# 4 distinct integers with prime sum for each triple

Here is a nice high school olympiad math problem:

Can you choose 4 distinct positive integers so that the sum of each 3 of them is prime? How about 5?

It looks that just by looking at reminders mod 2,3,6 is not enough.

One important issue to consider in choosing the $$4$$ integers is the parity. For the sum of each of them to be prime means that you must have $$2$$ or less being even as the sum of $$3$$ even numbers is even. If you have $$2$$ or $$1$$ even (and $$2$$ or $$3$$ odd) numbers, then choosing $$1$$ even and $$2$$ odd numbers makes the sum even. Thus, you need to have all $$4$$ numbers being odd.
Considering remainders modulo $$3$$, note if $$3$$ or $$4$$ of the numbers have the same remainder, the sum of $$3$$ of them would be a multiple of $$3$$ and, thus, not prime. As such, you need to have $$2$$ groups of $$2$$ different remainders modulo $$3$$.
One set which meets these requirements is $$\{1,3,7,9\}$$ giving sums, when taken $$3$$ at a time, of $$11,13,17,19$$. In this case, there are $$2$$ numbers each with a remainder of $$0$$ and $$1$$ modulo $$3$$.
For $$5$$ (or more) integers, once again, you can't have $$3$$ or more with the same remainder modulo $$3$$. However, with at most $$2$$ having any the same remainders modulo $$3$$, there must be at least one integer each with a remainder of $$0, 1$$ and $$2$$, so the sum of those $$3$$ would, once again, be a multiple of $$3$$, so it's not a prime. Thus, you can't have $$5$$ or more such integers where the sum of each subset of $$3$$ is a prime.
FYI, this property that a set of $$5$$ integers always has a subset of $$3$$ that sum to a multiple of $$3$$ is an example of the more general case of, for all $$n \in \mathbb{N}$$, that a set of $$2n - 1$$ integers always has at least one of subset of $$n$$ integers which sum to a multiple of $$n$$. For larger specific cases, there's Given 7 arbitrary integers,sum of 4 of them is divisible by 4 and Among any $$11$$ integers, sum of $$6$$ of them is divisible by $$6$$. At a math seminar I attended just under $$40$$ years ago, we were told the general case was not yet proven. I just did a brief online search but could not find whether or not this is still the case.