<s>[INST] an odd perfect number cannot be a prime number or a product of two prime numbers or power of prime number. how to prove :
an odd perfect number cannot be a prime number or a product of two prime numbers or power of prime number.
 [/INST] 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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