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Rezz5
Group Title
Describe in about 200300 words the RSA encryption algorithm,
Including:
*the method of encryption and decryption
*the three key theorems on which the algorithm depends
*the reason for the security of the algorithm
Thank you!
 one year ago
 one year ago
Rezz5 Group Title
Describe in about 200300 words the RSA encryption algorithm, Including: *the method of encryption and decryption *the three key theorems on which the algorithm depends *the reason for the security of the algorithm Thank you!
 one year ago
 one year ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
RSA i beleieve deals with 2 large primes and the totient function
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
im not about to type up 200 to 300 words on it tho, youll have to do that yourself
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
spose you are given a plaintext message: P we can encrypt it with a key "e" such that \(gcd(e,\phi(n))\) = 1 and n is the product of 2 large primes
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
this ensures us that a decryption key can be produced to uncode it all
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i had a powerpoit on it for the presentation i did in my number theory class; but the schools website is down at the moment
 one year ago

Rezz5 Group TitleBest ResponseYou've already chosen the best response.0
I understand, I have done this so fay from my notes but i dont understand what is what. SO 1) Choose two distinct primes, p and q 2) Find n such that n=pq 3) Use Q(n)= (p1)(q1) Q = phi, (*****) 4)Choose e such that 1<e<Q(n) e is publicly seen, and is relativly prime to Q(n) 5) Find d, which is multiplicatively inverse of e in Z _Q(n) i.e. de=1(mod(Q(n)) (****) d can be found using Euclidean algorithm and is a private key then i get lost, how do you decrypt the message? by using M, original message M=c^5 (mod(n))? where c is the coded message? (****) (****) which theorems are used here?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
one key aspect of the totient function is that it is a multiplicative function so long as the factors of a number are relatively prime.; there is a more complicated method for finding the phi value of any number. The phi value is simply the number of relatively prime values that are less the number; for example: the number 8: 1 2 3 4 5 6 7 8 ^ ^ ^ ^ there are 4 numbers less than 8 that are relatively prime to 8; (a,8)=1 the phi value of 8 is 4 the number 9: 1 2 3 4 5 6 7 8 9 ^ ^ ^ ^ ^ ^ i count 6 of them the phi value of 9 is 6 the phi value of a prime number is simply all the numbers less than it; phi(7) = 6, phi(13)=12, phi(17)=16 why? because a prime number is prime .... now to explain what it means to be a multiplicative function: f(ab) = f(a) * f(b) simple enough; and this is true for phi given that a and b are relatively prime. since 8 and 9 are relatively prime then, phi(72) = phi(8*9) = phi(8) phi(9) = 4 * 6 = 24 the reason for picking 2 large primes for the "n" is so that you can construct this phi value in a simple manner i believe
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
another thrm is that \[a^{\phi(n)}\cong a~(mod~n)\] proof: let "n" be a number with the set of relatively primes: {r1,r2,r3,...,r phi(n)}, since there are phi(n) relatively primes by definition. the product of these numbers is then: r1 r2 r3 ... r phi(n) = k therefore \(k\cong 1~(mod~n)\) lets take a number a such that (a,n) = 1; and multiply each r value in the set \[a r_1~ a r_2 ~ a r_3 ... ar_{ phi(n)} = a^{\phi(n)}k\] since (k,n)=1; and (a,n)=1; therefore a^(phi(n)) = 1 (mod n)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
how does this help us find an inverse?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[a*a^{\phi(n)1}=a^{\phi(n)}\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
Eulers thrm is the backbone of the RSA
 one year ago

Rezz5 Group TitleBest ResponseYou've already chosen the best response.0
aϕ(n)≅a (mod n), shouldnt this be aϕ(n)≅ 1 (mod n) ?
 one year ago
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