# 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
7 months, 1 week ago

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## Comments

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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.

- 7 months ago

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