For any $k\ge 1$ there is such an $n$ with exactly $k$ prime factors. For $k=1,\ n=3^t$ works for every $t\ge 1$. Take $t$ s.t. for $n_1=3^t,\ 2^{n_1}+1$ has a prime factor $p_2$ larger than $3$. Now take $n_2=n_1p_2$. Then $n_2|2^{n_1}+1|2^{n_2}+1$, and $2^{p_2}+1$ has a prime factor $p_3\not|n_2$. This is because $(2^{n_1}+1,2^{p_2}+1)=3,\ p_2\not|2^{p_2}+1$, and $2^{p_2}+1$ cannot be a power of $3$, since $p_2>3$ (I'm using the well known and easy to prove fact that the only positive integer solution $(a,b),a>1$ to $3^a=2^b+1$ is $(a,b)=(2,3)$). Then take $n_3=n_2p_3$. Just like above, we deduce that $2^{p_3}+1$ has a prime factor $p_4$ which is coprime to $n_3$, and take $n_4=n_3p_4$, and so on. $n_k$ will have exactly $k$ prime factors and will satisfy $n_k|2^{n_k}+1$.

Valentin Vornicu wrote: Does there exist a positive integer $n$ such that $n$ has exactly 2000 prime divisors and $n$ divides $2^n + 1$? You can find my paper for a general problem of this problem in This file

Let $N=2^n+1$. We will assume for the sake of contradiction that $n|N$. $2^n+1 \equiv 0$ (mod $n$) $\Rightarrow 2^n \equiv -1$ (mod $n$). So 2 does not divide $n$, and so $n$ is odd. Select an arbitrary prime factor of $n$ and call it $p$. Let's represent $n$ in the form $p^am$, where $m$ is not divisible by $p$. Note that $p$ and $m$ are both odd since $n$ is odd. By repeated applications of Fermat's Little Theorem: $N = 2^n+1 = 2^{p^am} + 1 = (2^{p^{a-1}m})^p + 1 \equiv 2^{p^{a-1}m} + 1$ (mod $p$) Continuing in this manner, and inducting on k from 1 to $a$, $2^{p^{a-k}m}+1 \equiv (2^{p^{a-k-1}m})^p + 1$ (mod $p$) $\equiv 2^{p^{a-k-1}m} + 1$ (mod $p$) So we have $N \equiv 2^m+1$ (mod $p$) Since $p$ is relatively prime to $m$, $N \equiv 1+1$ (mod $p$) $\equiv 2$ (mod $p$) Since $p$ is odd, $N$ is not divisible by $p$. Hence $N$ is not divisible by $n$. So we have a contradiction, and our original assumption was false, and therefore $N$ is still not divisible by $n$.

Hmmm... I made a mistake here somewhere but I do not see it.

Yes, since there are lots of integers $n$ such that $n$ divides $2^n + 1$, and the statement is true!

There exists a pretty beautiful generalization: " Let $s, a, b$ positive integers, such that $GCD(a,b) = 1$ and $a+b$ is not a 2-power. Show that there exists infinitely many $n \in N$ such that --- $n=p_1^{e_1} \cdot p_2^{e_2} \cdot p_3^{e_3} \cdots p_s^{e_s} \cdot$ is the canonical factoring of $n$. --- $n|(a^n+b^n)$ "

Valentin Vornicu wrote: Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$? Here is a solution that I don't think has been mentioned yet:

For any i, 2^3^i+1 is divisible by 3^i (the proof is easy with euler`s theorem+induction and maybe with primitive roots (2 is primitive root modulo 3^i for any i)). Hence, for i=1999, 3^1999 has 2000 divisors and it satisfies the asked in the problem. it is correct?? please , answer me. bye sorry for my english.

Well, the point is to find a number which has exactly $ 2000$ prime divisiors, whereas $ 3^{1999}$ has only one ($ 3$). But it is a very nice thought to look at powers of $ 3$, when the problem considers powers of $ 2$ (vide grobber's solution, which I do not completely understand yet, but it seems very nice and simple)

Answer: Yes. We say that $n$ is Korean if $n \mid 2^n+1$. First, observe that $n=9$ is Korean. Now, the problem is solved upon the following claim: Claim: If $n > 3$ is Korean, there exists a prime $p$ not dividing $n$ such that $np$ is Korean too. Proof. I claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$, which exists by Zsigmondy theorem. Obviously $p \neq 2$. Then: Since $p \nmid 2^{\varphi(n)}-1$ it follows then that $p \nmid n$. Moreover, $p \mid 2^n+1$ since $p \nmid 2^n-1$; Hence $np \mid 2^{np} + 1$ by Chinese Theorem, since $\gcd(n,p) = 1$. $\blacksquare$ EDIT: The version of the proof I posted four years ago was incorrect. This one should work.

