Prime Numbers

3.3 Recognizing smooth numbers 129
our list and 9699690 mod x = r. Thenr = ab, wherea = gcd(9699690,x)and
gcd(b, x) = 1. If the highest exponent on any prime in the prime factorization
of x is bounded above by 2e ,thengcd(r2emod x, x) is the 20-smooth part
of x. So in our case, we can take e =4,since224> 1008. Let us see what
happens for the number x = 1008. First, we have r = 714. Next we take
7142i mod 1008 for i =1, 2, 3, 4, getting 756, 0, 0, 0. Of course, we ought to
be smart enough to stop when we get the first 0, since this already implies
that 1008 is 20-smooth. If we apply this idea to x = 1004, we get r = 46, and
the requisite powers are 108, 620, 872, 356. We take gcd(356, 1004) and find
it to be 4. Surely this must be the long way around! But as we shall see, the
method scales beautifully. Further, we shall see that it is not interesting to
focus on any one number, but on all numbers together.
We form the product 9699690 of the primes up to 20 via a “product tree;”
see [Bernstein 2004e]. This is just the binary tree as so:
✟
✟❍9699690
✟
❍❍❍❍❍46189
✟
✟
✟
210
❅
❅
❅
❅
6
❅
2 ❅3
❅35 ❅
5 ❅7
143
❅
11 ❅13
❅323 ❅
17 ❅19
Product tree for P = {2, 3, 5, 7, 11, 13, 17, 19}
We start at the leaves, multiplying ourselves through the binary tree to the
root, whose label is the product P = 9699690 of all of the leaves.
We wish to find each residue P mod x as x varies over the numbers we are
examining for smoothness. If we do this separately for each x, sinceP is so
large, the process will take too long. Instead, we first multiply all the numbers
x together! We do this as with the primes, with a product tree. However, we
never need to form a product that is larger than P ; say we simply indicate such
large products with an asterisk. Let us consider the product tree T formed
from the numbers 1001, 1002,...,1008:
✟
✟❍∗
❍❍❍❍❍
✟
✟
✟
✟
∗
∗
❅
❅
❅
❅
1003002
❅
1001 ❅1002
❅1007012 ❅
1003❅1004 1011030 ❅1015056
❅
❅
1005 ❅1006 1007❅1008 Product tree T for X = {1001, 1002,...,1008}