Timeline for Special cases of Dirichlet's theorem
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2020 at 13:43 | comment | added | Jeff Strom | @S.SundaraNarasimhan I just fixed a typo in the inductive construction of $N_k$ | |
| Aug 31, 2020 at 13:42 | history | edited | Jeff Strom | CC BY-SA 4.0 |
fixed typos
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| Aug 31, 2020 at 3:27 | comment | added | Sundara Narasimhan | @JeffStrom Thank you for your prompt reply. I will look into it. | |
| Aug 30, 2020 at 1:30 | comment | added | Jeff Strom | @S.SundaraNarasimhan,@RP_ . I've filled in the details in the "generalized primes congruent to 3 mod 4" theorem | |
| Aug 30, 2020 at 1:29 | history | edited | Jeff Strom | CC BY-SA 4.0 |
provided detailed proof
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| Aug 29, 2020 at 6:03 | comment | added | Sundara Narasimhan | @JeffStrom I agree that you can find a $g$ such that $(g,p_i)=1$ using Chinese remainder theorem. Can you explain why we can find a $g$ such that $[g]\notin G$ simultaneously. | |
| Mar 16, 2017 at 1:48 | history | edited | Jeff Strom | CC BY-SA 3.0 |
added 33 characters in body
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| Mar 16, 2017 at 1:46 | comment | added | Jeff Strom | Since all the $p_i$ are prime to $n$ you can use the Chinese Remainder Theorem to produce $g$. | |
| Mar 16, 2017 at 1:17 | comment | added | R.P. | I have no doubt it is correct, but I find it hard to follow. Why is it obvious that the conditions on $g$ can be fulfilled? In particular it is not immediately clear to me that one condition does not exclude the other... | |
| Mar 16, 2017 at 1:03 | history | edited | R.P. | CC BY-SA 3.0 |
fixed tex
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| Mar 16, 2017 at 0:51 | history | answered | Jeff Strom | CC BY-SA 3.0 |