Quasiperfect Number is Square of Odd Integer

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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:

\(\ds \map {\sigma_1} n\) \(=\) \(\ds \map {\sigma_1} {2^k} \map {\sigma_1} {m^2}\) Divisor Sum Function is Multiplicative
\(\ds \) \(=\) \(\ds \paren {2^{k + 1} - 1} \map {\sigma_1} {m^2}\) Divisor Sum of Power of Prime
\(\ds \) \(=\) \(\ds 2 n + 1\) Definition of Quasiperfect Number
\(\ds \) \(=\) \(\ds 2^{k + 1} m^2 + 1\)

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:

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


Aiming for a contradiction, suppose $k > 0$.

Write:

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

But by First Supplement to Law of Quadratic Reciprocity (extended to Jacobi symbols):

$\paren {\dfrac {-1} {2^{k + 1} - 1} } = \paren {-1}^{\frac {2^{k + 1} - 1 - 1} 2} = \paren {-1}^{2^k - 1} = -1$


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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.

$\blacksquare$


Sources

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