Find those subsets S ⊂ Z+ such that all but finitely many sums of elements from S (possibly with repetitions) are composite numbers.

To find the subsets S⊂Z+S \subset \mathbb{Z}^+ such that all but finitely many sums of elements from SS (possibly with repetitions) are composite numbers, we need to identify sets SS whose sums, when large enough, are not prime.

Let’s denote the sums of elements from SS as the set TT. We seek SS such that only finitely many elements in TT are prime.

Observation and Strategy:

  1. Single Element Case:
    • If SS consists of only one element, say S={n}S = \{ n \}, then sums are of the form n,2n,3n,…n, 2n, 3n, \ldots. For these sums to be composite for large values, nn must be such that knkn is composite for all large kk. This is satisfied if n≥2n \geq 2, as knkn for k>1k > 1 will always be composite.
  2. Multiple Elements:
    • If SS contains more than one element, the sums are more complex. To ensure all but finitely many sums are composite, the elements in SS should form sums that are not often prime. This can be achieved if all the elements in SS are such that their linear combinations (sums) are highly composite.
    • For example, if SS contains numbers that have a common factor greater than 1 (e.g., S={4,6,8}S = \{ 4, 6, 8 \}), then any sum of these numbers is also divisible by this common factor, thus making them composite except for possibly the first few terms.

Let’s formalize these observations:

Case 1: Single Element n≥2n \geq 2

Any S={n}S = \{ n \} where n≥2n \geq 2 results in sums knkn for k∈Z+k \in \mathbb{Z}^+. For k≥2k \geq 2, knkn is composite. Thus, only the first sum (which is nn itself) may be prime (if nn is prime), but all larger sums are composite.

Case 2: Multiple Elements with a Common Factor Greater than 1

If SS has a common factor d>1d > 1, say S={d⋅a1,d⋅a2,…,d⋅ak}S = \{ d \cdot a_1, d \cdot a_2, \ldots, d \cdot a_k \} where aia_i are positive integers, then any sum of elements from SS will be of the form d⋅(sum of ai)d \cdot (\text{sum of } a_i). Since d≥2d \geq 2, sums will be composite for sufficiently large values.

Conclusion:

The sets S⊂Z+S \subset \mathbb{Z}^+ such that all but finitely many sums of elements from SS are composite numbers include:

  1. Any set SS containing a single element n≥2n \geq 2.
  2. Any set SS where all elements share a common factor greater than 1.

Thus, subsets SS with these properties will ensure that all but finitely many sums are composite.