1. In RSA, we know that factoring n allows you to recover φ(n). Show it works the other way as well: given φ(n), give an algorithm to efficiently recover the factors of n.
2. Here is one way in which RSA can be misused: suppose you have n = pq for distinct large primes p and q, and distinct encryption exponents e1, e2 where gcd(e1, e2) = 1. You publish n along with e1, e2. Show that if an adversary has C1 = M e1 mod n and C2 = M e2 mod n, then she can recover M. (Note: M is the same for C1 and C2.)
3. What irrational is represented by [1,3,3,3,3,...]? Show your work.
4. Let n be the product of distinct primes p,q and let e,d be inverses mod φ(n). Here d is less than 1/3 * n1/4 and p is between q and 2q. Factor n for the following values:
n = 151339355784268862864759120910925660361478568759617718849686393875476226860927728820162298798444560267328429 e = 47011760371307472336991404805847545822065047648754062545767018099994341894667675275660856278086265516059587