Pollard's $(p-1)$-Method

Let $N$ be an integer that we wish to factor. Choose a positive integer $B$ (usually $\leq 10^6$ in practice). The Pollard $(p-1)$-method hunts for prime divisors $p$ of $N$ such that $p-1$ is $B$-power-smooth. Here is the strategy. Suppose that $p\mid N$ and $a>1$ is an integer that is prime to $p$. By Fermat's Little Theorem,

$$a^{p-1}\equiv 1\pmod{p}.$$

Assume that further that $p-1$ is $B$-power-smooth and let $m=\lcm(1,2,3,\ldots B)$. Then $B\mid m$, so $p-1\mid m$, and so

$$a^m \equiv 1 \pmod{p}.$$

Thus

$$p\mid \gcd(a^m-1,N) > 1.$$

Usually $\gcd(a^m-1,N)<N$ also, and when this is the case we have split $N$. In the unlikely case when $\gcd(a^m-1,N)=N$, then $a^m\equiv 1\pmod{q^r}$ for every prime power divisor of $N$. In this case, repeat the above steps but with a smaller choice of $B$ (so that $m$ is smaller). Also, it's a good idea to check from the start whether or not $N$ is not a perfect power $M^r$, and if so replace $N$ by $M$.

In practice, we don't know $p$. We choose a $B$, then an $a$, cross our fingers, and proceed. If we split $N$, great! If not, increase $B$ or change $a$ and try again.

For fixed $B$, this algorithm works when $N$ is divisible by a prime $p$ such that $p-1$ is $B$-power-smooth. How many primes $p$ have the property that $p-1$ is $B$-power-smooth? Is this very common or not? Using the above two functions, we find that roughly 15% of primes $p$ between $10^{15}$ and $10^{15}+10000$ are such that $p-1$ is $10^6$ power-smooth.

\\ Count the number of B-power-smooth numbers an interval.
{cnt(B)= s=0;t=0; 
   for(p=10^15, 10^15+10000,  
      if(isprime(p), 
         t++;if(ispowersmooth(p-1,B),s++)
      )
   ); 
   s/t*1.0
}
? cnt(10^6)
%5 = 0.1482889733840304182509505703

Thus the Pollard $(p-1)$-method with $B=10^6$ is blind to 85% of the primes around $10^{15}$. There are nontrivial theorems about densities of power-smooth numbers, but I will not discuss them today.