Let $n$ [odd perfect number] has only two odd prime divisors, say $p, q$, then sum of all divisors of $n$ is no more than $n\times\left(\sum_{i=0}^{\infty} \sum_{j=0}^{\infty}p^{-i}q^{-j}\right)\leq n\times\left(\sum_{i=0}^{\infty} \sum_{j=0}^{\infty}3^{-i}5^{-j}\right)=\frac{15n}{8}<2n.$

If there would be such numbers, then let it be $n=p^r*q^s$, because $n=p^r$ is clearly not respecting the condition. I'll denote $S(n)=sum$ $of$ $divisors$ $of$ $n$. So $S(n)=\frac{p^{r+1}}{p-1}*\frac{q^{s+1}}{q-1}$. After easy computations we get that $2p^{r+1}q^{s}+2p^{r}q^{s+1}-p^{r+1}-q^{s+1}-2p^{r}q^{s}+1=0$, which is false from parity. Am I wrong, because it seems like I didn't use that $n$ is odd nowhere except in the end?

@above You've used the wrong expression for sum of divisors; it should be $S(n)=\frac{p^{r+1}-1}{p-1}\cdot \frac{q^{s+1}-1}{q-1}.$

See wikipedia. It states with a reference that it has been proven that any odd perf number has more than 10 prime factors.

Let $n=p_1^{\alpha_1}p_2^{\alpha_2}$ then $\sigma(n)= \frac{p_1^{\alpha_1+1}-1}{p_1-1} \cdot \frac{p_2^{\alpha_2+1}-1}{p_2-1}=2p_1^{\alpha_1}p_2^{\alpha_2}$. This implies $\frac{p_1^{\alpha_1+1}-1}{p_1-1}=2p_2^{\alpha_2}$ and $\frac{p_2^{\alpha_2+1}-1}{p_2-1}= p_1^{\alpha_1}$ with $\alpha_1 \equiv 1 \pmod{4}$. Substituting and we get $$p_1^{\alpha_1}(p_2-1)= p_2 \cdot \frac{p_1^{\alpha_1+1}-1}{2(p_1-1)}-1.$$This implies $$p_2= \frac{2(p_1^{\alpha_1}-1)}{p_1^{\alpha_1}-p_1^{\alpha_1-1}-\cdots - p_1-1}.$$But $p_1^{\alpha_1}-p_1^{\alpha_1-1}-\cdots - p_1-1= p_1^{\alpha_1}- \frac{p_1^{\alpha_1}-1}{p_1-1}$ so $p_2=1+ \frac{p_1^{\alpha_1+1}-2p_1+1}{p_1^{\alpha_1+1}-2p_1^{\alpha_1}+1}= 2+ \frac{2(p_1^{\alpha_1}-p_1)}{p_1^{\alpha_1+1}-2p_1^{\alpha_1}+1}$. Note that if $\alpha_1 \ge 2$ then $2(p_1^{\alpha}-p_1) \ge p_1^{\alpha_1+1}-2p_1^{\alpha_1}+1$ or $(4-p_1)p_1^{\alpha}-2p_1 \ge 1$. This happens only when $p_1 \le 4$ so $p_1=3$. Combine with $\alpha_1 \equiv 1 \pmod{4}$ we find $p_1^{\alpha_1+1}-2p_1^{\alpha_1}+1\equiv 4 \pmod{8}$ and $4 \mid p_1^{\alpha_1}-1$ implies $2 \mid p_2$ or $p_2=2$, a contradiction.

@shinichiman could you explain more clearly what you did above? especially the final part. How does this imply that $\alpha_1=1$ or $p_2=2$? Thanks.

iamsomekelp wrote: @shinichiman could you explain more clearly what you did above? especially the final part. How does this imply that $\alpha_1=1$ or $p_2=2$? Thanks. Thanks for the notice, iamsomekelp. I edited my ugly solution.

Sorry for reviving this thread but i guess i have another solution On the contrary assume that we have an odd perfect number consisting of two prime divisors p and q. $n=p^rq^s$ $\sigma(n)= \frac{p^{r+1}-1}{p-1} \cdot \frac{q^{s+1}-1}{q-1}=2p^{r}q^{s}$

Now we can see $p^{r+1}-1=2q^s(p-1)$ And $q^{s+1}-1=p^r(q-1)$ $\implies q^{s+1}\equiv 1\pmod{p}-(1)$ $\implies p^{r+1}\equiv 1\pmod{q}-(2)$ As $(p^r+p^{r-1}+\cdots +1)=2q^s$ $$\implies 2q^s\equiv 1\pmod{p}-(3)$$Similarly $$p^r\equiv 1\pmod{q}-(4)$$From $(4)$ and $(2)$ $\implies p\equiv 1\pmod{q}$ As $(p,2)=1$ In$(3)$ we get $q^s\equiv 2^{-1}\pmod{p}$ And from$(1)$,$q\equiv 2 \pmod{p}$ But this gives a contradiction as $q=pl+2$ where $l>0$ And $p=qy+1$ where$y>0$ Which is a contradiction.

