The question is essentially in the title.
I'm working with polynomial-time reductions, but I figure this portion of the question is most relevant to Mathematics, instead of Computer Science.
Essentially, in the partition problem, the goal is to partition a multiset of integers M into two subsets, $S_1 $and $S_2$, such that $\sum_{e \in S_1} e = \sum_{e \in S_2} e$. Of course, not all multisets can do this $\{2, 5\}$ for example, and some have more than one unique solution, eg $\{1, 1, 1, 2, 2, 3\}$.
Is it mathematically correct to say that partitioning $M + M$ into subsets $S_i$ for $i \in \{1, 2, 3, 4\}$, that is to say duplicating the elements in $M$ and doubling the subsets to partition, will result in four subsets of equal sum, iff the original set could be 2-partitioned?
I suppose a more rigorous way to express this would be:
If $f(s, k) = $ set $s$ can be partitioned to $k$ subsets, such that all the subsets have integers with equal sum:
$f(S,2) \iff f(S + S, 4)$
Or more generalizably,
$f(S, 2^x) \iff f(\sum_{i=1}^{x+1} S, 2^{x+1})$
I'm doubting that the generalizable case is true, but is the 2->4 case true? And if not, as a side question, does anyone know how to find some set $M$ such that $f(S,2) \iff f(M, 4)$?
I apologize if this question is unclear or obvious, my background is not entirely mathematical.
No
Take $M=\{2,9,10,10,14,15\}$. Then $|M|=60$. But you can't split $M$ into two equal parts, because you need to put $15$ somewhere and can't obtain 15 with $\{2,9,10,10,14\}$ to obtain $30$.
But $M+M$ can be split into 4 equal subsets $\{15,15\}$ $\{14,14,2\}$ $\{10,10,10\}$ and $\{10,9,9,2\}$
I think you can prove that no such example can use less than 6 elements for $M$.
Another example with all different values $\{2,6,11,12,14,15\}$.