#1

Does anyone know how RSA encryption works? I have a problem for class and I cant get it :/

The textbook RSA encryption scheme is deterministic (if the same message m is encrypted twice, then we get the same ciphertext). Consider instead the following

scheme. Let (e, n) be an RSA public key, with n = pq, and let (d, p, q) be the secret key, with ed = 1 mod Phi(n). To encrypt a message m epsilon{0,1,2,...,n-1}, compute a random r EpsilonZ group n and one of the following encryption pairs (all operations are modulo n):

(a) [a="r^e,B = m + r"]. To decrypt, compute B - A^d.

(b) [a="r,B = (m + r)^e"]. To decrypt, compute B^d - A.

Do any of these encryption pairs improve the security of textbook RSA? Why?

I know, I know its complicated, but any help would be greatly appreciated

The textbook RSA encryption scheme is deterministic (if the same message m is encrypted twice, then we get the same ciphertext). Consider instead the following

scheme. Let (e, n) be an RSA public key, with n = pq, and let (d, p, q) be the secret key, with ed = 1 mod Phi(n). To encrypt a message m epsilon{0,1,2,...,n-1}, compute a random r EpsilonZ group n and one of the following encryption pairs (all operations are modulo n):

(a) [a="r^e,B = m + r"]. To decrypt, compute B - A^d.

(b) [a="r,B = (m + r)^e"]. To decrypt, compute B^d - A.

Do any of these encryption pairs improve the security of textbook RSA? Why?

I know, I know its complicated, but any help would be greatly appreciated

#2

huh?

#3

(a) [a="r^e,B = m + r"]. To decrypt, compute B - A^d.(b) [a="r,B = (m + r)^e"]. To decrypt, compute B^d - A.

Do any of these encryption pairs improve the security of textbook RSA? Why?

This one improves it because it is the second answer and not the first, which is (a).

Good luck figuring this out TS

#4

i dont know why you would post this on UG expecting an answer, unless you're trying to look pretentious

#5

He really needs a new amp.

Seriously... I thought algebra and geometry were bad. That crypto stuff is really heavy. Good luck!

Seriously... I thought algebra and geometry were bad. That crypto stuff is really heavy. Good luck!

#6

What are the symbols, numbers, vectors, matrices, tensors, what?

#7

Lol thanks anyway guys, and I tried the pit cuz well maybe someone knew something and could help?

#8

a. d.n.p.

b. q.b.e

q. No, because it can be determined that d=q by anyone born after 1945, when that particular encryption became obsolete.

b. q.b.e

q. No, because it can be determined that d=q by anyone born after 1945, when that particular encryption became obsolete.

#9

Does anyone know how RSA encryption works? I have a problem for class and I cant get it :/

The textbook RSA encryption scheme is deterministic (if the same message m is encrypted twice, then we get the same ciphertext). Consider instead the following

scheme. Let (e, n) be an RSA public key, with n = pq, and let (d, p, q) be the secret key, with ed = 1 mod Phi(n). To encrypt a message m epsilon{0,1,2,...,n-1}, compute a random r EpsilonZ group n and one of the following encryption pairs (all operations are modulo n):(a) [a="r^e,B = m + r"]. To decrypt, compute B - A^d.

(b) [a="r,B = (m + r)^e"]. To decrypt, compute B^d - A.

Do any of these encryption pairs improve the security of textbook RSA? Why?

I know, I know its complicated, but any help would be greatly appreciated

#10

The best encryptions have no formulae.

#11

a. d.n.p.

b. q.b.e

q. No, because it can be determined that d=q by anyone born after 1945, when that particular encryption became obsolete.

What? how could it become obsolete in 1945 when it was first theorized in 1978 and only implimented in the last 10 years?

#12

its really not nearly as complicated as it seems. It took me around 20-30 minutes to code it up. depends though on what language you want to code it in.

python probably the easiest route.

python probably the easiest route.

#13

Cryptography?

#14

I thought cryptology had something to do with tomb raider...

#15

I'm not trying to code it though :/ all I want to know is why they are more secure the the original cipher method

#16

Actually never mind, I think i figured it out

#17

What school do you go to? Hacker University?