Every superperfect number(except $2$) is a square number?

Question

I recently read about superperfect numbers: $σ^2(n) = 2n$, where $σ(n)$ is the divisor function.

I saw that the first few numbers were: $2, 4, 16, 64, 4096, 65536, 262144$, which are all square-numbers(except $2$). Source: A019279.

Is there a solid proof to this sentence, or is this true in average?

I tried to use the attitude of square numbers, that they have odd numbers of divisors, but I can't really move forward.

Any help appreciated.

Answer 1

It is known that every super-perfect number but $2$ is a perfect square.

The even super-perfect number are all even powers of $2$, more specifically they are $2^k$ such that $2^{k+1}-1$ is a prime, which implies in particular that $k+1$ is prime and thus $k$ is even or $1$.

For an argument see Super Perfect numbers (Thanks to ET93 for the link!)

It is conjectured that no odd super-perfect numbers exists. But if one exists it is known that it would be a square and $n$ or $\sigma(n)$ would be divisible by at least three primes.

That it is a perfect square was shown by Kanold in "Über "Super perfect numbers". Elemente der Mathematik 24 (1969): 61-62..

Also, see the Math World page of super-perfect numbers for an overview.

Answer 2

If $n$ is an even superperfect number then $n$ must be a power of $2$, $n=2^k$, such that $2^{k+1}-1$ is a Mersenne prime. In particular with the exception of $k+1=2$ which gives the counterexample $n=2$ all $k+1$ are odd primes and this means $n=2^k$ is a perfect square.

It is not known whether there are any odd superperfect numbers. An odd superperfect number $n$ would have to be a square number such that either $n$ or $\sigma(n)$ is divisible by at least three distinct primes.There are no odd superperfect numbers below $7\times 10^{24}$.