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IRL ABooklet

Combinatorics Difficulty 5.0 AIME Prove it Ireland

Find the smallest number of colours needed for colouring all positive integers in such a way that any two integers that differ by a prime number have different colours.

Solution

Assume such a colouring exists and consider the numbers 1,3,6,81, 3, 6, 8. Any two of these numbers differ by a prime number: 3−1=23 - 1 = 2, 6−1=56 - 1 = 5, 8−1=78 - 1 = 7, 6−3=36 - 3 = 3, 8−3=58 - 3 = 5 and 8−6=28 - 6 = 2. Thus, at least four colours are needed.

To prove that four colours are sufficient, we pick four colours and number them 0,1,2,30, 1, 2, 3. Colour number ii is applied to all integers nn that satisfy n≡i(mod4)n \equiv i \pmod 4. The difference of any two integers of the same colour is then divisible by 44, hence not a prime.

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