4.4. Suppose that Alice and Bob communicate using the RSA PKC. This means that Alice has a public modulus NA = PAqA, a public encryption exponent eд, and a private decryption exponent dд, where PA and q are primes and eд and dд satisfy eдdд = 1 (mod (pa − 1)(qa − 1)). Similarly, Bob has a public modulus NB = PBqB, a public encryption expo- nent eg, and a private decryption exponent dB. In this situation, Alice can simultaneously encrypt and sign a message in the following way. Alice chooses her plaintext m and computes the usual RSA ciphertext ев c=m²B (mod NB).
4.4. Suppose that Alice and Bob communicate using the RSA PKC. This means that Alice has a public modulus NA = PAqA, a public encryption exponent eд, and a private decryption exponent dд, where PA and q are primes and eд and dд satisfy eдdд = 1 (mod (pa − 1)(qa − 1)). Similarly, Bob has a public modulus NB = PBqB, a public encryption expo- nent eg, and a private decryption exponent dB. In this situation, Alice can simultaneously encrypt and sign a message in the following way. Alice chooses her plaintext m and computes the usual RSA ciphertext ев c=m²B (mod NB).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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[Algebraic Cryptography] How do you solve this? ty

Transcribed Image Text:4.4. Suppose that Alice and Bob communicate using the RSA PKC. This
means that Alice has a public modulus NA = PAqA, a public encryption
exponent eд, and a private decryption exponent dд, where PA and q are
primes and eд and dд satisfy
едdд = 1 (mod (pa − 1)(qa − 1)).
Similarly, Bob has a public modulus NB = PBqB, a public encryption expo-
nent eg, and a private decryption exponent dB.
In this situation, Alice can simultaneously encrypt and sign a message in
the following way. Alice chooses her plaintext m and computes the usual RSA
ciphertext
c = mB (mod NB).
She next applies a hash function to to her plaintext and uses her private
decryption key to compute
s = Hash (m)dA (mod NA).
She sends the pair (c, s) to Bob.
Bob first decrypts the ciphertext using his private decryption exponent dB,
mcdB (mod NB).
He then uses Alice's public encryption exponent eд to verify that
Hash (m) SA (mod NA).
Explain why verification works, and why it would be difficult for anyone
other than Alice to send Bob a validly signed message.
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