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}^+S⊂Z+ such that all but finitely many sums of elements from $S$ (possibly with repetitions) are composite numbers, we need to identify sets $S$ whose sums, when large enough, are not prime.

Let’s denote the sums of elements from $S$ as the set $T$. We seek $S$ such that only finitely many elements in $T$ are prime.

Observation and Strategy:

1. Single Element Case:
   • If $S$ consists of only one element, say $S=\{n\}$, then sums are of the form $n,2n,3n,\dots$. For these sums to be composite for large values, $n$ must be such that $kn$ is composite for all large $k$. This is satisfied if $n\ge 2$, as $kn$ for $k>1$ will always be composite.
2. Multiple Elements:
   • If $S$ contains more than one element, the sums are more complex. To ensure all but finitely many sums are composite, the elements in $S$ should form sums that are not often prime. This can be achieved if all the elements in $S$ are such that their linear combinations (sums) are highly composite.
   • For example, if $S$ contains numbers that have a common factor greater than 1 (e.g., $S=\{4,6,8\}$), 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 $n\ge 2$

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

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

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

Conclusion:

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

1. Any set $S$ containing a single element $n\ge 2$.
2. Any set $S$ where all elements share a common factor greater than 1.

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