We enunciate two results that we can prove using FLT. Theorem 1 Let p and (.1 be two prime numbers such that p 7E q. Let n = pq and 60 = gb(n) = (10 1)(Q 1). If h 2 0 then bka'\"r1 E b (mod n), for (ill b E Z. Theorem 2 (RSA) Let p and q be two prime numbers such that p % q. Let n = pq and 60 = (p 1)(q1). Let e, d E N be two positive integers such that (e d) E 1 (mod 50). Let F : Jn > J\" and G : J\" > J\" be the functions given by F (m) = (me) mod n, G (c) 2 (ed) mod n. Then both functions are bijectiue and G = F '1. This is c=F(m) 4=>m= G(c). The pair (n, e) is called the public key and (n, d) the private key. As an example, we can consider p = 5, q = 11, n = 55 and 60 = 40. Choosing e = 7 and d = 23, we have that e, d > 0 and we can verify that (e d) mod 60 = 161 mod 40 = 1. This implies that (e ct) E 1 (mod 60). Using the RSA theorem, we conclude that the following functions are inverses of each other: F : J\" > Jn and G : 1,, > Jn, where F (m) = m7 mod 55, G(c) = C23 mod 55. We have, for instance, F(2) = 18 and C(18) = 2. Also, F(8) = 2 and G(2) = 8. 7. Consider the RSA theorem. Fixing p, q, n 2 pg and the public key 8, assume that we can nd two different natural numbers d1 and d2 such that (6 031) E 1 (mod 60) and (6 (12) E 1 (mod 60). Using these numbers, we dene two functions GI : J > J1.1 and Gg : J\" > J\