I feel like it isn't interesting to make any new remark on this problem, but anyway I'm posting my approach too. If we show that for any positive integer k, there exists a positive integer n with exactly k distinct prime divisors such that n | 2^n + 1, then we are done, since the problem asks us to examine a special case, more exactly k = 2000. Furthermore, we can even show that we can find these n's divisible by a power of 3, which will help us on our proof. We use induction on k. k = 1, we can choose n(1) = 3, which clearly satisfies the conditions. Assume that k >= 1, and there exists n(k) = 3^t * m, where gcd(3, m) = 1, and m has exactly (k - 1) distinct prime divisors. So, we have n(k) | 2^(n(k)) + 1. Before generating n(k+1) from n(k), let us look at the number 3n(k), which clearly has k distinct prime divisors. 2^(3n(k)) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1). Since we must have n(k) always odd because of the fact that n(k) = 1 (mod 2), we deduce that 3 | (2^(2n(k)) - 2^(n(k)) + 1), so we have that 3n(k) | 2^(3n(k)) + 1. It is enough to find a prime p, such that p | 2^(3n(k)) + 1 and p doesn't divide (2^(n(k)) + 1), which could guarantee us that p doesn't divide n(k) and consequently, we could generate n(k + 1) = 3n(k)*p, which could clearly work by observing that 2^(3n(k)*p) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1)*A. But, since we can pick up this prime p by Zsigmondy, we are done. Note that in case of not using Zsigmondy, we can observe that gcd(a^2 - a + 1, a + 1) = gcd(3, a + 1) = 1 if a is not 2 mod 3 and 3 if a = 3k + 2. But if a = 3k + 2, then a^2 - a + 1 is divisible by 3 but not by 9, so we could pick up any prime p that divides (a^2 - a + 1) / 3.

Let $n=\prod_{i=1}^{2000}{p_i}$ and wlog let $p_1<p_2<\dots<p_{2000}$.I dscribe a process on how to construct such an $n$.By the problem we have $2^{2n} \equiv 1\pmod{p_1} \Rightarrow ord_{p_1}{2} |2n$ and so by the minimality of $p_1$ we get that $ord_{p_1}=1$ or $ord_{p_1}=2$.In the first case we get $p_1|1$ which is absurd while in the second case we get $p_1|2^2-1=3 \Rightarrow \boxed{p_1=3}$.Similarly it is easy to note that $ord_{p_2}{2}|2n$ and so by the choice of $p_2$ we get $ord_{p_2}{2}|2*3=6$.Clearly there exists such a prime such that $ord_{p_2}{2}=6$(By Zsigmondy!!)In general we have $2^{2n} \equiv 1\pmod{p_k} \Rightarrow ord_{p_k}2|p_1p_2\dots p_{k-1}$.Clearly there exists a prime such that $ord_{p_k}2=p_1p_2\dots p_{k-1}$(Again Zsigmondy!!).So we are done!!!(In fact by this procedure we may fix any number of primes).

Since $N=3^k$ all satisfay the condition.Now,it is enough to prove that numbers of the form $N=2{}^3{}^k+1$ have infinitely many primes dividing them,but this is easy to prove,since we have for $n<m$ $2{}^3{}^n+1$ divides $2{}^3{}^m+1$ so suppose opposite.Now,we just need to prove that $a+1$ and $a^3+1$ can't have identical sets of primes for $a>2$,and this is true because $GCD(a+1,a^2-a+1)$ is at most $3$ and $a^2-a+1>3$ for $a>2$ so we are done.

