Exercise 2 (15 points). Notice that, when n is an odd number, n > 1, then 2] E Z,. Let f be a function f : {nEN | nodd AND n > 1} - N defined as follows: f(n) is the order of [2] modulo n. Prove that, for h, k odd, h, k > 1, and god(h, k) = 1, we have f(hk) = 1cm(f (h), f(k))