Public Key Homework Problems

1. Public Key Homework Problems • The ciphertext 5859 was obtained from the RSA algorithm using m = 11413 and e = 7457. Using the factorization 11413 = 101113, find the plaintext.  = 100112 = 11200 d = e-1 mod  = e-1 mod 11200 = 5793 decoded message = 58595793 mod 11413 = 910 • All calculations done using Maple: > 100*112;11200 > 7457^(-1) mod 11200;5793 > 5859^5793 mod 11413;910 > 5859^%% mod 11413;910

2. Public Key Homework Problems 2 Let p be a prime and n and x be integers. Prove by induction on n thatif s = pn, then xsp x. Base case (n = 1) If n = 1, then s = p and xs = xpp x by the corollary to Fermat’s Little Theorem Inductive case: Show true for s = pk implies true for s = pk+1 Assume that the result holds for s = pk, namely xs p x If t = pk+1, then t = (pk)p = sp and hence xt = (xs)p p xsCorollary to Fermat’s Little Theorem p xInductive Hypothesis

3. Public Key Homework Problems 3 Let p be a large prime. Suppose you encrypt a message x by computing y = xe mod p for some exponent e. • Explain why you would never choose e to be a power of p. By problem 2, if e is a power of p, then xe = x mod p, so the ciphertext would be the same as the plaintext. • Assuming e is not a power of p, how do you find a decryption exponent d such that yd = x? Write e = st where s is a power of p and t  p. That is, remove all factors of e equal to p to get t. Then, setting d = t-1 mod p, we have: yd = (xe)d = (xst)d = (xs)tdp xtd (by problem 2) and, since td p 1, xtd p x1 = x.

4. Public Key Homework Problems • Naïve Nelson uses RSA to decrypt a single ciphertext c, corresponding to message n. His public modulus is m and his public encryption exponent is e. Since he feels guilty about using his system only once, he agrees to decrypt any message that anyone sends him, as long as it is not c, and return the decrypted message to the sender. Evil Eve sends him the ciphertext 2ec. Explain how this allows Eve to find n. (2ec)d = (2e)dcd  2n, so Evil Eve just divides by 2 to obtain n.