Math 403: Perfect Numbers
Euclid (circa 300 BC) and the classical Greek mathematicians were interested in numbers they
called perfect numbers, defined as those positive integers which are the sum of their proper positive
divisors (the sum of the positive divisors excluding the number itself). The number 6 is the first
perfect number, since 1, 2, 3 are all proper positive divisors of 6, and 6 = 1 + 2 + 3. One can check
that 28, 496, and 8128 are the next perfect numbers; these were known to the ancient Greeks as
well. Note that 6 = 21 (22 − 1), 28 = 22 (23 − 1), 496 = 24 (25 − 1), and 8128 = 26 (27 − 1). In the
Elements, Euclid proves that the formula 2n−1 (2n − 1) results in a perfect number if 2n − 1 is prime.
Throughout the discussion below, we need the finite sum S = 1 + a + a2 + · · · + ak . Can you find
a formula for S? Hint: Consider aS − S. Another fact we need is the Fundamental Theorem of
Arithmetic (or unique-prime-factorization theorem), which states that every natural number greater
than 1 can be written as a unique product of prime numbers: If N > 1 is a positive integer, then
mr
1 m2
we can write N = pm
1 p2 · · · pr , where the pj are the prime divisors of N with multiplicities mj .
For example, 24 = 23 · 3, 25 = 52 , and 300 = 22 · 3 · 52 .
1. Show that 23 (24 − 1) is not perfect.
2. Prove Euclid’s theorem: If p = 2n − 1 is prime, then 2n−1 p is perfect. Hint: What are the
proper positive divisors of 2n−1 p?
3. A theorem by Leonhard Euler (Swiss, 1707 − 1783) and Leonard E. Dickson (American,
1874 − 1954) says that the formula 2n−1 (2n − 1) includes all even perfect numbers. Prove this
theorem. Hint: Assume 2n q is perfect, where q > 1 is odd and n > 0. Use the definition of
perfect, and the proper positive divisors of q.
4. Prove that no power of a prime number can be perfect (pn cannot be perfect for p prime and
n ∈ N).
5. Prove that there are no odd perfect numbers of the form pq, where p and q are distinct prime
numbers.