Question

Assuming that nn is an even perfect number, say n=2k1(2k1),n=2^{k-1}\left(2^{k}-1\right), prove that the product of the positive divisors of nn is equal to nk;n^{k} ; in symbols,

dnd=nk\prod_{d | n} d=n^{k}

Solution

Verified

Assume that nn is an even perfect number given by n=2k1(2k1)n=2^{k-1}(2^k-1), where 2k12^k-1 is a prime. So that,

Πdnd=Πj=0k12jΠm=0k12m(2k1)=(2k1)k(21+2+...+(k1))2=(2k1)k2k(k1)=(2k1(2k1))k=nk\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*}

Hence, the proof!\textbf{Hence, the proof!}

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