Timeline for Is knowing the private key of RSA equivalent to the factorization of $N$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 24, 2017 at 20:02 | history | protected | e-sushi | ||
| Jan 24, 2017 at 2:33 | answer | added | 111 | timeline score: 7 | |
| Jan 23, 2017 at 5:26 | answer | added | Carl Knox | timeline score: 6 | |
| May 11, 2014 at 4:35 | vote | accept | T.B | ||
| May 6, 2014 at 15:23 | history | tweeted | twitter.com/#!/StackCrypto/status/463700578884124672 | ||
| May 6, 2014 at 10:12 | comment | added | fgrieu♦ | Knowing $(N,e,d)$, including with big $e$, allows finding the factorization of $N$, by heuristic or deterministic methods pointed in Samuel Neves's nice answer. If $e$ is unknown but small, we can enumerate small values to find $e$, using $(x^d)^e\equiv x\pmod N$ for arbitrary $x$ such as 2 as a test of having reached the right $e$; and then are back to the same problem [reposted with fix]. | |
| May 6, 2014 at 9:14 | history | edited | Maarten Bodewes♦ | CC BY-SA 3.0 |
secret -> private and language
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| May 6, 2014 at 8:47 | answer | added | Samuel Neves | timeline score: 12 | |
| May 6, 2014 at 8:17 | comment | added | T.B | @CodesInChaos Do you mean that knowing $(N,e,d)$ if $e$ is big, then we cannot still factor $N$? | |
| May 6, 2014 at 6:44 | comment | added | CodesInChaos | If you know $e$, $d$ and $n$, you can efficiently factor $n$. Just knowing $d$ is not enough if $e$ is big. | |
| May 6, 2014 at 6:35 | history | asked | T.B | CC BY-SA 3.0 |