There are no Odd Unitary Perfect Numbers
Theorem
No unitary perfect numbers exist which are odd.
Proof
Let $n$ be an odd number with prime decomposition $n = p_1^{a_1} \cdots p_k^{a_k}$.
By Sum of Unitary Divisors of Integer, the sum of its unitary divisors is $\ds \prod_{1 \mathop \le i \mathop \le k} \paren {1 + p_i^{a_i} }$.
To be a unitary perfect number, this must be equal to $2 n$.
Since each $p$ is odd, each $1 + p_i^{a_i}$ is even.
Hence $\ds \prod_{1 \mathop \le i \mathop \le k} \paren {1 + p_i^{a_i} }$ is divisible by $2^k$.
Since $n$ is odd, $2 n$ is divisible by $2$ but not $4$.
Thus $k = 1$.
So $n$ is a prime power.
By Sum of Unitary Divisors of Power of Prime, the sum of its unitary divisors is $1 + n$.
This cannot be equal to $2 n$, since $n > 1$.
Therefore there are no unitary perfect numbers which are odd.
$\blacksquare$
Sources
- 1994: Richard K. Guy: Unsolved Problems in Number Theory (2nd ed.)
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $87,360$