Show that no proper divisor of a perfect number is perfect.
proof. let n be perfect, and let be divisors of n. Now, .
Now, what should I do to get O(d) not equal to 2d for all ds?
Try proving the contrapositive: if d is perfect and n=kd is a multiple of d, then n is not perfect. To see this, notice that if d is the sum of its proper divisors, say , then . So n is the sum of some of its proper divisors. But n also has other divisors that are not multiples of k (for example, 1). Therefore the sum of the proper divisors of n is greater than n.