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Are there finitely many $(n,k) \in \mathbb{N}^2$ with $2^n-1=p_1p_2\cdots p_k$ ?

$p_1=3,p_2=5 , ...,p_k$ are consecutive odd primes in ascending order.
An example is when $n=4, k=2$:
$2^4-1=3\cdot 5=p_1p_2$
Are there finitely many $n$?
I tried to use Zsigmondy's theorem without success.
Thanks in advance!


marked as duplicate by user9072, მამუკა ჯიბლაძე, Stefan Kohl, Dan Petersen, Alexey Ustinov Oct 10 '15 at 14:31

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$\let\dvds\mid$Yes. By Zsigmondy's theorem, $2^{12}-1$ has some prime divisor $p_s$ not dividing $2^i-1$ for $i<12$ (in fact, $p_s=13$). Now, if $2^n-1=p_1p_2\dots p_k$ with $k\geq s$, then $p_s\dvds 2^n-1$, so $12\dvds n$ and hence $3^2\dvds 2^n-1$, which is impossible. Thus only the cases with $k<s$ are left.

  • $\begingroup$ I see, thanks. One minor remark - one hardly needs any theorems nowadays to check that $13$ does not divide $2^i-1$ for $i<12$ :) $\endgroup$ – მამუკა ჯიბლაძე Oct 10 '15 at 11:12
  • 2
    $\begingroup$ Surely. I just wanted to emphasize that we do not need to determine this prime explicitly. $\endgroup$ – Ilya Bogdanov Oct 10 '15 at 11:13

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