Week+5+and+6

Try any of the first 12 odd problems at the end of the chapter. Also, try these.

p192 6.8 1. c=5859, n=11413(=101*113), e=7469, m=? first, find d such that de=1mod(p-1 * q-1) d(7467)=1 mod (100*112) --> d(7467)+n(11200)=1 use extended algorithm to find 3(7467)-2(11200)=1, so **d=3** m=c^d mod n, so m=5859^3mod11413 using maple, m=1415

3. c=75, n=437, e=3, m=8 or 9 ? c=m^e mod n so 75=8^3 mod 437 or 9^3 mod 437 8^3 = 512 = 75 mod 437 9^3 = 729 = 292 mod 437 so, m=8

5. p=large prime, m=message x, e=encryption exponent, how do we find d? we want d such that d*e=1mod(p-1) d*e+(p-1)*k=1 for some integer k use extended algorithm to find d and k for the known e and p

7. Not sure if this one is right or not.... original message=c^d mod n  eve sends c*2^e, so nelson decrypts (c*2^e)^d mod n = (2^ed)(c^d) mod n so to get original message, eve just divides by 2^ed? user:bea001

1. What's so bad if P is composite? Give a precise answer.

2. If we found the factoring for PQ, how could we hack the RSA using PQ? This is an essential question. If you have the factorization of PQ, you can find phi from (P-1)(Q-1). e is already public knowledge, so you have enough information to use the Euclidean Algorithm to find d and y. Once you have d, you can decrypt the ciphertext to get the original message. user:meganternes 3. Let's follow up on Ian's observation that d + PQ for the decryption exponent will work. IS this true? c^(d+PQ) = c^(d)*c^(PQ). That c^(PQ)=0 mod PQ, so it would decrypt to the same as just regular c^d. So yes, if you find a d that works, d+PQ will also work. user:kml003 4. Fix the Maple program or write one that actually uses some number theory to perform the Extended Euclidean algorithm.

The personal problems for grades are just a day away.