Timeline for Calculating RSA Public Modulus from Private Exponent and Public Exponent
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
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| Jul 1, 2020 at 7:58 | history | edited | kodlu | CC BY-SA 4.0 |
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| Jun 30, 2020 at 8:44 | comment | added | fgrieu♦ | Remark that if $d=e^{-1}\bmodφ(n)$ as customary, then it holds $e\,d-1=k\,φ(n)$ for some positive $k<e$. And then for usual $e$ that leaves few choices for $k$, thus for $φ(n)$. | |
| Jun 30, 2020 at 4:05 | history | answered | kodlu | CC BY-SA 4.0 |