4 (a)
When , m-Hilbert numbers are all positive integers (since , which covers all positive integers). The 1-Hilbert prime is then almost identical to what we normally define as a “prime number” (greater than 1 and not factor in smaller positive integers). The prime factorization of positive integers in the usual sense satisfies the unique factorization theorem. Therefore, the m=1 case must be uniquely decomposable.
When , the m-Hilbert numbers are all the odd numbers. A 2-Hilbert prime is a number in the set of odd numbers that cannot be multiplied by a smaller set of odd factors. In other words, it is an “irreducible element in the set of odd numbers”.
But we know that all odd numbers in the usual factorization of integers factor into only a few odd primes. Since 2 is not in this set, there is no problem of multiple factorization due to “missing 2”; within the set of odd numbers, the factorization still corresponds only to the same set of odd primes, and remains unique. Thus, the case m=2 is still uniquely factorized.
4 (b)
The 4-Hilbert numbers and 5-Hilbert numbers do not factor uniquely. The 4-Hilbert numbers are “closer” to having unique factorization in the sense that the failure is more structured, involving primes from a single residue class. In contrast, the 5-Hilbert numbers involve more complex interactions between different residue classes, leading to a higher degree of non-unique factorization. Higher values of m can exhibit even worse factorization with more distinct factorizations.