1. Using the fact that 2018 = 2 x 1009 and that 1009 is prime, prove that there is no n E N such that (n) = 2018. Hint: Suppose 1009 is a factor of o(n). Where can this factor come from? 2. Prove that for all me N, there are at most finitely many n E N such that o(n) = m. Hint: Try to bound the prime factors of n in terms of m. 3. Prove that (n) is even for all n 3. Hint: Start with the case when n is a prime power.