from itertools import chain, cycle

def factors_of(n):
    f = 2
    for step in chain([0,1,2,2], cycle([4,2,4,2,4,6,2,6])):
        f += step

        if f*f > n:
            if n != 1:
                yield n
            break

        if n%f == 0:
            yield f
            while n%f == 0:
                n /= f

def is_prime(n):
    '''proves primality of n using Lucas test.'''
    x = n-1

    if n%2 == 0:
        return n == 2

    factors = list( factors_of(x))

    b = 2
    while b < n:
        if pow( b, x, n ) == 1:

            for q in factors:
                if pow( b, x/q, n ) == 1:
                    break
            else:
                return True

        b += 1

    return False

My confusion was over the word “any” in

“… if b^((n-1)/q) != 1 (mod n) for *any* prime divisor q of n-1, n is prime.”

At first, I read the phrase to mean if I found any one q for which the non-congruence was true, then n is prime. However, the correct reading is that the non-congruence holds no matter which q you try. Therefore, you have to try every q to prove n is prime. Perhaps using “every” instead of “any” would be clearer.

By the way, your version of is_prime incorrectly reports that 2 is composite. You should change the test at lines 22 and 23 to read:

if n%2 == 0:
    return n == 2

Don’t feel bad; Jon Bentley got that wrong in one of his books, so you’re in good company.

Studying the proof at the mathpages.com link, given above, I think the algorithm should be to find a b so that b^(x-1) == 1 (mod n) then, if b^((n-1)/q) != 1 (mod n) for *all* prime divisors q of n-1, n is prime. If b^((n-1)/q) == 1 (mod n) for some q, then try a new value for b. This matches the pseudocode at wikipedia.

My python code is below:

from itertools import chain, cycle

def factors_of(n):
    f = 2
    for step in chain([0,1,2,2], cycle([4,2,4,2,4,6,2,6])):
        f += step

        if f*f > n:
            if n != 1:
                yield n
            break

        if n%f == 0:
            yield f
            while n%f == 0:
                n /= f

def is_prime(n):
    '''proves primality of n using Lucas test.'''
    x = n-1

    if n == 2 or n%2 == 0:
        return False

    factors = list( factors_of(x))

    b = 2
    while b < n:
        if pow( b, x, n ) == 1:

            for q in factors:
                if pow( b, x/q, n ) == 1:
                    break
            else:
                return True

        b += 1

    return False

import Data.Bits
import Data.List

tdFactors :: Integer -> [Integer]
tdFactors n = f n [2..floor . sqrt $ fromIntegral n] where
    f _ []     = []
    f r (x:xs) | mod r x == 0 = x : f (div r x) (x:xs)
               | otherwise    = f r xs

expm :: Integer -> Integer -> Integer -> Integer
expm b e m = foldl' (\r (b', _) -> mod (r * b') m) 1 .
             filter (flip testBit 0 . snd) .
             zip (iterate (flip mod m . (^ 2)) b) .
             takeWhile (> 0) $ iterate (`shiftR` 1) e

isPrime :: Integer -> Bool
isPrime n = f 2 $ tdFactors (n - 1) where
    f _ []     = False
    f b (q:qs) | expm b (n - 1)         n /= 1 = f (b + 1) (q:qs)
               | expm b (n - 1 `div` q) n /= 1 = True
               | otherwise                     = f b qs