Post #43
(ISO)
» Mon May 21, 2018 11:59 am
I think it can be done if we split the proof up into odd and even perfect numbers.
1) Let n be an odd perfect number with k divisors s_1 < s_2 < ... < s_k that is refactorable. Then n = s_1 + s_2 + ... + s_{k - 1} and n = ak for some integer a. Since n is odd, a and k have to be odd, and so do all s_i. But if k is odd, k - 1 is even. An even sum of odd integers is even, so we have that n is even, contradicting the hypothesis.
2) Let n be an even perfect number. Then n is of the form (2^k - 1)2^{k - 1} with 2^k - 1 prime, and n has 2k divisors. Observe that since 2^k - 1 is prime, so is k. Therefore, if 2k divides n = (2^k - 1)2^{k - 1}, then either k = 2, or k = 2^k - 1 (the only prime divisors of n). If k = 2, n = 6, which is not refactorable, and there is no prime k > 1 so that k = 2^k - 1.
You don't have ambiguity; you have options.