Prime Numbers

140 Chapter 3 RECOGNIZING PRIMES AND COMPOSITES
Thus there are just as many solutions to a m ≡ 1(modn) asthereareto
a m ≡−1(modn), and the lemma is proved. ✷
Proof of Theorem 3.5.4. From Lemma 3.5.8 and (3.6), it will suffice to show
that S(n)/ϕ(n) ≤ 1/4 whenever n is an odd composite that is greater than 9.
From Lemma 3.5.9, we have
ϕ(n) 1
=
S(n) 2
p a−1 p − 1
2ν(n)−1 gcd(t, p − 1) ,
p a n
where the notation p a n means that p a is the exact power of the prime p
in the prime factorization of n. Each factor (p − 1)/(2 ν(n)−1 gcd(t, p − 1)) is
an even integer, so that ϕ(n)/S(n) is an integer. In addition, if ω(n) ≥ 3, it
follows that ϕ(n)/S(n) ≥ 4. If ω(n) =2andn is not squarefree, the product
of the various p a−1 is at least 3, so that ϕ(n)/S(n) ≥ 6.
Now suppose n = pq, wherep < q are primes. If 2 ν(n)+1 |q − 1, then
2 ν(n)−1 gcd(t, q − 1) ≤ (q − 1)/4 andϕ(n)/S(n) ≥ 4. We may suppose then
that 2 ν(n) q − 1. Note that n − 1 ≡ p − 1(modq − 1), so that q − 1doesnot
divide n − 1. This implies there is an odd prime dividing q − 1toahigher
power than it divides n − 1; that is, 2 ν(n)−1 gcd(t, q − 1) ≤ (q − 1)/6. We
conclude in this case that ϕ(n)/S(n) ≥ 6.
Finally, suppose that n = p a ,wherea ≥ 2. Then ϕ(n)/S(n) =p a−1 ,so
that ϕ(n)/S(n) ≥ 5, except when p a =9. ✷
3.5.1 The least witness for n
We have seen in Theorem 3.5.4 that an odd composite number n has at least
3n/4 witnesses in the interval [1,n − 1]. Let W (n) denote the least of the
witnesses for n. ThenW (n) ≥ 2. In fact, for almost all odd composites, we
have W (n) = 2. This is an immediate consequence of Theorem 3.4.2. The
following theorem shows that W (n) ≥ 3 for infinitely many odd composite
numbers n.
Theorem 3.5.10. If p is a prime larger than 5, then n =(4 p +1)/5 is a
strong pseudoprime base 2, so that W (n) ≥ 3.
Proof. We first show that n is a composite integer. Since 4 p ≡ (−1) p ≡−1
(mod 5), we see that n is an integer. That n is composite follows from the
identity
4 p +1=(2 p − 2 (p+1)/2 + 1)(2 p +2 (p+1)/2 +1).
Note that 2 2p ≡−1(modn), so that if m is odd, we have 2 2pm ≡−1(modn).
But n − 1=2 2 t,wheret is odd and a multiple of p, the latter following from
Fermat’s little theorem (Theorem 3.4.1). Thus, 2 2t ≡−1(modn), so that n
is a strong pseudoprime base 2. ✷
It is natural to ask whether W (n) can be arbitrarily large. In fact, this
question is crucial. If there is a number B that is not too large such that every