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Meta Crypto SE
Meta Crypto SE
2:37 AM
0
Q: Flagging conspiratorial rants

forestOn Sec.SE, it is not uncommon for people to flag (and delete) posts that spout conspiracies and do not give a real answer. I came across a question like that here and flagged it. The result was "helpful". Later, I came across another answer in the same vein and flagged it as well. That flag was d...

support flagging
 
 
9 hours later…
Niing
Niing
11:53 AM
 
 
2 hours later…
DannyNiu
DannyNiu
2:12 PM
可怜的保罗。。
 
SEJPM
SEJPM
@DannyNiu please only post english here.
 
Niing
Niing
3:02 PM
@SEJPM Hello, sorry for the ping... But I think I ask you here will be more suitable...
 
SEJPM
SEJPM
@Niing well then, go ahead
 
Niing
Niing
How could I make sure that $m^e > N$ will always happen? How about $m$ is too small so that $m^e > N$ didn't hold?
Thank you so much...
I really appreciate your time.
 
SEJPM
SEJPM
@Niing in any serious implementation you will pad your message $m$
so that the padded $m$ is about the same size as $n$
 
Niing
Niing
Sorry if the text is too small...
My textbook use $00$ to encrypt 'A', so did you meant that the segment should be as large as $n$?
So I should use some 'non-sense non-zero' value used as the padding?
 
SEJPM
SEJPM
@Niing you should use OAEP as your padding.
(for serious applications)
(for non-serious, PKCS#1 v1.5 padding should also do)
 
Niing
Niing
3:11 PM
This is my first time know about OAEP... But thank you!
 
SEJPM
SEJPM
(which essentially does some random padding on the high bytes of the padded message)
 
Niing
Niing
So there won't be a situation that after the padding $m^e$ still less than $N$?
 
SEJPM
SEJPM
@Niing exactly, even with $e=3$
 
Niing
Niing
I was thinking the case that I, myself, use $00$ as padding then after the message is padded $m^e$ is still less than $N$.
Oh thank you so much
I know too little about RSA... It's pretty sad that I thought I knew about RSA and gave myself a proof but I done it wrong...
The book I'm reading is Discrete Mathematics and It's Application, by Rosen btw...
Do you have any recommend book about RSA?
 
SEJPM
SEJPM
@Niing not about RSA specifically, but a good book on modern crypto (a bit formal, but I hope it won't scare you), is Introduction to Modern Cryptography by Katz and Lindell
2
 
Niing
Niing
3:20 PM
to SEJPM: Thank you! I really appreciate your time!
I will consider it
 
SEJPM
SEJPM
@Niing if you hover over one of my messages, there's a little arrow on the right, you can use that to answer to a message
it will highlight the referenced message and give the recipient an acoustical ping
 
Niing
Niing
@SEJPM Thank you, I never noticed that...
@SEJPM So if I don't apply the OAEP, or PKCS#1 v1.5, then what Nova said in the comment will happen, I meant that someone could just calculate the e-th root of $c$ and decrypt my message?
 
Meta Crypto SE
Meta Crypto SE
1
Q: How should conspiracy answers be treated?

SEJPMThere recently was a Meta Q&A where a discrepancy in the handling of so called "tin-foil hatted" answers was brought up, i.e. answers which claim that e.g. the NSA can break AES easily or answers that the NSA is behind a certain weak cryptography standard. This Q&A has shown that even within the...

discussion featured answers
 
SEJPM
SEJPM
@Niing yes, if your message is somewhat small and / or you use a small public exponent
 
Niing
Niing
3:35 PM
@SEJPM Wow that's cool
I'm taking some note about all your words...
@SEJPM If you have time, help me check about everything I done wrong: either the math part or idea part... I'm trying to fix it now.https://math.stackexchange.com/q/2703182/390226
 
SEJPM
SEJPM
@Niing well, you have successfully shown that if two ciphertexts are equal in RSA, their respective (padded) messages must be equal as well, which is to be expected
 
Niing
Niing
3:51 PM
@SEJPM In the comment, I don't understand why calculate $m^{e^{-1}}$ requires the factorization of $n$
 
SEJPM
SEJPM
@Niing it can be shown that computing $m^{e^{-1}}$ is equivalent to factoring $n$ (that is if you can do either efficiently, you can do the other)
 
Niing
Niing
@SEJPM Thank you... Wow you know about everything...
 
SEJPM
SEJPM
(also finding $\varphi(n)$ from $n$ is equivalent to factoring)
 
Niing
Niing
4:05 PM
@SEJPM It's me again... So instead of "in RSA the plaintext $M$ to be encrypted into $C$ should satisfy $C=M^e>n$, this is required so if you were attacker you won't be able to recover the plaintext by simply solving $C^{1/e}=M$ using just the $e$ in public key $(e, n)$.", I should say "$C=M^e>n$ promise that it's difficult to find M" or something?
 
 
2 hours later…
Niing
Niing
5:43 PM
@SEJPM: I've updated it again if you want to take a look. I mentioned you in my new post if you don't like it I can delete that part! link
 
 
2 hours later…
SEJPM
SEJPM
7:26 PM
@Niing ... by simply solving $C^{1/e}=M$ using standard real-valued e-th roots ...
@Niing the proof looks right to me
 
Q-Club
Q-Club
8:18 PM
@DannyNiu Is someone sad or poor?
 
 
3 hours later…
Niing
Niing
11:45 PM
Good morning here..
 
Niing
Niing
11:55 PM
@SEJPM Is the third approach in your answer called the Wiener's attack? I haven't learnt it though... But it's cool! (I just choose this link, but it's not about proof)
 

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