4 The divisor-closed multiplier
For a finite set \(S\) of natural numbers put \(M(S) = \max (3, \max S)\) and
\[ C(S) \; =\; 2 \cdot M(S)! \, . \]
\(C(S)\) is even and positive, and for every prime \(q \in S\) one has \((q-1)\mid C(S)\).
Proof
\(q \le M(S)\), so \(q - 1\) is a positive integer at most \(M(S)\) and therefore divides \(M(S)!\).
Let \(m = C(S)\cdot 2^t\). Every prime \(q\) dividing \(m\) satisfies \((q-1)\mid m\).
Proof
If \(q = 2\) this is trivial. Otherwise \(q\) divides \(M(S)!\), hence \(q \le M(S)\), and as before \((q-1) \mid M(S)! \mid m\).