You want to color the numbers \(\{1,2,3,4,5\}\) with two colors, red and blue. The condition is that there are no three numbers of the same color \(x,y,z\) that satisfy \(x+y=z\) (repeating numbers is allowed if equality requires it). Is it possible to do it?
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Impossible coloring from 1 to 5
It looks like a tiny, almost harmless coloring. Precisely for this reason it is surprising: in such a small group there are already inevitabilities that intuition does not usually see in time.
Hints
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- The most dangerous obstacles are very short sums: 1+1=2, 1+2=3, 2+2=4, 1+4=5 and 2+3=5.
- Start assigning color to 1 and then force 2 and 3 to avoid single-color trios.
- You will see that the 5 is trapped: either color creates a forbidden sum.
Solution
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Answer: No, it is impossible. Without loss of generality, color 1 red. 1. Like $1+1=2$, the 2 cannot be red. Then 2 is blue.
- Like $2+2=4$, 4 cannot be blue. Therefore 4 is red.
- Like $1+4=5$, 5 cannot be red. Therefore 5 is blue. Now look at number 3: - If 3 is red, then $1+3=4$ gives a forbidden red triple.
- If 3 is blue, then $2+3=5$ gives a forbidden blue triple. The 3 can be neither red nor blue: contradiction. Therefore there is no valid coloring for $\{1,2,3,4,5\}$.
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