Show that one can color the set of integers with four colors: blue, red, yellow and purple, such that for any four numbers with the same colors $a, b, c, d$ (not necessarily distinct, not all four of them equals to zero) we have $3a - 2b \neq 2c - 3d$

My first idea was linking the problem with Four color theorem but it doesn't seem to relate to the problem. Then I tried to color the integers considering modulo $n$, but still failed in the case $a \equiv b \equiv c \equiv d \equiv 0 \pmod n$. Any idea?

  • $\begingroup$ $3(a+d)\neq2(b+c).$ $\endgroup$ – Lucian Nov 6 '15 at 20:15
  • $\begingroup$ Can you please tell me something less obvious? $\endgroup$ – primitiveroot Nov 7 '15 at 16:51
  • $\begingroup$ Please clarify what coloring means here. $\endgroup$ – Peter Nov 7 '15 at 18:56
  • $\begingroup$ It means that we color every integer with one of four colors, for example 1 is blue, 2 is red, 3 is yellow, 4 is purple etc. $\endgroup$ – primitiveroot Nov 8 '15 at 2:12
  • $\begingroup$ It looks like this question fits into the framework of Ramsey sequences on the integers. For example, there is a theorem of Rado which implies that there is a coloring with a finite number of colors that satisfies your conditions. $\endgroup$ – Zur Luria Nov 8 '15 at 7:40

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