Timeline for Is it possible to validate a Public Key in RSA?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2015 at 6:18 | history | edited | otus |
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| Aug 10, 2015 at 18:45 | history | post merged (destination) | |||
| Aug 10, 2015 at 15:47 | comment | added | CodesInChaos |
There are efficient algorithms with which the owner of the key can prove that e and phi(n) are co-prime.
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| Aug 10, 2015 at 13:33 | answer | added | fgrieu♦ | timeline score: 6 | |
| Feb 10, 2012 at 8:29 | history | tweeted | twitter.com/#!/StackCrypto/status/167887909398134784 | ||
| Jan 16, 2012 at 8:14 | answer | added | hrishikeshp19 | timeline score: 3 | |
| Jan 11, 2012 at 1:45 | vote | accept | yydl | ||
| Jan 10, 2012 at 15:10 | answer | added | PulpSpy | timeline score: 11 | |
| Jan 10, 2012 at 9:05 | comment | added | j.p. | There are (rarely used) variants of RSA with more than two prime factors, so as long as the modulus $m$ and the public exponent $e$ are coprime, you can encrypt data with them. A sensible sanity check would be to verify that (1) gcd($m$, $e$) = 1; (2) $m$ does not have small prime divisors (test division with primes up to $2^{16}$) and (3) (only if you want to be extra sure) use Lenstra's elliptic curve factorization algorithm to check that $m$ does not have smallish factors. If anyone is able to decrypt, you cannot check. | |
| Jan 10, 2012 at 6:33 | answer | added | Jim McKeeth | timeline score: 3 | |
| Jan 10, 2012 at 5:26 | history | asked | yydl | CC BY-SA 3.0 |