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A super-perfect number is a number with $\sigma(\sigma (n))=2n$.

How can I prove that every even super perfect number is from the form $n=2^k$ when $2^{k+1}-1$ is prime.

I tried every way please give me some guidance

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I wanted to ask what is "super perfect numbers"... –  DonAntonio Nov 27 '12 at 4:26
    

1 Answer 1

up vote 4 down vote accepted

Let $p$ be a prime. We know that $\sigma (p^\alpha)= \dfrac{p^{\alpha+1} -1}{p-1}$.

Now, let $m=p^\alpha r$ be an integer with $p\not| r$.
If $p \not| m$ then $\sigma(p m) = \sigma(p) \sigma(m) =(p+1) \sigma(m) > p \sigma(m)$.
If $p | m$ then we have $\sigma(pm) = \sigma (p^{\alpha +1})\sigma(r)= \dfrac{p^{\alpha +2} -1} {p-1} \sigma(r) >\dfrac{p^{\alpha +2} -p} {p-1} \sigma(r)= p\sigma(p^\alpha)\sigma(r) = p \sigma(m)$.

So, for any product $ab$ we have $\sigma(ab) \ge a \sigma(b)$ and equality holds exactly for $a=1$.


Now, we can conclude: Let $n= 2^ku$ with $u$ odd and $k \ge 1$.

We have $\sigma(n)= (2^{k+1}-1) s$ where $s=\sigma(u) \ge u$.

Therefore $\sigma(\sigma(n))= \sigma((2^{k+1}-1)s) \ge \sigma((2^{k+1}-1) s \ge 2^{k+1} u = 2n$.

So, for a super-perfect number, equality has to hold in this chain of inequalities.

The first inequality is true iff $s=1$ and the second inequality is true iff $2^{k+1}-1$ is prime (because the two trivial divisors already give $2^{k+1}$ and the number is clearly bigger than 1, this is where we use $k \ge 1$) and $s=u$.

So, in total $u=1$ and $2^{k+1}-1$ is prime and $n=2^k$.



Just to add a word of motivation:

Perfect numbers have the property $\sigma(n)=2n$, so with $\rho(n)=\sigma(n)-n$, this becomes $\rho(n)=n$.

The concatenation that is expected to stay the same size "on average" would be $\rho(\rho(n))=n$ which is related to the search for amicable numbers.

Now, instead we apply $\sigma$ twice, so we can expect that the number grows too much most of the time. This motivates the search for an inequality chain of the type $\sigma(\sigma(n))\ge 2n$ which then turns out to only work for even $n$.

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Well, the average value of $\rho(n)$ is $(\pi^2/6-1)n$, so I don't know that I'd say that $\rho(\rho(n))$ is $n$ on average -- I would expect it to be more like 0.3n to 0.5n. –  Charles Nov 28 '12 at 16:28
    
@Charles: I know that it is not the mean, I'm open to a suggestion of a better formulation. My point is that if you start with the concept of perfect numbers (and abundant numbers, etc), you will expect that concatenation of sigmas to be too large and search for an inequality. This is more linked to "median" than to "mean", but motivations is not about being the exact median, either. –  Phira Nov 28 '12 at 17:54
    
I think that finding the right formulation of 'staying the same size' is hard and interesting. In my past experience I find $\rho$ an awkward function and much prefer $\sigma_{-1}$ which has quite elegant properties. But the inner $\rho$ is forced on us in this problem, which causes some difficulties since $\rho(n)$ can't be assumed to act like ordinary numbers. –  Charles Nov 28 '12 at 18:24

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