For any $K$, there is an $n$ with exactly $K$ prime divisors for which $n|2^n+1$. The "exactly $K$" part seems to be makeup. Indeed it is, because if $n|2^n+1$, where $n=q^\alpha \cdot m$ with $q$ the greatest prime divisor of $n$, then $m|2^m+1$ as well. Indeed, it suffices to prove that $m|2^n+1\implies m|2^m+1$. Indeed, if we put $d=ord_m(-2)$ ($m$ is odd), then whenever $m|2^\ell+1$ for odd $\ell$, we have $(-2)^\ell\equiv 1\pmod m$ and hence $d|\ell$. Since $n|2^n+1$, $n$ must be odd, but then $d|n$. Yet by Euler-Fermat, $d|\varphi(m)$, which is relatively prime to $q^\alpha$ because we chose $q$ to be larger than any prime divisor of $m$, so it is larger than the prime divisors of $\varphi(m)$ as well. But $gcd(q^\alpha,d)=1$ means that $d|n$ implies $d|m$, hence $(-2)^m\equiv 1\pmod m$, i.e. $m|2^m+1$ ($m$ is odd). Set $f(n)=2^n+1$. Note that $n|f(n)$ implies $f(n)|f(f(n))$, because if $f(n)=n\cdot N$, then $f(f(n))=2^{f(n)}+1=(2^n)^N+1$ is divisible by $2^n+1$ as $N$ is odd. In fact, if $p|f(n)$, so by Lifting the Exponent, $$v_p(f(f(n)))=v_p((2^n)^N+1)=v_p(2^n+1)+v_p(N)\le 2\cdot v_p(f(n)).$$ Therefore $$\prod_{p|f(n)}p^{v_p(f(f(n)))}|f(n)^2.$$

), so if $f(n)\ge 4$, then $f(f(n))$ has a prime divisor that doesn't divide $f(n)$. But meanwhile, it carries on all prime divisors of $f(n)$ because $f(n)$ divides it. Therefore in the sequence $f(3),f(f(3)),\dots$, each term divides the next term (by induction, from $3|f(3)=9$), and each term is at least $4$ (trivial induction), hence by our previous observation, the number of prime divisors of the terms strictly increases. All we need is to take a term, say the $K$'th term, which has at least $K$ prime divisors, and chop off the largest prime divisor from its factorization until exactly $K$ prime divisors are left.

EDIT: I got it.

Does this work? Define the sequence $n_1=9$ and $n_{k+1}=n_k p_k$ where $p_k$ is the primitive prime factor of $2^{n_k}+1$.For example $n_2=171$

Please check my solution

Wait... isn't this trivial by LTE? Since $k=v_3(3^k) \le k+1=v_3(2+1)+v_3(3^{k+1})$ and so $3^k | 2^{3^k}+1$ for any $k$. So set $n=3^{2000}$ which has $2000$ prime factors? (Since it is not mentioned if the prime factors have to be distinct)

ok ngl idt i understood what i was doing back then skull heres the formally written; i reported my old post to not double post

We consider an $n$ such that $n |2^n+1$. We claim that for all such $n \neq 3$ there exists a prime $p$ such that $pn | 2^{pn} + 1$. We can see that this obviously means that the answer is $\boxed{\text{yes}}$. We start this inductive building with $n = 9$. We can see that from Zsigmondy's there must be a prime $p$ such that $p \not | 2^n + 1$ and $p |2^{pn} + 1$. $\blacksquare$

Yes there does. We may construct one as follows: Define a sequence $a_i$ such that $a_1=9$ and $a_{i+1} \mid 2^{a_1a_2\dots a_i}+1$ and $a_{i+1} \nmid 2^j+1$ for $1 \le j < a_1a_2\dots a_i$ and $a_i$ prime for $i \ge 2$ (existence guaranteed by Zsigmondy). Take $n=a_1a_2\dots a_{2000}$ to finish, which works since $a_i$ are distinct, none of them are $3,$ and $9 \mid 2^9+1.$

Pain and suffering. We will induct. For the base case take $n = 9$ which indeed satisfies $n \mid 2^n + 1$ and has exactly one prime factor. Now assume that we have found $n$ such that $n \mid 2^n + 1$. Then we wish to find an odd prime $p$ not dividing $n$ such that $$np \mid 2^{np} + 1$$To do this we take $p$ as a primitive prime divisor of $2^{2n} - 1$, which is guaranteed to exist by Zsigmondy. Clearly this $p$ is relatively prime to $n$ as $n \mid 2^{\phi(n)} - 1$ yet $p \nmid 2^{\phi(n)} - 1$ from Zsigmondy. Also $p \mid 2^n + 1$ as we must have $p \nmid 2^n - 1$, once again from Zsigmondys.

