Timeline for Is there any way to solve this equation without knowing the inverse modulo? [closed]
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Jan 29, 2020 at 17:14 | history | closed |
Emil Jeřábek Ben Barber user44191 Felipe Voloch Max Alekseyev |
Not suitable for this site | |
| Jan 28, 2020 at 21:33 | comment | added | none | Since p is prime, each element except 1 has a multiplicative inverse, so k can be anything you want, and $m=c\cdot k^{-1}$ and you can compute $k^{-1}$ with the extended Euclidean algorithm. Usually you'd use exponentiation by repeated squaring whose inverse is discrete log, or point multiplication on an elliptic curve by repeated doubling, etc. I downvoted because this is not a research math question. | |
| Jan 28, 2020 at 20:46 | answer | added | Nemo | timeline score: 0 | |
| Jan 28, 2020 at 18:59 | comment | added | Aravind A | @Nate Eldredge Thankyou, your help is very much appreciated! | |
| Jan 28, 2020 at 18:39 | comment | added | Nate Eldredge | Sorry, I'm not really interested in discussing this any further. | |
| Jan 28, 2020 at 18:18 | comment | added | Aravind A | @Nate Eldredge Thankyou very much ! But can you please tell me if a same key k is used for different m . How can an attacker exploit the key repetition to break this one time pad ? Is there any "mathematical exploit" to this construction ? | |
| Jan 28, 2020 at 18:00 | comment | added | Nate Eldredge | This is the wrong site to ask about that, and I am the wrong person. But it would appear to have all the same advantages and disadvantages as any other one-time pad. Whether that's "good" depends on your requirements. It certainly doesn't seem to be any significant improvement on what's well known. | |
| Jan 28, 2020 at 17:47 | comment | added | Aravind A | @Nate Eldredge Can this be used as a good crypto system ? | |
| Jan 28, 2020 at 17:27 | comment | added | Nate Eldredge | Neither quantum computers nor any other kind can break a one-time pad. | |
| Jan 28, 2020 at 15:19 | comment | added | Aravind A | @Emil Jeřábek supports Monica Thankyou very much ! Is this easy for a Quantum computer to break ? | |
| Jan 28, 2020 at 15:16 | comment | added | Emil Jeřábek | It’s just as good and just as bad as any other one-time pad. (The most common construction uses XOR, which is slightly easier to implement, but mathematically speaking any abelian group will work just the same, such as the one you use.) | |
| Jan 28, 2020 at 14:54 | comment | added | Aravind A | @Emil Jeřábek supports Monica Can this be used as a good crypto system ? | |
| Jan 28, 2020 at 14:52 | vote | accept | Aravind A | ||
| Jan 28, 2020 at 14:40 | review | Close votes | |||
| Jan 29, 2020 at 17:14 | |||||
| Jan 28, 2020 at 14:23 | comment | added | Emil Jeřábek | Without knowing $k$, $m$ can be literally anything. So discrete log is not going to help you, the problem is simply impossible. | |
| Jan 28, 2020 at 14:21 | answer | added | JMP | timeline score: 3 | |
| S Jan 28, 2020 at 14:10 | history | suggested | Daniele Tampieri | CC BY-SA 4.0 |
Math Jaxed+ minor typo removal
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| Jan 28, 2020 at 13:15 | review | First posts | |||
| Jan 28, 2020 at 14:17 | |||||
| Jan 28, 2020 at 13:15 | review | Suggested edits | |||
| S Jan 28, 2020 at 14:10 | |||||
| Jan 28, 2020 at 13:11 | history | asked | Aravind A | CC BY-SA 4.0 |