The exponent on a Mersenne prime must also be prime. To illustrate, consider 215-1. Now 15 is not prime, infact it is 3×5, so replace 23 with 8, and write 85-1. This is divisible by 8-1, just as xn-1 is divisible by x-1.
If p is a Mersenne prime, say 2k-1, then consider n = p×2k-1. Let's compute σ(n). There is one instance of p, so the first component of σ is 1+p. We know this better as 2k. The second factor of n is 2, possibly raised to a high power. Hence the second component of σ becomes 1+2+4+8+16+…+2k-1. This is equal to 2k-1. We know this better as p. Therefore σ(n) = 2k×p, which is twice n, hence n is a perfect number. Perfect numbers include 6, 28, and 496, corresponding to the first three Mersenne primes.
conversely, let n be a perfect even number. Let n contain k-1 powers of 2. Thus σ(n) includes 2k-1, which we will call j. If σ(n) = 2n, then j, an odd number, also divides n. If j is prime, it contributes 1+j, or 2k, to σ(n). This makes j a Mersenne prime, and n is a perfect even number as described above. If j is anything else, a power of a prime, or a product of multiple primes, σ(j) exceeds j+1. With σ(j) > 2k, n becomes abundant. Any additional factors, besides 2 and j, will also make n abundant. Therefore all perfect even numbers have been characterized. They correspond one to one with the Mersenne primes.
No perfect odd numbers have been found.
Consider the number 2(2h×n)+1, where n is odd, and n > 1. If x is 2(2h), then our fermat prime can be written as xn+1, which is divisible by x+1. To be prime, n must equal 1, whence the exponent is a power of 2. Review the 5 examples above. Each is 2 to a power of 2, plus 1. In fact, these are the only known Fermat primes, and we believe there are no others.