# proof of the correspondence between even 2-superperfect numbers and Mersenne primes

Statement. Among the even numbers, only powers of two $2^{x}$ (with $x$ being a nonnegative integer) can be 2-superperfect numbers (http://planetmath.org/SuperperfectNumber), and then if and only if $2^{x+1}-1$ is a Mersenne prime. (The default multiplier $m=2$ is tacitly assumed from this point forward).

Proof. The only divisors of $n=2^{x}$ are smaller powers of 2 and itself, $1,2,\ldots,2^{x-1},2^{x}$. Therefore, the first iteration of the sum of divisors function is

 $\sigma(n)=\sum_{i=0}^{x}2^{i}=2^{x+1}-1=2n-1.$

If $2n-1$ is prime, that means its only other divisor is 1, and thus for the second iteration $\sigma(2n-1)=2n$, and is thus a 2-superperfect number. But if $2n-1$ is composite then it is clear that $\sigma(2n-1)>2n$ by at least 2. So, for example, $\sigma(8)=15$ and $\sigma^{2}(8)=24$, so 8 is not 2-superperfect. One more example: $\sigma(16)=31$ and since 31 is prime, $\sigma^{2}(16)=32$.

Now it only remains to prove that no other even number $n$ can be 2-superperfect. Any other even number can of course still be divisible by one or more powers of two, but it also must be divisible by some odd prime $p>2$. Since the sum of divisors function is a multiplicative function, it follows that if $n=2^{x}p$ then $\sigma(n)=\sigma(2^{x})\sigma(p)$. So, if, say, $p=3$, it is clear that $(2^{x+3}-4)>2^{x+1}3$, and that on the second iteration this value that already exceeded twice the original value will be even greater. For example, $12=2^{2}3$, and $\sigma(12)=2^{5}-4$ which is greater than $2^{3}3$ by 4. With any larger $p$ the excess will be much greater.

Title proof of the correspondence between even 2-superperfect numbers and Mersenne primes ProofOfTheCorrespondenceBetweenEven2superperfectNumbersAndMersennePrimes 2013-03-22 17:03:48 2013-03-22 17:03:48 PrimeFan (13766) PrimeFan (13766) 5 PrimeFan (13766) Proof msc 11A25