# Does there always exist a monochromatic solution to ma+mb = nc+nd when m,n are coprime and N is coloured using 4 colours?

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.

• Why 4 colors specifically? Do you know the answer when you use fewer or more than 4 colors? – Wojowu Feb 22 at 18:39
• This might be relevant sfu.ca/~vjungic/RamseyNotes/sec_RadoThm.html; proposition 4.4.11 – Vlad Matei Feb 23 at 6:53
• @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. – Yaakov Baruch Feb 23 at 7:36
• 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 $3a+3b=2c+2d.$ – Aditya Guha Roy Feb 23 at 8:02
• My comment was meant to show that for sufficiently many colors it is easy. It does not provide an answer for $4$ colors. – Vlad Matei Feb 23 at 8:31