Quasiperfect Number is Square of Odd Integer

Theorem
Let $n$ be a quasiperfect number.

Then:
 * $n = \paren {2 k + 1}^2$

for some $k \in \Z_{>0}$.

That is, a quasiperfect number is the square of an odd integer.

Proof
By definition of quasiperfect number:
 * $\map {\sigma_1} n = 2 n + 1$

where $\map {\sigma_1} n$ denotes the divisor sum of $n$.

That is, $\map {\sigma_1} n$ is odd.

Then from Divisor Sum is Odd iff Argument is Square or Twice Square:

$n$ is either square or twice a square.

Suppose $n = 2^k m^2$ is a quasiperfect number, where $m$ is odd and $k \in \Z_{\ge 0}$.

Then:

Hence we have:
 * $\paren {2^{k + 1} - 1} \divides \paren {2^{k + 1} m^2 + 1}$

Since:
 * $\paren {2^{k + 1} - 1} \divides \paren {2^{k + 1} m^2 - m^2}$

we have:
 * $\paren {2^{k + 1} - 1} \divides \paren {1 + m^2}$

$k > 0$.

Write:
 * $m^2 \equiv -1 \pmod {2^{k + 1} - 1}$

But by First Supplement to Law of Quadratic Reciprocity (extended to Jacobi symbols):
 * $\paren {\dfrac {-1} {2^{k + 1} - 1} } = \paren {-1}^{\frac {2^{k + 1} - 1 - 1} 2} = \paren {-1}^{2^k - 1} = -1$

Hence $-1$ is not a quadratic residue modulo $2^{k + 1} - 1$, contradicting the above.

Therefore we must have $k = 0$.

In this case, $n = m^2$, an odd square.