• Contributors
• |
• History
• |
• Req. Co-author
• |
• Watch
• |
• Suggest Correction |
• Comment

Lemma   An even number is perfect if and only if it equals $2^{k-1}(2^k-1)$ for some integer $k>1$ and $2^k-1$ is prime.

Proof. Let $\sigma$ denote the sum of divisors function. Recall that this function is multiplicative.

Necessity: Let $p=2^k-1$ be prime and $n=2^{k-1}p$ . We have that \begin{eqnarray*} \sigma(n) & = & \sigma(2^{k-1}p)\\ & = & \sigma(2^{k-1}) \sigma(p)\\ & = & (2^k-1)(p+1)\\ & = & (2^k-1)2^k\\ & = & 2n, \end{eqnarray*}which shows that $n$ is perfect.

Sufficiency: Assume $n$ is an even perfect number. Write $n=2^{k-1}m$ for some odd $m$ and some $k>1$ . Then we have $\gcd(2^{k-1},m)=1$ . Thus,$$ \sigma(n)=\sigma(2^{k-1}m)=\sigma(2^{k-1})\sigma(m)=(2^k-1)\sigma(m).$$ Since $n$ is perfect, $\sigma(n)=2n$ by definition. Therefore, $\sigma(n)=2n=2^km$ . Piecing together the two formulas for $\sigma(n)$ yields$$ 2^km=(2^k-1)\sigma(m).$$ Thus, $(2^k-1)\mid 2^km$ , which forces $(2^k-1)\mid m$ . Write $m=(2^k-1)M$ . Note that $1\le M<m$ . From above, we have: \begin{eqnarray*} 2^km & = & (2^k-1)\sigma(m) \\ 2^k(2^k-1)M & = & (2^k-1)\sigma(m) \\ 2^kM & = & \sigma(m) \end{eqnarray*}Since $m\mid m$ by definition of divides and $M\mid m$ by assumption, we have$$ 2^kM=\sigma(m)\geq m+M=2^kM,$$ which forces $\sigma(m)=m+M$ . Therefore, $m$ has only two positive divisors, $m$ and $M$ . Hence, $m$ must be prime, $M=1$ , and $m=(2^k-1)M=2^k-1$ , from which the result follows.

perfect number is owned by Warren Buck, Kimberly Lloyd.