Timeline for Can Shor's algorithm compromise RSA when both the public and private key are secret?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2016 at 19:53 | vote | accept | Mathew | ||
| Jun 11, 2016 at 21:03 | comment | added | Mathew | sorry, forgot to mention that, yes I would be using the same modulus | |
| Jun 11, 2016 at 21:02 | comment | added | poncho | @mathew: actually, that doesn't work with RSA unless K1 and K2 share the same modulus. | |
| Jun 11, 2016 at 21:00 | comment | added | Mathew | just to clarify, what I mean by this is that if you encrypt plain text P with key K1 and then key K2 you should get the same cipher text as if you were to encrypt P with K2 then K1. I think this property should be true for RSA because encrypting with RSA (if I understand correctly) is just raising the plain text to an exponent. order does not matter for exponents so it shouldn't matter for RSA either | |
| Jun 11, 2016 at 20:57 | comment | added | Mathew | The reason why I want to use RSA, even though both private and public keys are secret, is because of a property I think RSA has, that being that order does not matter in RSA encryption. I am going to post anther question about this but my 40 minute time limit is still making me wait | |
| Jun 11, 2016 at 20:50 | comment | added | SEJPM | Note (to anybody interested): $P^e-C$ will likely be much larger than the actual public key and would require a much larger quantum computer to be factored, so there may be a time period where we can factor 2048-bit numbers but not encryptions with $e=F_4=65537$ which would have $65537\times 2048=134219776$ bits. | |
| Jun 11, 2016 at 20:41 | history | answered | poncho | CC BY-SA 3.0 |