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This is a question that has been bothering me for a while. If anybody can help me solve this, it would take a great burden off my chest. Thank you very much!

Suppose $n$ is a natural number and $2^n-1$ is prime. Prove that $2^{n-1}(2^n-1)$ is perfect.

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I'm not sure why there are two different proofs. Would you be able to explain to me? Thank you so much! – Michael Feb 26 '13 at 21:08
Because the statement goes backwards as well, you're interested in the first implication. – L. F. Feb 26 '13 at 21:31
up vote 4 down vote accepted

The sum of divisors of a number with prime factorization $N=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ is $\sigma(N)=\frac{p_1^{a_1+1}-1}{p_1-1}\cdot \frac{p_2^{a_2+1}-1}{p_2-1}\cdots \frac{p_k^{a_k+1}-1}{p_k-1}$. So here we obtain $$\sigma(N)=\sigma(2^{n-1}p)=\frac{2^n-1}{2-1}\cdot \frac{p^2-1}{p-1}=(2^{n}-1)(p+1)=p\cdot 2^n=2N $$ as desired.

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Multiplicative function (+1) :) – Cortizol Feb 26 '13 at 21:02
Thank you very much for your time! – Michael Feb 26 '13 at 21:06

Euler's way: Define the sum of divisors function $\sigma(n)$. It is easy to show that $\sigma(m n) = \sigma(m) \sigma(n)$ if $m$ and $n$ are relatively prime. Now $n$ is perfect if $\sigma(n) = 2 n$.

Apply the above to $2^{n - 1} (2^n - 1)$ As $2^n - 1$ is prime, $\sigma(2^n - 1) = 2^n$:

$$ \sigma(2^{n - 1} (2^n - 1)) = \sigma(2^{n - 1}) \cdot \sigma(2^n - 1) = \left( \sum_{0 \le k \le n - 1} 2^k \right) 2^n = \left( 2^n - 1 \right) \cdot 2^n $$
Twice the number we started with.

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I really appreciate your help and I understand it now! – Michael Feb 26 '13 at 21:05

The sum of the divisors of $2^{n-1} (2^n-1)$ is

$$(2 + 2^2 + \ldots+2^{n-2}) (2^n-1) = 2^{n-1} (2^n-1)$$

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Thank you for the time to help me! – Michael Feb 26 '13 at 21:05

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