We claim that $n=3^\theta\cdot p_1p_2\cdots p_{2000}$ works, where $p_i$ are primitive prime divisors of some term in the sequence $2^{3^1}+1, 2^{3^2}+1, 2^{3^3}+1,\ldots$ which by Zsigmondy they exists; note that $$2^{3^x}+1\mid 2^{3^\theta\cdot p_1p_2\cdots p_{2000}}+1 \hspace{0.5cm} \text{for enough big} \hspace{0.11cm}\theta$$and also $3^\theta\mid 2^{3^\theta\cdot p_1p_2\cdots p_{2000}}+1$ by LTE; so $3^\theta\cdot p_1p_2\cdots p_{2000}\mid 2^{3^\theta\cdot p_1p_2\cdots p_{2000}}+1$ , as desired.

The answer is $\boxed{\text{yes}}$. Let $p_x$ be a prime that divides $2^{3^x}+1$ but not $2^{3^{i}}+1$ for $i<x$. This prime will always exist by Zsigmondy, meaning our desired number is \[n=3^{2000}p_2p_3\cdots p_{2000}.\] Seeing that this works is trivial, so we are done.

We note $n_1=3^2$ satisfies the condition. We aim to find a prime $p$ such that $pn_i$ satisfies the condition and $\gcd(p,n_i)=1$, so it suffices to have \[2^{pn} \equiv -1 \pmod{p} \implies 2^{2n} \equiv 1, 2^n \not\equiv 1 \pmod{p}.\] This primitive root of 2 modulo $p$ exists by Zsigmondy, as desired. Hence we can induct to find $n_{2000}$ with the desired 2000 prime divisors. $\blacksquare$

Overkill? In general, I claim that there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$. I prove this by claiming that if there exists a positive integer $n$ with $k$ distinct integer divisors in the form of \[n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k},\]such that $p_1<p_2\dots<p_k$, then there exists a prime $p_{k+1}>p_k$ such that for all positive integers $e_{k+1}$, it is true that $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the problem condition, where we can make $m$ any positive integer we want. C1: First, before we begin, I claim that if any prime $q\mid n$, then $nq$ also satisfies problem conditions. This will then prove the part where we can make $m$ whatever we want. This is because \[n\mid 2^n+1 \iff \nu_q(n)\leq \nu_q(2^n+1),\]and by LTE (which we can use, since $n\mid 2^n+1$ implies that $q\mid 2^n+1$), this gives us that \[\nu_q(nq)=1+\nu_q(n)\leq 1+\nu_q(2^n+1)=\nu_q(2^{nq}+1),\]which means that $nq$ also satisfies the problem conditions, as desired. Therefore, by induction, we also get that $nq^m$ also satisfies problem conditions. C2: Now, I claim that there exists a prime $p_{k+1}$ such that for all positive integers $e_{k+1}$, $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the conditions. For simplicity, from here on out, I will refer to $p_{k+1}$ as $r$ and $p_k$ as $q$. Note that this implies that \[r\mid 2^n+1 \iff ord_r(2)\mid 2n,\]and since $ord_r(2)\mid \phi(r)=r-1$, we get that $ord_r(2)\mid \gcd(2n,r-1)$. First, I claim that there always exists an $r$ that divides $2^{q^c}+1$ for some $c$ we can choose such that $r>q$. This is clearly true by Zsigmondy's theorem. I now claim that if we take this $r$, it is true that $rn$ also satisfies problem conditions. This, combined with (C1), will prove our master claim, which states that $nr^e_{k+1}$ satisfies problem conditions for any $e_{k+1}$. To prove this, first make sure that our previous $n$ is divisible by the $q^c$ we chose. We can do this by making another $n$ that satisfies problem conditions by multiplying it by a power of $q$, which we can do by (C1). Next, by LTE, we have, \[\nu_r(rn)=1\leq \nu_r(2^{rn}+1)=\nu_r(2^n+1)+1,\]which we can do since we know that $q^c \mid n$ and $r\mid 2^{q^c}+1$, which gives that $r\mid 2^n+1$. This means that $rn$ satisfies problem conditions, since $rn\mid 2^{rn}+1$, which combined with (C1), proves our inductive step claim. Finally, to complete our induction, note that $3\mid 2^3+1$. Therefore, there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$, finishing the problem.

