Cryptanalysis of RSA Factorization - Library(ISI Kolkata) - Indian ...

Chapter 2: Mathematical Preliminaries 14
Correctness of RSA algorithm
Here we present the proof of the correctness of the RSA algorithm, that is c d mod
N = m. We know that ed ≡ 1 (mod φ(N)). So, we can write ed = 1+kφ(N) for
some integer k. Let us first assume that m ∈ Z ∗ N (integers less than and co-prime
to N). Then,
c d ≡ m ed (mod N)
≡ m 1+kφ(N) (mod N)
≡ m·(m φ(N) ) k (mod N)
≡ m (mod N) (as m φ(N) ≡ 1 (mod N) by Euler’s Theorem [126]).
Now assume that m ∈ Z N \ Z ∗ N , that is gcd(m,N) > 1. If m ≡ 0 (mod p) and
m ≡ 0 (mod q), then m ≡ 0 (mod N). Then c d ≡ m ed ≡ m ≡ 0 (mod N). On
the other hand, let us assume without loss of generality that m ≡ 0 (mod p) and
m ≢ 0 (mod q). Then c d ≡ m ed ≡ m ≡ 0 (mod p). So, p divides m ed − m. We
also have
m ed ≡ m 1+kφ(N) (mod q)
≡ m·m k(p−1)(q−1) (mod q)
≡ m·(m q−1) k(p−1)
≡ m (mod q)
(mod q)
as m q−1 ≡ 1 (mod q).
So, q divides m ed −m. Hence, pq = N divides m ed −m i.e., m ed ≡ m (mod N).
Example of RSA Cryptosystem
Let us present a toy example to illustrate the basic operations.
Example 2.3. Bob chooses two primes p = 653,q = 877, and calculates N =
pq = 572681, φ(N) = (p−1)(q−1) = 571152. Suppose that Bob picks an integer
e = 13 as the encryption exponent. Now he has to find the decryption exponent d
which is e −1 in Z φ(N) . One can check that 13×395413 ≡ 1 (mod 571152). Hence,
the RSA parameters for Bob are
• public key: (13,572681), and