p_i }$, i.e. non-zero. to satisfy non-divisibility. | |
| Jul 12, 2017 at 17:01 | comment | added | mesel | In first comment, I showed that $k(n)\leq n-1$. | |
| Jul 12, 2017 at 16:59 | comment | added | user56983 | $i\leq k(n)$, $p_n$ uniquely determined is what you specified in the question. | |
| Jul 12, 2017 at 16:52 | comment | added | mesel | I did not understand what you mean, did you read the first comment ? | |
| Jul 12, 2017 at 16:49 | comment | added | user56983 | $p_n = p_{k(n)}$ precisely when $n - k(n) = 0$. | |
| Jul 12, 2017 at 16:40 | comment | added | mesel | I think the reason is the following; You are claiming there are infinitely many prime number satisfying the congruence relation, which is true. But, $p_n$ is bounded by $k(n)$. More specifically, $$p_n \leq p_{[k(n)]}2^{n-k(n)}$$ Ofcourse, there are better bounds also. Thus, the question turns to be whether there is a unique solution satisfying the inequality. | |
| Jul 12, 2017 at 16:19 | comment | added | mesel | If I am wrong, let me know I claimed that $k(n)\leq n-1$. Now we know all primes $p_1,...,p_{n-1}$. Then $p_n\leq 2p_{n-1}$. Now, since I know $x_1$ and $x_{n-1}$, and $p_1=2$ we know that $p_n\equiv t \ mod \ 2p_{n-1}$. Since $p_n\leq 2p_{n-1}$, $p_n=t$. | |
| Jul 12, 2017 at 15:09 | history | answered | user56983 | CC BY-SA 3.0 |