Final answer:
To find all positive integers 'm' satisfying Φ(m) = 16, we look at the possible forms of 'm' and find that the only positive integer that meets the condition is m = 34.
Step-by-step explanation:
The student has asked to find all positive integers m such that Φ(m) = 16, where Φ denotes Euler's totient function. The totient function Φ(m) counts the positive integers up to m that are relatively prime to m. To solve this, we can look at the properties of Φ(m) and realize that since 16 is a power of 2, m could be a power of 2, twice an odd prime, or four times an odd number. The only powers of 2 less than 16 are 1, 2, 4, and 8, so these can be eliminated as they do not satisfy the condition. The next option we consider is twice an odd prime (2p), where p itself is a prime. The totient of 2p is simply Φ(2p) = p - 1. However, for Φ(2p) to be 16, p - 1 must be 16, which would make p equal to 17. So, one solution is m = 2×17 = 34. Additionally, for four times an odd number (4q), Φ(4q) = 2(q - 1) provided q - 1 is an odd prime. Solving 2(q - 1) = 16 gives us q - 1 = 8, thus q = 9 which is not a prime number. So there are no solutions in this case. Therefore, the only positive integer satisfying Φ(m) = 16 is m = 34.