Timeline for Least simultaneous quadratic non-residue
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 30, 2015 at 22:36 | comment | added | Brando | Unfortunately, this argument seems to break down when there are three primes $p,q,r$. What could happen that $n_p=n_q<n_r$. Then if $(n_r|p)=-1$ and $(n_r|q)=1$ I don't see the same trick working. | |
| Nov 30, 2015 at 3:28 | comment | added | Brando | No, I am thinking that $p$ and $q$ are very near in size, which prohibits this obstruction. | |
| Nov 30, 2015 at 3:26 | comment | added | Lucia | The only fly in the ointment is if $n_p=q$ or $n_q=p$. But the OP is perhaps not interested in these pathological cases. | |
| Nov 30, 2015 at 3:25 | comment | added | Brando | Aha I suspected it should be simple! | |
| Nov 30, 2015 at 3:25 | vote | accept | Brando | ||
| Nov 30, 2015 at 3:24 | vote | accept | Brando | ||
| Nov 30, 2015 at 3:25 | |||||
| Nov 30, 2015 at 3:16 | history | answered | Fan Zheng | CC BY-SA 3.0 |