We claim that the answer is yes. We will induct to show that if there exists such an $n$ with $x$ distinct prime factors, then there also exists such an $n$ with $x+1$ distinct prime factors. By repeating this logic we can find an $n$ with exactly $2000$ distinct prime factors. Our base case is $n=9$, which works since $9\mid 513$. By Zsigmondy, for any $n>3$, there exists a prime $p$ dividing $2^n+1$ that does not divide any $2^k+1$ for smaller $k$. Since this implies that $2$ has order $2n$ modulo $p$, the prime is at least $2n+1$ and thus cannot divide $n$. Thus we know that $pn\mid 2^n+1$ and consequently $pn \mid 2^{pn}+1$. $pn$ has $1$ more distinct prime factor than $n$, so the induction is complete.

Do you realise that this post was made 20 years ago.

Overkill proof while forgetting about Zsigmondy: Define two sequences as follows: $n_0 = 1$; $p_i$ is (a) the smallest prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$, or (b) if that doesn't exist, the smallest prime factor of $2^{n_i} + 1$; and $n_{i+1} = n_ip_i.$ Notice that each $n_i$ is divisible by all of the previous ones, and that all $n_i$ and $p_i$ are odd. First, we show that all $n_i$ satisfy $n_i \mid 2^{n_i} + 1$. We proceed by induction. We can see that $n_0 = 1$ works, so assume $n_i$ works (and all the ones before it). We want to prove $n_{i+1} \mid 2^{n_{i+1}} + 1$. Note that if a prime $p$ divides $n_{i+1}$, then $p = p_j$ for some $j$ satisfying $0 \leq j \leq i$. This also means that $p_j$ is a factor of $2^{n_j} + 1$. Then, by LTE, we have \[ \nu_{p_j}(2^{n_{i+1}} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}\left(\frac{n_{i+1}}{n_j}\right) = \nu_{p_j}(n_{i + 1}) + \nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j). \]However, the strong inductive hypothesis implies $\nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j) \geq 0$, so we have $\nu_{p_j}(2^{n_{i+1}} + 1) \geq \nu_{p_j}(n_{i+1})$. As this is true for all $p$, the inductive step is complete. Next, we claim that eventually, the number of distinct prime factors of $n_i$ is always one more than the number of distinct prime factors of $n_{i-1}$. This is equivalent to showing that eventually, there always exists a prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$. Suppose, for some $i$, that every prime factor of $2^{n_i} + 1$ is also a factor of $n_i$. Then, let $p$ be a prime factor of $n_i$, and pick the minimal $j$ such that $p_j = p$. This minimality implies that $\nu_{p_j}(n_j) = 0$. We have \[ \nu_{p_j}(2^{n_i} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i) - \nu_{p_j}(n_j) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i). \]We can raise $p_j$ to the power of both sides to get \[ {p_j}^{\nu_{p_j}(2^{n_i} + 1)} = {p_j}^{\nu_{p_j}(2^{n_j} + 1)} {p_j}^{\nu_{p_j}(n_i)}. \]Doing this for every prime factor $p$ of $n_i$ (notice that $j$ is now a one-to-one function of $p$) and multiplying the resulting equations, we get \[ 2^{n_i} + 1 = n_i \prod_p p^{\nu_p(2^{n_{j(p)}} + 1)}. \]For all $p$, we have $p^{\nu_p(2^{n_{j(p)}} + 1)} \leq 2^{n_{j(p)}} + 1$. Thus, \[ 2^{n_i} + 1 \leq n_i \prod_p (2^{n_{j(p)}} + 1). \]Since $j$ is one-to-one, every $j(p)$ is unique and in the set $\{0, 1, \ldots, i-1\}$. Therefore, we have the inequality \[ 2^{n_i} + 1 \leq n_i \prod_{j=0}^{i-1} (2^{n_j} + 1). \]Taking the log base $2$ of both sides, we have \[ n_i < \log _2 (2^{n_i} + 1) \leq \log _2 n_i + \sum _{j=0}^{i-1} \log_2(2^{n_j} + 1) < \log_2 n_i + \sum _{j=0}^{i-1} (n_j + 1) = \log_2 n_i + i + \sum_{j=0}^{i-1}n_j. \]Now, in order to achieve a bound on $n_i$, we notice that $p_i \geq 3$ for all $i$, so therefore, $n_i \geq 3^i$. It is then easy to see that $\sum_{j=0}^{i-1}n_j \leq \frac12 n_i$ for all $i \geq 1$. Then, \[ n_i < \log_2 n_i + i + \frac12 n_i \leq \log_2 n_i + \log_3 n_i + \frac12 n_i \iff n_i < 2(\log _2 n_i + \log _3 n_i). \]This inequality obviously cannot be satisfied as $n_i$ grows large. Thus, eventually, we must have that there exists a prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$. Hence, there must eventually exist an $n$ in our sequence with exactly 2000 distinct prime factors.

