Help: Pollard's p-1, correct way to build M

I need some help regarding one of the steps in Pollard's p-1.

What I understood: We are assuming that there exists \(p\), a prime factor of \(n\), such that \(p-1\) is \(B\)-powersmooth for some \(B\) (not \(B\)-smooth!) and that we're trying to build a number \(M\) (to be used as an exponent) such that \((p-1)|M\). Most commonly this is done in two ways:

  1. \(M = B!\)
  2. \(M = \prod {q^a} \), for all primes \(q <= B\) and \(a\) sufficiently large.

Let's focus on the second. My confusion comes from the value of a. Using the main assumption we made (\(p-1\) is \(B\)-powersmooth) it seems clear that \(a = \lfloor log_q{B} \rfloor\). However, some sources (including Wikipedia) use \(a = \lfloor log_q{n} \rfloor\). Since we can assume \(n >> B\) this works too but makes \(a\) unnecessarily large.

At first I thought it's just an error on Wikipedia and I made a correction, but I found several revisions in page history that change this both ways (\(B\) to \(n\) and back), with no real discussion, which makes me unsure. Can someone explain why \(n\) makes more sense than \(B\) here, or confirm that \(n\) was just an error?

Note by Nikola Jovanović
4 months ago

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The confusion seems to be whether we're assuming \(p-1\) is \(B\)-powersmooth or \(B\)-smooth, right? If it's \(B\)-powersmooth, then we can use the smaller bound, but if it's \(B\)-smooth, then we have to use the larger one.

The Wikipedia page cites the smaller bound, but the example (\(n=299,\) \(B=5\)) appears to use the larger bound.

Patrick Corn - 3 months, 3 weeks ago

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