Cube of Prime is Multiplicatively Perfect
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Theorem
Let $n \in \Z_{>0}$ be a positive integer which is the cube of a prime number.
Then $n$ is multiplicatively perfect.
Proof
Let $n = p^3$ where $p$ is prime.
From Cube of Prime has 4 Positive Divisors, the positive divisors of $n$ are:
- $1, p, p^2, p^3$
Thus the product of all the divisors of $n$ is:
- $1 \times p \times p^2 \times p^3 = p^6 = n^2$
Hence the result, by definition of multiplicatively perfect.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $33$