是否存在一个恰有2000个素因子的正整数$n$，满足$n \mid 2^n + 1$?

是否存在一个恰有2000个不同素因子的正整数$n$，满足$n \mid 2^n + 1$

The answer is yes. We will construct a positive integer $n = p_1^2 p_2p_3 \cdots p_{2000}$ where $p_1 < p_2 < \cdots < p_{2000}$ are distinct primes such that $n \mid 2^n + 1$. First, we set $p_1 = 3$. For $i \ge 2$, let $q_i = 3^2 \cdot p_2 \cdots p_3 \cdots p_{i - 1}$. Then, we shall construct $p_i$ recursively by picking a primitive divisor of $2^{2q_i} - 1$, which exists by Zsigmondy's theorem. Now, we show that $n \mid 2^n + 1$. It suffices to show that for $1 \le i \le 2000$, we have $\nu_{p_i}(2^n + 1) \ge \nu_{p_i}(n)$. For $i = 1$, we note that by LTE, $$\nu_3(2^n + 1) = \nu_3(2 + 1) + \nu_3(n) = 1 + \nu_3(n) > 1.$$For $i > 1$, we note that by construction, $\text{ord}_{p_i}(2) = 2q_i$. Thus, $2^{q_i} \equiv -1 \pmod {p_i}$, so by LTE, we have $$\nu_{p_i}(2^n + 1) = \nu_{p_i}((2^{q_i})^{n/q_i} + 1) = \nu_{p_i}(2^{q_i} + 1) + \nu_{p_i}(n/q_i) \ge 1 + \nu_{p_i}(n) > \nu_{p_i}(n),$$so we are done.

Lemma. If $a\mid 2^a+1$ and $b\mid 2^b+1$, where $a=3^\alpha a'$, $b=3^\beta b'$, $3 \nmid a', b'$ and $a'$ and $b'$ are coprime, then $ab\mid 2^{ab}+1$. Proof. Obviously, $a$ and $b$ are odd. $$2^{ab}\equiv (-1)^b\equiv -1 \pmod{a}$$$$2^{ab}\equiv (-1)^a\equiv -1 \pmod{b}$$By LTE, $v_3(2^{ab}+1)=v_3(ab)+1=\alpha + \beta + 1$. Finally, as $a', b', 3^{\alpha+\beta}\mid 2^{ab}+1$ and all three numbers are pairwise coprime, $ab=a'b'3^{\alpha+\beta}\mid 2^{ab}+1$. $\square$ We construct a sequence of primes $p_2, p_3, \dots, p_{2000}$ in the following manner. Let $p_2=19$ (so that $p_2\mid 2^{3^2}+1$). Inductively assume we have constructed $p_2, p_3, \dots, p_{i}$, for some $i\ge 2$, such that: $\bullet$ all $p_j$ with $2\le j\le i$ are distinct, $\bullet$ $p_j>3$ and $p_j\mid2^{3^{i}}+1$ for each $2\le j\le i$. As $2^{3^{i+1}}+1=(2^{3^i}+1)(4^{3^i}-2^{3^i}+1)$ and the greatest common divisor of the terms in the parenthesis is $3$, we conclude that $4^{3^i}-2^{3^i}+1$ has all its prime divisors distinct from $p_j$ for $2\le j\le i$. As $9\mid 2^{3^i}+1$ and $4^{3^i}-2^{3^i}+1>3$, it has a prime factor $p_{i+1}>3$, so the inductive claim holds for $i+1$ as well. Having constructed $p_2, p_3, \dots, p_{2000}$, let $n_i=3^i p_i$ for each $2\le i\le 2000$. By LTE, $v_3(2^{n_i}+1)=i+1$. By Fermat's Little Theorem, $2^{n_i}\equiv 2^{3^i p_i}\equiv 2^{3^i}\equiv -1\pmod{p_i}$. Therefore, $n_i = 3^i p_i\mid 2^{n_i} + 1$ for each $i$. Let $n=n_2n_3\dots n_{2000}$. By repeated application of the lemma, $n\mid 2^n + 1$, and by construction, $n$ has $2000$ distinct prime divisors.