Infinitely many pseudoprimes
By Chris Caldwell
Let a > 1 be an integer. Recall that Fermat's (little) theorem shows that if p does not divide a, then ap-1 ≡ 1 (mod p). m is a probable-prime base a (an a-PRP) if am-1 ≡ 1 (mod m). "Most" such numbers m are prime, but if they are not, they are called pseudo-primes base a. The purpose of this note is to prove the following:
- Let a be any integer greater than one. There is an infinite number of pseudoprimes with respect to the base a.
- Proof: We will follow Ribenboim [HW79 p72].
Choose any prime p which does not divide a(a2-1) (this condition can be weakened to p which does not divide a2-1, see [CP2001 thm 3.3.4]) and then define
m = (a2p-1)/(a2-1).
m is divisible by (ap-1)/(a-1), so m is composite. Also this m is different for each choice of p, so we need only show m is an a-PRP to complete the proof.
Rewrite the expression for m above:
(a2-1)(m-1) = a(ap-1-1)(ap+a).
Clearly the rightmost term is even and a2-1 divides ap-1-1 (as p-1 is even). Also, Fermat's theorem tells us that ap-1-1 is divisible by p, so
2p(a2-1) divides (a2-1)(m-1)
or just 2p divides m-1. Finally, by the definition of m we have a2p = 1 + m(a2-1) or just a2p ≡ 1 (mod m). It follows that am-1 is also one modulo m, completing the proof. ∎