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This question already has an answer here:

how to prove :

an odd perfect number cannot be a prime number or a product of two prime numbers or power of prime number.

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marked as duplicate by Lord_Farin, Julian Kuelshammer, Dan Rust, Amzoti, Shuhao Cao Jun 9 '13 at 15:03

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    $\begingroup$ What?${}{}{}{}$ $\endgroup$ – Git Gud Mar 24 '13 at 19:01
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    $\begingroup$ Please write the question in your own language below the english version. With a bit of luck someone will help you translate. $\endgroup$ – Git Gud Mar 24 '13 at 19:07
  • $\begingroup$ I think the question is to prove that an odd perfect number cannot be a prime power or a product of two prime powers. $\endgroup$ – Librecoin Mar 24 '13 at 19:08
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Let $\sigma$ denote the sum-of-divisors function. A positive integer $N$ is said to be perfect if $\frac{\sigma(N)}{N}=2$.

Let $p^\alpha$ be an odd prime power and suppose that $N=p^\alpha$ is an odd perfect number. Since $\sigma(p^\alpha)=1+p+p^2+\dots+p^\alpha=\frac{p^{\alpha+1}-1}{p-1}$, we can observe that

$$\sigma(p^\alpha) = \frac{p^{\alpha+1}-1}{p-1} < \frac{p^\alpha p}{p-1} \implies \frac{\sigma(p^\alpha)}{p^\alpha} < \frac{p}{p-1} \le \frac{3}{2} < 2$$

since $p \ge 3$, so $N$ must be deficient.

Now, let $p^\alpha$ and $q^\beta$ be two odd prime powers with $p<q$ and suppose that $N=p^\alpha q^\beta$ is an odd perfect number. Noting that $\sigma$ is multiplicative, we obtain

$$\sigma(p^\alpha q^\beta) = \frac{p^{\alpha+1}-1}{p-1} \cdot \frac{q^{\beta+1}-1}{q-1} < \frac{p^\alpha p}{p-1} \cdot \frac{q^\beta q}{q-1} \implies \frac{\sigma(p^\alpha q^\beta)}{p^\alpha q^\beta} < \frac{p}{p-1} \cdot \frac{q}{q-1} \le \frac{3}{2} \cdot \frac{5}{4} = \frac{15}{8} < 2$$

since $p \ge 3$ and $q \ge 5$, so again $N$ must be deficient.

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  • $\begingroup$ what's deficient mean? $\endgroup$ – djechlin Jan 28 '15 at 3:16
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Mostly you push symbols around with the intent of getting a nice factorization:

$$(p+1)(q+1)=2pq \iff 2=(p-1)(q-1),$$

$$\frac{p^{a+1}-1}{p-1}=2p^a \iff p^{a+1}-2p+1=0 \iff p^2-2p+1=(p-1)^2\le0.$$

We also assume $a\ge1$ so $p^{a+1}\ge p^2$ for the prime-power case (since the $p^1$ case is easy).

You should be familiar with $\sigma_1$'s explicit formula $\displaystyle\sigma_1\left(\prod_p p^{v_p}\right)=\prod_p\frac{p^{v_p+1}-1}{p-1}$.

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