## Related questions with answers

Question

Assuming that $n$ is an even perfect number, say $n=2^{k-1}\left(2^{k}-1\right),$ prove that the product of the positive divisors of $n$ is equal to $n^{k} ;$ in symbols,

$\prod_{d | n} d=n^{k}$

Solution

VerifiedAssume that $n$ is an even perfect number given by $n=2^{k-1}(2^k-1)$, where $2^k-1$ is a prime. So that,

$\begin{align*} \Pi_{d\mid n}d&=\overset{k-1}{\underset{j=0}\Pi}2^j\overset{k-1}{\underset{m=0}\Pi}2^m(2^k-1)\\ \\ &=(2^k-1)^k\big(2^{1+2+...+(k-1)}\big)^2\\ \\ &=(2^k-1)^k2^{k(k-1)}\\ \\ &=\big(2^{k-1}(2^k-1)\big)^k\\ \\ &=n^k \end{align*}$

$\textbf{Hence, the proof!}$

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