6 The Carlitz escape argument
Let \(p\) be an odd prime and \(m\) an even index with \((p-1)\nmid m\). Then \(m' := m \bmod (p-1)\) satisfies \(0 {\lt} m' {\lt} p-1\), \(m'\) is even, and \(m \equiv m' \pmod{p-1}\). In particular \(2 \le m' \le p-3\).
The residue is nonzero precisely because \((p-1)\nmid m\), and it inherits parity from \(m\) since \(p-1\) is even.
An odd prime \(p\) is irregular if and only if \(p \mid \operatorname {num}(B_m/m)\) for some positive even index \(m\).
If \(p\) is irregular, theorem 1.2 produces a witness \(B_{2k}\) in Kummer’s range; since \(2k \le p-3 {\lt} p\), the index is a unit modulo \(p\) and the divisibility passes to the divided numerator.
Conversely, suppose \(p\mid \operatorname {num}(B_m/m)\) with \(m\) positive and even. By theorem 2.4, necessarily \((p-1)\nmid m\). Let \(m'\) be the positive residue of lemma 6.1. The full Kummer congruence (theorem 5.3) gives \(B_m/m \equiv B_{m'}/m' \pmod p\), so \(p\mid \operatorname {num}(B_{m'}/m')\) by lemma 2.2, hence \(p \mid \operatorname {num}(B_{m'})\) by lemma 2.3. Since \(2 \le m' \le p-3\) and \(m'\) is even, this is a Bernoulli witness in Kummer’s range, and \(p\) is irregular by theorem 1.1.
For every finite set \(S\) of natural numbers there are \(M\) and a prime \(p\) such that \(M\) is even and positive, \((q-1)\mid M\) for every prime \(q \in S\), the prime \(p\) is odd, \(p\mid \operatorname {num}(B_M/M)\), and \(p \notin S\).
Put \(M = C(S)\cdot 2^t\) with \(t\) chosen by corollary 3.3 so that \(|B_M/M| {\gt} 1\). By lemma 2.1 some prime \(p\) divides \(\operatorname {num}(B_M/M)\). By theorem 2.4, \((p-1)\nmid M\); in particular \(p \ne 2\). If \(p\) were in \(S\), then \((p-1)\mid C(S)\mid M\) by lemma 4.2, a contradiction; so \(p \notin S\). The divisibility \((q-1)\mid M\) for \(q \in S\) is again lemma 4.2.
For every finite set \(S\) of natural numbers there is an irregular prime \(p \notin S\).
Take the prime \(p\) and the index \(M\) of theorem 6.3. Then \(p\) is irregular by theorem 6.2 and \(p \notin S\).
There are infinitely many irregular primes.
By theorem 6.4 no finite set contains all irregular primes, so the set is infinite by lemma 1.3.
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