<s>[INST] A Problem Concerning Odd Perfect Number Briefly, prove that every odd number having only three distinct prime factors cannot be a perfect number.
I know there are results much stronger than the one above, but I am looking for an answer without computer cracking (which means the computation can be carried out by a person), thanks.
 [/INST] Here is a proof, perhaps not the simplest. Let $n$ be an odd perfect number with three distinct prime factors. As observed in the comments, $n$ is of the form $3^a5^bp^c$, where $p\in\{7,11,13\}$.
It is a simple fact (observed by Euler) that exactly one of the exponents $a$, $b$, $c$ is odd. Let $q^r\parallel n$ be the corresponding prime power (i.e. $r$ is odd), then $q+1$ divides $\sigma(q^r)$, hence also $\sigma(n)=2n$. Therefore $(q+1)/2$ divides $n/q^r$, which forces $q\not\in\{3,7,11,13\}$, i.e. $q=5$. Hence $b$ is odd, while $a$ and $c$ are even. Now we can rewrite the equation $\sigma(n)=2n$ as
$$ 3^{a-1}5^bp^c=(1+3+\dots + 3^a)(1+5^2+\dots + 5^{b-1})(1+p+\dots+p^c),$$
where the middle sum on the right hand sum only contains even powers of $5$. In particular, $5$ divides $1+3+\dots + 3^a$ or $1+p+\dots+p^c$. The first possibility is excluded by $a$ even. Similarly, the second possibility is excluded for $p\in\{7,13\}$ by $c$ even, whence $p=11$. Now the condition becomes
$$ 3^{a-1}5^b11^c=(1+3+\dots + 3^a)(1+5^2+\dots + 5^{b-1})(1+11+\dots+11^c).$$
Looking at this equation modulo $4$, we get $3\equiv (b+1)/2\pmod{4}$, i.e. $b\equiv 5\pmod{8}$. In particular, $b\geq 5$. Combining this with
$$ \frac{5}{3}=\left(1+\frac{1}{3}+\dots+\frac{1}{3^a}\right)\left(1+\frac{1}{5^2}+\dots+\frac{1}{5^{b-1}}\right)\left(1+\frac{1}{11}+\dots+\frac{1}{11^c}\right),$$
we get that
$$ \frac{5}{3}>\left(1+\frac{1}{3}+\dots+\frac{1}{3^a}\right)\left(1+\frac{1}{5^2}+\frac{1}{5^4}\right)\left(1+\frac{1}{11}+\frac{1}{11^2}\right).$$
This shows that $a<4$, i.e. $a=2$. Hence $\sigma(3^a)=13$ divides $\sigma(n)=2n=2\cdot 3^a5^b11^c$, which is a contradiction.
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