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Let $m \ge 2 ,n \ge 2$ be positive integers which are coprime (that means that the greatest common divisor of $m,n$ is $1$). Is it possible to paint the set $\mathbf{N}$ of all natural numbers using $4$ colours such that there are no positive integers $a,b,c,d$ of the same colour satisfying $ma+mb = nc+nd?$

The answer is YES for the specific case when $m=3, n=2.$ For instance if $h \in \mathbf{N}$ is of the form $h = 5^m (5k+i)$ for $i \in \{1,2,3,4 \}$ then if we paint $h$ using the $i$ th color then by chasing the coefficient of $i$ it is easy to see that we do not have a monochromatic solution to the equation $3a+3b=2c+2d.$

In fact if one goes through the proof for this particular case then it seems to imply a much weaker relation where $4$ gets replaced by a larger number of colors.

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    $\begingroup$ Why 4 colors specifically? Do you know the answer when you use fewer or more than 4 colors? $\endgroup$ – Wojowu Feb 22 at 18:39
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    $\begingroup$ This might be relevant sfu.ca/~vjungic/RamseyNotes/sec_RadoThm.html; proposition 4.4.11 $\endgroup$ – Vlad Matei Feb 23 at 6:53
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    $\begingroup$ @VladMatei The last proposition there (Rado's Theorem) implies the claim even with any finite number of colors: take $(c_1,c_2,c_3,c_4)$ in the paper equal to the $(m,m,-n,-n)$ in the question. Even though the question seems more suitable for math.stackexchange.com you probably should make this into an answer. $\endgroup$ – Yaakov Baruch Feb 23 at 7:36
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    $\begingroup$ I doubt that. For the case when $m=3,n=2$ we do have a coloring. For instance if $n = 5^m (5k + i)$ for $i \in \{1,2,3,4 \}$ then we paint the natural number $n$ with the $i$ th color. It is easy to see that for this coloring we do not have a monochromatic solution to $3a+3b=2c+2d.$ $\endgroup$ – Aditya Guha Roy Feb 23 at 8:02
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    $\begingroup$ My comment was meant to show that for sufficiently many colors it is easy. It does not provide an answer for $4$ colors. $\endgroup$ – Vlad Matei Feb 23 at 8:31

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