Let ${x_1, x_2, x_3, . . . , x_n}$ be a set of $n$ distinct positive integers, that the sum of any $3$ of them is a prime. What is the max of $n$?

Question

Let ${x_1, x_2, x_3, . . . , x_n}$ be a set of $n$ distinct positive integers, such that the sum of any $3$ of them is a prime number. What is the maximum value of $n$?

This question is from the NZMO for year $2021$.

I began this problem by knowing that obviously $n=3$ is possible.

Then we show that $n = 4$ is possible with an example. The example ${x_1, x_2, x_3, x_4} = {1, 3, 7, 9}$ satisfies the problem because:

• $1 + 3 + 7 = 11$ is prime,

• $1 + 3 + 9 = 13$ is prime,

• $1 + 7 + 9 = 17$ is prime, and

• $3 + 7 + 9 = 19$ is prime.

We now have to prove that $n ≥ 5$ is impossible. I am stuck here and I do not know how to continue calculating the question to get an answer.

However, I am thinking that any set ${x_1, x_2, x_3, . . . , x_n}$ such that the sum of any $3$ of them is a prime number to consider the three “pigeonholes” modulo $3$; the residue $0, 1$ and $2$.

If anyone can help solve this problem and explain it to me, I would appreciate it a lot. Thank you very much.

FYI, there's a general result that, for among any $2m-1$ integers, there's always a subset of $m$ of them which sum to an integral multiple of $m$. With $m=3$, this means $n\ge 5$ is impossible, as you surmised. A couple of questions here which deal with that specific case are How do I prove that among any $5$ integers, you are able to find $3$ such that their sum is divisible by $3$? and ... — John Omielan, Aug 17 '24 at 12:39
(cont.) Prove that for any set of 5 integers, there is at least one subset of 3 integers whose sum is divisible by 3. Also, I've collected various posts regarding this general result in my question and answers at Proving that among any $2n - 1$ integers, there's always a subset of $n$ which sum to a multiple of $n$. — John Omielan, Aug 17 '24 at 12:40
FYI, in $1980$ in Canada, the Euclid Math Contest (targeting grade $12$ students) had a $2n−1$ question with $n=3$, i.e., that among any $5$ integers, prove there's always a subset of $3$ which sum to an integral multiple of $5$. Note I wrote this contest myself then. Also, that same year, the Descartes Math Contest (targeting grade $13$ students, & used to help to determine entrance scholarships, but this no longer exists now since grade $13$ stopped existing in Ontario) had a similar question for $n=4$, i.e., among any $7$ integers, there's a subset of $4$ summing to a multiple of $7$. — John Omielan, Aug 17 '24 at 13:07

score 4 · Accepted Answer · answered Aug 17 '24 at 06:04

You're very much on the right track with the modulo 3 pigeon holes.

If there are numbers in each hole, then one number from each hole will sum to be divisible by 3 (and strictly larger than 3).

If there are three numbers in any hole, the sum of three of them will likewise be divisible by 3.

So we can use at most 2 of the holes, and the holes can have at most 2 numbers each. That gives us a maximum of 4.

Let ${x_1, x_2, x_3, . . . , x_n}$ be a set of $n$ distinct positive integers, that the sum of any $3$ of them is a prime. What is the max of $n$?

1 Answers1