Set $m=p_1 p_2 ... p_n$. It's well knwon that $2^{p_i}+1 | 2^m+1$. If $p,q$ are different odd primes, we have that $\gcd(2^p+1,2^q+1) | \gcd(2^{2p}-1,2^{2q}-1) = 2^2-1=3$. For all primes $p,q>3$ we get that $\frac{2^p+1}3, \frac{2^q+1}3$ are coprime (and clearly $>0$). We set $s_i=\frac{2^{p_i}+1}3$. For each $p_i$ we choose a prime $q_i$ with $q_i|s_i$. Then we set $r_i = \frac{s_i}{q_i}$. Now $r_i \neq q_i$ because otherwise $2^{p_i}+1=2q_i^2$, impossible $\mod 4$. Consider the set $S_i := \{1,q_i,r_i,s_i\}$: As we've seen, it really has four different elements, and elements from different sets are coprime. Thus we get a total number of $4^n$ different products $\prod x_i$ with $x_i \in S_i$. And every such product divides $2^m+1$.

It can be proven that number of divisors of this number is not less than $ 2^{2^{n-1}}$.

baysa wrote: It can be proven that number of divisors of this number is not less than $ 2^{2^{n - 1}}$. Really? Can you prove that? If that is true, it's very interesting.

Let \[ \frac {2^{p_i} + 1}{3} = q{}_{i{}_1}^{l_{i1}}..q{}_{i{}_m{}_i}^{l{}_im{}_i}. \] Let $ d|p_1...p_n$ divisor. Exist at least one prime divisor $ q_d|C_{2d}(2) = \prod{}_{d{}_1|d}(2^{d/d_1} + 1)^{\mu (d_1)}$. Therefore \[ \tau (2^{p_1...p_n} + 1)\ge 2^{\tau (p_1...p_n) - 1} = 2{}^{2{}^n - 1}. \]

See here, page 7 http://reflections.awesomemath.org/2008_2/cyclotomic.pdf Hello ZetaX , isn't it possible for $ s_i$ to be a prime. Forgive me if im wrong

ZetaX wrote: Set $m=p_1 p_2 ... p_n$. It's well knwon that $2^{p_i}+1 | 2^m+1$. If $p,q$ are different odd primes, we have that $\gcd(2^p+1,2^q+1) | \gcd(2^{2p}-1,2^{2q}-1) = 2^2-1=3$. For all primes $p,q>3$ we get that $\frac{2^p+1}3, \frac{2^q+1}3$ are coprime (and clearly $>0$). We set $s_i=\frac{2^{p_i}+1}3$. For each $p_i$ we choose a prime $q_i$ with $q_i|s_i$. Then we set $r_i = \frac{s_i}{q_i}$. Now $r_i \neq q_i$ because otherwise $2^{p_i}+1=2q_i^2$, impossible $\mod 4$. Consider the set $S_i := \{1,q_i,r_i,s_i\}$: As we've seen, it really has four different elements, and elements from different sets are coprime. Thus we get a total number of $4^n$ different products $\prod x_i$ with $x_i \in S_i$. And every such product divides $2^m+1$. si is prime number isnt it

I think i have found another solution to this exercise based only on Zgimondy theorem

my solution is a little bit long but the main idea that for each $2^{p_i}+1$ can be divided by two distincts primes wich are either distinct with other primes wich divide $2^{p_k}+1$ such that k>i

Maybe I'm making a mistake, but can't Zsigmondy give us an even stronger estimate than baysa's? If we let $a_1, a_2, \dots, a_{2^n}$ be, in order, the $2^n$ divisors of $p_1 p_2 \dots p_{n}$ (there are exactly $2^n$ since the $p_i$ are distinct) by Zsigmondy's Theorem (since $p_i > 3$ we don't have to worry about the exceptional case) \[2^{a_i} + 1 | 2^{p_1p_2\dots p_n} + 1\] has a primitive prime divisor $q_i$; that is \[q_i | 2^{a_i} + 1\] and \[q_i \nmid 2^{a_j} + 1\] for any $j < i$. Then \[q_1q_2 \dots q_{2^n} | 2^{p_1p_2\dots p_n} + 1\] where the $q_i$ are distinct, so that this number has at least $2^{2^n}$ divisors.

what is the baysa theorem

I was referring to the user baysa's previous post in this thread.