Here are the statements. I have several ideas on how to go about proving them, but I couldn't develop those ideas fully. I'd like to ask for some comments/hints.
(i) Show that whenever the natural numbers $\Bbb N$ are finitely coloured, one can find monochromatic $x,y,z \in \Bbb N$ such that $x+y=z$.
(ii) Show that whenever the natural numbers $\Bbb N$ are finitely coloured, one can find distinct monochromatic $x,y,z \in \Bbb N$ such that $x+y=z$.
In my proof attempts I use the infinite Ramsey theorem, which is as follows:
Theorem 1.1. For any positive integer $k$ and and $r$-colouring of $\binom{\Bbb{N}}{k}$, there exists an infinite set $S \subseteq \Bbb N$ for which $\binom{S}{k}$ is monochomatic.
Here of course $k =1$.
My first idea is to use induction on the number of infinite monochromatic sequences, which is at least 1 by the infinite Ramsey theorem. If there's only one such sequence, then it contains a number $u$ that is greater than any number in any other sequences. Then we can have $x = u, y = u, z = 2u$ (or $x = u, y = 2u, z = 3u$ for part (ii)). .
If, otherwise, there are $k$ infinite monochromatic sequences, we can go "color-blind", considering the colors $k-1$ and $k$ to be the same, thus reducing the number of infinite monochromatic sequences by 1. By the induction hypothesis, a monochromatic triple $\{x,y,z\}$ exists, and in particular it must belong to the sequence $S_{k-1} \cup S_k$ (as otherwise the merging or splitting of the two sequences $S_{k-1} \text{ and } S_k$ doesn't affect the membership status of $x,y,z$ and we would be done). But here I can't go further to show that $\{x,y,z\}$ must be either in $S_{k-1} \text{ or } S_k$.
The second idea comes from the observation that, assuming the contradiction, if a sequence contains some members $x \text{ and } y$, then it cannot contains $2x, 2y \text{ and } x+y$. The monochromatic sequences are infinite, the number of colors is finite, and so I think the pigeonhole principle can somehow be used to show that some appropriate multiples of numbers or sums of them would end up in one of the sequences.
Edit: it has been pointed out to me that the triples I seek are called Schur's triples. The proof presented uses a different version of Ramsey's theorem, but I'm still interested in whether it's possible to prove this using my first idea above. I feel like it's come quite close.
Referring to i) only, what you are after are monochromatic Schur triples. (A Schur triple is a triple $(a,b,c)$ with $c=b+a$.) It was shown by Schur that for any fixed $r$ there exists some $s(r)$ such that if the integers $1,2,...,s(r)$ are colored in $r$ colors, then one of the color classes contains a monochromatic triple. This is, in fact, an easy consequence of the Ramsey theorem: assuming that every positive integer $z$ is assigned the color $c(z)\in\{1,\dotsc,r\}$, consider the complete graph on $n$ vertices with the edge joining vertices $i$ and $j$ assigned the color $c(|i-j|)$. If $n$ is large enough, then (by the Ramsey theorem) this graph contains a monochromatic triangle, and it is very easy to see that the vertices of this triangle correspond to a Schur triple.