I am trying to show that the following statement is false: whenever $\mathbb{N}$ is finitely coloured by $c: \mathbb{N} \to \{1,\ldots,k\}$ (in the sense of Ramsey theory), there exists an infinite set $x_1
So, I tried to find a colouring $c$ for which there is clearly no such monochromatic set, but I'm not having any luck so far by attempting to pick a colouring for which $c(x_i + 2x_j) \neq c(x_j + 2x_i)$. The closest I think I've got is by choosing $c(z)$ to be the parity of the largest power of $2$ which divides $z$: for example, we colour $1$ "EVEN", 2 "ODD", 4 EVEN, 6 ODD, 17 EVEN. This does indeed give $x_i + 2x_j$ and $x_j + 2x_i$ different colours, unless the powers of 2 which divide them are at most one apart; e.g. $x_i=2^ap,\,x_j = 2^aq$ or $x_j = 2^{a \pm 1}q$ with $p,\,q$ odd. In the former case, this gives you the same colour, and in the latter case you necessarily can't say what the colour is because you end up trying to count the power of 2 dividing $2^{a+1}(p+q)$, and $p+q$ is some even number.
I'm not sure where I'm going wrong: what I'm trying to do is produce a colouring such that for any $x_i$ and $x_j$ in our set, $x_i + 2x_j$ and $x_j + 2x_i$ are differently coloured. In fact we only actually need to find 2 elements which are differently coloured, not necessarily every pair, so maybe my attempt to prove this is overkill. For example, I tried pursuing my above colouring, assuming all $x_i$ are of the form $2^a s$ or $2^{a+1}t$ for odd $s,\,t$ but was unable to derive a contradiction. I think I might be looking at the wrong colouring - could anyone please suggest how to prove that the statement is false? Many thanks for your help.