The Boolean algebra generated by sets of prime divisors of the numbers $2^n-1$

Question

Let $\Pi$ be the set of odd prime numbers and let $\mathcal P(\Pi)$ be the Boolean algebra of subsets of $\Pi$.

For a number $x$ denote by $\Pi(x)$ the set of odd prime divisors of $x$.

Problem. Does each singleton $\{p\}\subset \Pi$ belong to the Boolean algebra generated by the family $\{\Pi(2^n-1):n\ge 2\}$ in $\mathcal P(\Pi)$?

We can also ask a stronger

Question. Is it true that for every odd prime number $p$ there exit numbers $n_1,\dots,n_k$ such that $\{p\}=\bigcap_{i=1}^k\Pi(2^{n_i}-1)$?

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Added in Edit. The number $p=5$ mentioned in the answer of Ofir Gorodetsky gives a counterexamle to Question but not to Problem as $\{5\}=\Pi(2^4-1)\setminus\Pi(2^2-1)=\{3,5\}\setminus\{3\}$.

Answer 1

$2^{11}-1=2047=23\cdot 89$ and neither prime divides a smaller such number. So $23 \mid 2^n-1$ exactly when $n=11m.$ Similarly $89 \mid 2^n-1$ exactly when $n=11m.$ So $\{23,89\}$ is in the Boolean Algebra but neither of its singleton subsets is.

Here is a list for $n \leq 33$ of the set of primes which divide $2^n-1$ but not any $2^m-1$ for $1 \lt m \lt n$. These sets (excluding $\{\}$) are the atoms of the Boolean Algebra:

$[2, \left\{ 3 \right\} ],[3, \left\{ 7 \right\} ],[4, \left\{ 5 \right\} ],[5, \left\{ 31 \right\} ],[6, \left\{ \right\} ],[7, \left\{ 127 \right\} ],[8, \left\{ 17 \right\} ],[9, \left\{ 73 \right\} ],[10, \left\{ 11 \right\} ],$

$[11, \left\{ 23,89 \right\} ],[ 12, \left\{ 13 \right\} ],[13, \left\{ 8191 \right\} ],[14, \left\{ 43 \right\} ],[15, \left\{ 151 \right\} ],[16, \left\{ 257 \right\} ],[ 17, \left\{ 131071 \right\} ],$$[18, \left\{ 19 \right\} ],[19, \left\{ 524287 \right\} ],[20, \left\{ 41 \right\} ],[21, \left\{ 337 \right\} ],[22, \left\{ 683 \right\} ],[23, \left\{ 47,178481 \right\} ],$

$[24, \left\{ 241 \right\} ],[25, \left\{ 601,1801 \right\} ],[26, \left\{ 2731 \right\} ],[27, \left\{ 262657 \right\} ],[28, \left\{ 29,113 \right\} ],$

$[29, \left\{ 233,1103,2089 \right\} ] ,[30, \left\{ 331 \right\} ],[31, \left\{ 2147483647 \right\} ],[32, \left\{ 65537 \right\} ],[33, \left\{ 599479 \right\} ]$

Answer 2

No. The gcd (greatest common divisor) of $\{ 2^{n_i}-1\}_{i=1}^{k}$ can be seen to be $2^{\gcd(n_1,\ldots,n_k)}-1$, by an application of the Euclidean algorithm. If $p$ divides this expression, then $\gcd(n_1,\ldots,n_k)$ is divisible by the order of $2$ in the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^{\times}$. In general, $2^{\mathrm{ord}_p(2)}-1$ may have prime factors other than $p$.

For instance, if $p=5$ and $p$ divides all of $\{ 2^{n_i}-1\}_{i=1}^{k}$ then each $n_i$ is divisible by $\mathrm{ord}_p(2) = 4$, and so each $2^{n_i}-1$ is divisible by $2^4-1=15= 3 \times 5$.

In fact, the answer to your question is positive only for primes $p$ such that $p^k$ is of the form $2^q-1$ for some $k$, where $q$ is a prime. (For $k=1$ these are known as Mersenne primes.)

Answer 3

For example, if $2^n-1$ is divisible by 43, then $n$ is divisible by 14 and therefore $2^n-1$ is divisible by 127. Thus 43 is a counterexample.

Answer 4

This was already answered in comments, so I'm posting cw.

The Boolean subalgebra $A$ defined by the OP being generated by finite subsets, it is atomic. The atoms are precisely the (finite) fibers of the function $f$ mapping the odd prime $p$ to the order $f(p)$ of $2$ modulo $p$; it is given by OEIS:A014664.

And indeed $f$ is not injective, so $A$ does not include all singletons. For instance $f(23)=f(89)=11$ (it was mentioned in a comment by Gehrard Paseman).

Hence the problem (and hence the question) has a negative answer. However the question already has a simpler negative answer, as the singleton $\{5\}$ (which belongs) to the Boolean algebra $A$ is not in the smaller set of subsets defined in the question using only finite intersections, see Ofir's answer.