Timeline for Covering system of congruences with specific properties?
Current License: CC BY-SA 4.0
20 events
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| Aug 1, 2018 at 8:21 | history | edited | asad | CC BY-SA 4.0 |
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| Jul 31, 2018 at 23:45 | comment | added | Gerry Myerson | Any thoughts on the answers that have been posted, asad? | |
| Jul 30, 2018 at 23:03 | answer | added | Gerry Myerson | timeline score: 3 | |
| Jul 30, 2018 at 20:17 | history | edited | asad | CC BY-SA 4.0 |
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| Jul 30, 2018 at 19:39 | history | edited | asad | CC BY-SA 4.0 |
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| Jul 30, 2018 at 19:38 | comment | added | asad | @JoeSilverman, You are right, as those example that I put are known, but what I am looking for is the condition about quadratic non-residue relation that I imposed above. I added the condition more than one modulus in the question. | |
| Jul 30, 2018 at 17:38 | comment | added | Joe Silverman | @asad If you want to allow the $n_i$ to be equal, then the examples that you gave are misleading ,since they use different moduli. If you allow the moduli coincide, then by your definition a covering system includes the trivial system $$0\bmod n,\;1\bmod n,\;2\bmod n,\cdots,\;n-1\bmod n.$$ | |
| Jul 30, 2018 at 17:19 | comment | added | Philipp Lampe | There is no finite covering system. On the contrary, suppose that $(a_i,n_i)$ with $i\in[1,r]$ is a finite covering system of the odd non-square integers. Then no $n_i$ can be a power of $2$. Hence every $n_i$ has an odd divisor $d_i>1$. Fix an odd common multiple $l$ of $\{\,d_i\mid i\in [1,r]\}$ that is not a perfect square. Then some congruence must cover $l$ but $(l/n_i)=0$ for all $i$. | |
| Jul 30, 2018 at 16:42 | history | edited | asad | CC BY-SA 4.0 |
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| Jul 30, 2018 at 16:39 | comment | added | asad | @JoeSilverman, thanks, but I supposed that of course $r\geq2$ and also possible to have repeated moduli. | |
| Jul 30, 2018 at 15:39 | history | edited | Joe Silverman | CC BY-SA 4.0 |
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| Jul 30, 2018 at 15:30 | comment | added | Joe Silverman | You've said that $n_i\le n_{i+1}$, which would allow using a modulus more than once. I'm going to change it to $n_i\le n_{i+1}$, and also fix your formatting. | |
| Jul 30, 2018 at 13:58 | history | edited | asad | CC BY-SA 4.0 |
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| Jul 30, 2018 at 13:42 | comment | added | asad | @GerryMyerson, what if we consider to cover non perfect square odd positive integers? | |
| Jul 30, 2018 at 13:32 | comment | added | Gerry Myerson | Then some congruence has to cover 9. | |
| Jul 30, 2018 at 13:31 | comment | added | asad | @GerryMyerson, what if we assume odd numbers greater than 1? | |
| Jul 30, 2018 at 13:30 | history | edited | asad | CC BY-SA 4.0 |
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| Jul 30, 2018 at 13:26 | comment | added | Gerry Myerson | Some congruence has to cover the number 1, so it has to be $1\bmod n_i$. | |
| Jul 30, 2018 at 13:14 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
removed (covering) tag; related discussion on meta: https://meta.mathoverflow.net/questions/3545/how-should-questions-about-various-meanings-of-coverings-be-tagged
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| Jul 30, 2018 at 13:00 | history | asked | asad | CC BY-SA 4.0 |