Perfect Number is Primitive Semiperfect
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Theorem
Let $n \in \Z_{>0}$ be a perfect number.
Then $n$ is also a primitive semiperfect number.
Proof
Let $n$ be perfect.
From Divisor of Perfect Number is Deficient, all divisors of $n$ are deficient.
But from Semiperfect Number is not Deficient, it follows that the divisors of $n$ cannot be semiperfect.
Hence the result, by definition of primitive semiperfect number.
$\blacksquare$