There is no set of exactly five elements $A \subset \mathbb{N}$ such that sum of any distinct three is prime.

Question

Each the five elements belong to any of the modulo classes of $3$, i.e. $P=\{3k+1:k\in\mathbb{N} \cup \{0\}\}\bigcap A$

$ \ Q=\{3k:k\in\mathbb{N} \setminus\{ 0\}\}\bigcap A$

$R=\{3k+2:k\in\mathbb{N} \cup \{0\}\}\bigcap A$.

If any of the three sets contain $3$ or more elements, the case is resolved. [since their sum would be divisible by $3$. ]

$\mathbf{Case \ 2:}$

One of the three sets contain at least $2$ elements. Now, the other two contains

$2$ and $1$ elements $\mathbf{OR}$ $3$ and $0$ elements. The latter case has been solved.

We discuss the case when the configuration is $2-2-1$:

$|P|=2, |Q|=2,|R|=1$: Pick one element from each set. $3 \mid Sum$.

$|P|=1, |Q|=2, |R|=2$: Similar

$|P|=2, |Q|=1,|R|=2$: Similar.

Is it correct? Kindly Verify.

Answer 1

The idea is right, and most of it looks correct to me.

When you say

One of the three sets contain at least $2$ elements. Now, the other two contains

$2$ and $1$ elements $\mathbf{OR}$ $3$ and $0$ elements.

The class you mention in the first sentence could have three, four or five elements (you say at least $2$). In that case what you've written in the second sentence isn't true.

More generally, I would prefer the proof to be clearer on the fact that these are the two cases you're considering:

• There is a class that has no elements from $A$. In this case, one of the other classes must have at least three
• There is at least one element from each class in $A$

You've said as much in your answer, but it's a bit hidden away. Making this separation clearer, and proving the result separately in each case makes the proof easier to mentally compartmentalise and read.