Prime Numbers

134 Chapter 3 RECOGNIZING PRIMES AND COMPOSITES
Korselt felt sure that no examples could possibly exist, and developed the
criterion as a first step toward proving this.
Proof. First, suppose n is a Carmichael number. Then n is composite. Let p
be a prime factor of n. Fromp n ≡ p (mod n), we see that p 2 does not divide
n. Thus,n is squarefree. Let a be a primitive root modulo p. Sincea n ≡ a
(mod n), we have a n ≡ a (mod p), from which we see that a n−1 ≡ 1(modp).
But a (mod p) has order p − 1, so that p − 1 divides n − 1.
Now, conversely, assume that n is composite, squarefree, and for each
prime p dividing n, wehavep − 1 dividing n − 1. We are to show that a n ≡ a
(mod n) for every integer a. Sincen is squarefree, it suffices to show that
a n ≡ a (mod p) for every integer a and for each prime p dividing n. So suppose
that p|n and a is an integer. If a is not divisible by p,wehavea p−1 ≡ 1(modp)
(by (3.3)), and since p − 1 divides n − 1, we have a n−1 ≡ 1(modp). Thus,
a n ≡ a (mod p). But this congruence clearly holds when a is divisible by p,
so it holds for all a. This completes the proof of the theorem. ✷
Are there infinitely many Carmichael numbers? Again, unfortunately for
primality testing, the answer is yes. This was shown in [Alford et al. 1994a].
P. Erdős had given a heuristic argument in 1956 that not only are there
infinitely many Carmichael numbers, but they are not as rare as one might
expect. That is, if C(x) denotes the number of Carmichael numbers up to the
bound x, thenErdős conjectured that for each ε>0, there is a number x0(ε)
such that C(x) >x 1−ε for all x ≥ x0(ε). The proof of Alford, Granville, and
Pomerance starts from the Erdős heuristic and adds some new ingredients.
Theorem 3.4.7. (Alford, Granville, Pomerance). There are infinitely many
Carmichael numbers. In particular, for x sufficiently large, the number C(x)
of Carmichael numbers not exceeding x satisfies C(x) >x 2/7 .
The proof is beyond the scope of this book; it may be found in [Alford et al.
1994a].
The “sufficiently large” in Theorem 3.4.7 has not been calculated, but
probably it is the 96th Carmichael number, 8719309. From calculations in
[Pinch 1993] it seems likely that C(x) >x 1/3 for all x ≥ 10 15 .Alreadyat
10 15 , there are 105212 Carmichael numbers. Though Erdős has conjectured
that C(x) > x 1−ε for x ≥ x0(ε), we know no numerical value of x with
C(x) >x 1/2 .
Is there a “Carmichael number theorem,” which like the prime number
theorem would give an asymptotic formula for C(x)? So far there is not even
a conjecture for what this formula may be. However, there is a somewhat
weaker conjecture.
Conjecture 3.4.1 (Erdős, Pomerance). The number C(x) of Carmichael
numbers not exceeding x satisfies
as x →∞.
1−(1+o(1)) ln ln ln x/ ln ln x
C(x) =x