Prove or Disprove :

There exists $A\subset\mathbb{N}$ with exactly FIVE elements, such that sum of any three elements of $A$ is a prime number.

I don't have any hint or insight about proving or disproving the above statement. I even don't know whether the statement is true or not.

Can anyone guide me how to approach this problem ?


1 Answer 1


Consider the residues of the elements $\pmod 3$. Note that if we have $a_1 \equiv 0 \pmod 3, a_2 \equiv 1 \pmod 3, a_3 \equiv 2 \pmod 3$, then we are done - their sum is divisible by 3. So only two of the residues $\pmod 3$ can be present.

But by the pigeonhole principle, that means that at least $\left\lceil\frac{5}{2}\right\rceil = 3$ of these elements have the same residue $\pmod 3$. Then their sum is divisible by $3$. Hence, no such set exists.

(To be a little more precise, since the $a_i$'s are distinct, their sum must be at least $6$ if you take $0 \notin \mathbb{N}$. If you consider $0 \in \mathbb{N}$, then you could technically have $0,1,2$ all be elements to get this sum. In this case, we would then have $a_4 + 0 + 1$ and $a_4 + 0 + 2$ are both prime, which is a contradiction as one is even and strictly greater than $2$.)

  • $\begingroup$ Short and sweet. $\endgroup$
    – 123
    Aug 10, 2017 at 15:43

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