For the sake of contradiction, assume that $n$ has at most $2$ odd prime divisors. There are two cases; Case 1: $n$ has only one prime divisor, i.e $n = p^a$. This is simple because we require $$p^{a+1} - 1 = 2p^a(p-1) \implies p \mid p^{a+1} - 1$$which is impossible. Case 2: $n$ has two prime divisors, i.e $n = p^aq^b$. WLOG, we may assume $2< p < q$. This case gives $(p^{a+1} - 1)(q^{b+1} - 1) = 2p^aq^b(p-1)(q-1)$, which upon simplification, yields $$p^{a+1} + q^{b+1} -1 = 2p^{a+1}q^b + 2p^aq^{b+1} - p^{a+1}q^{b+1} - 2p^aq^b$$ Suppose that $p > 3$, then $$2p^{a+1}q^b + 2p^aq^{b+1} - p^{a+1}q^{b+1} - 2p^aq^b < 4p^aq^b+1 - p^{a+1}q^{b+1} - 2p^aq^b < 4p^aq^{b+1} - p^{a+1}q^{b+1} < 0$$ Whereas $p^{a+1} + q^{b+1} -1$ is clearly positive. This is a contradiction. Now suppose $p = 3$. If $q > 5$, then $q > 2p$, which means $$2 \cdot 3^{a+1}q^b < 3^aq^{b+1} \implies 2 \cdot 3^{a+1}q^b + 2 \cdot 3^aq^{b+1} < 3 \cdot 3^a q^{b+1} \implies 2 \cdot 3^{a+1}q^b + 2 \cdot 3^aq^{b+1} - 3^{a+1}q^{b+1} - 2 \cdot 3^aq^b < 3 \cdot 3^a q^b - 3^{a+1}q^{b+1} < 0$$ again, a contradiction. The only remaning case is when $p = 3, q = 5$. Left as an exercise.

First we set $n=p^a$, to get that $p^{a+1}-1=2(p-1)p^a$, but since we have that $p \nmid p^{a+1}-1$, we can't have that the conjecture holds. Now we set $n=p^{a}q^b$. We have that: $$2p^aq^b=\frac{p^{a+1}-1}{p-1}.\frac{q^{b+1}-1}{q-1}$$Since we can't have that $p$ divides $\frac{p^{a+1}-1}{p-1}$ we have that: $$\frac{q^{b+1}-1}{q-1} = 2p^a \; \; \text{and} \; \; \frac{p^{a+1}-1}{p-1} = q^b$$of course we claim this factor of $2$ WLOG. Now we focus on the second expression. Since we have that $q^{b+1}=q^b.q$, our second expression gets turned into (after simplification): $$qp^{a+1}=(2p^a-1)(p+q-1)$$Since we have that $p \nmid 2p^a-1$ we have that $2p^{a}-1=q$, since this expression must always be greater that $1$. Thus we have that $p+q-1=p^{a+1}$, which of course implies that $p^a(p-2)=p-2$. Since $p>2$ we must have that $p^{a}=1$, but $a> 0$ so we get a contradiction. Thus we have that $n$ must have at least $3$ distinct prime divisors.

We consider case for number of prime divisors of $n$. C1: If $n = p^k$ then sum of divisor is $\frac{p^{k+1}-1}{p-1}$ which should be equal to $2p^k$ its falls down to $2p^k-1 = p^{k+1}$ which is not true. C2: For $n = p^m.q^n$ we get number of divisor = $ \frac{p^{m+1}-1}{p-1}. \frac{q^{n+1}-1}{q-1}$ now if its equal to $2.p^mq^n.$ $$\frac{p^{m+1}-1}{p-1}. \frac{q^{n+1}-1}{q-1} = 2.p^m.q^n$$$$(p^{m+1}-1)(q^{n+1}-1) = 2.p^m.q^n(p-1)(q-1)$$$$2p^m.q^n(p+q-1) + 1 = p^{m+1}.q^{n+1} + p^{m+1} + q^{n+1}$$ Quote: Now we will prove $2(p+q-1) p^m.q^n + 1 =< (pq) p^{m}.q^{n}$ which give contradiction as $p^{m+1} + q^{n+1}$ very large number (>2). Now $2(p+q-1) < pq \implies 2<(p-2)(q-2)$ which is always true as $p,q$ are distinct primes and no one of them equal to $2$(they are odd perfect number) which give us $2(p+q-1) p^m.q^n < (pq) p^{m}.q^{n} \implies 2(p+q-1) p^m.q^n +1 < = (pq) p^{m}.q^{n}$ $\blacksquare$