Timeline for On square root modulo $2^t-1$
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 9 at 6:50 | comment | added | J. W. Tanner | For $a=1$, take $x\equiv\pm1$; for $a=2, $ take $x\equiv2^{(t+1)/2}$ for $t$ odd and no solutions for $t $ even; for $a=3$, no solutions for $t>2$; for $a=4$, take $n=\pm2$; for $a=5 $ or $6$, no solutions for $t>1$ | |
| Jul 9 at 6:30 | answer | added | Mark Schultz-Wu | timeline score: 6 | |
| Jul 9 at 5:57 | comment | added | Turbo | There may be another method for special numbers. | |
| Jul 9 at 5:37 | comment | added | Turbo | That is why I said factorization of $2^t-1$ is not given to us. | |
| Jul 9 at 5:28 | history | asked | Turbo | CC BY-SA 4.0 |