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.

$\blacksquare$