Monochromatic solutions to $x_{1}+x_{2}+\cdots+ x_{m}=x_{m+1}$ What is the least $n$ such that any $2$-coloring of $[n]$ has a monochromatic solution to the equation
$$x_{1}+x_{2}+\cdots+ x_{m}=x_{m+1}.$$ 
I think the answer is $m^{2}+m-1$. I have a coloring of $[m^{2}+m-2]$ with no monochromatic solution. So $n\geq m^{2}+m-1$. I'm looking for the other inequality. But I have no idea how to approach the problem. Any hints would be very nice. Thanks. 
 A: (Below, intervals denote integer intervals, that is, $[a,b]=\{n\in\mathbb N\mid a\le n\le b\}$, and the colors are red and blue.)
To see the lower bound, we can consider the coloring where $[1,m-1]\cup[m^2,m^2+m-2]$ is colored red, and $[m,m^2-1]$ is colored blue. This coloring gives us no monochromatic solutions to $$x_1+\dots+x_m=x_{m+1}. $$
To see that any $2$-coloring of $[m^2+m-1]$ gives us a monochromatic solution, proceed by contradiction, and consider a putative counterexample. 


*

*We may assume $\color{red}{1}$ is red, so $\color{blue}{m}=\color{red}{1}\cdot m$ is blue and $\color{red}{m^2}=\color{blue}m\cdot m$ is red. 

*Since $\color{red}{1},\color{red}{m^2}$ are red, then $\color{red}{1}\cdot(m-1)+\color{red}{m^2}=\color{blue}{m^2+m-1}$ is blue. 

*Since $\color{blue}{m},\color{blue}{m^2+m-1}=\color{blue}{m}\cdot(m-1)+(\color{red}{2m-1})$ are blue, then $\color{red}{2m-1}$ is red, and since $\color{red}{2m-1}=\color{red}{1}+\color{blue}{2}\cdot(m-1)$, then $\color{blue}{2}$ is blue, which gives us that $\color{blue}{2}\cdot(m-1)+\color{blue}{m}=\color{red}{3m-2}$ is red.

*Since $\color{red}{1},\color{red}{3m-2}$ are red, then $\color{red}{1}\cdot(m-1)+\color{red}{3m-2}=\color{blue}{4m-3}$ is blue. But since $\color{blue}{4m-3}=\color{red}{3}\cdot(m-1)+\color{blue}{m}$, then $\color{red}{3}$ is red, and therefore $\color{blue}{m+2}=\color{red}{1}\cdot(m-1)+\color{red}{3}$ is blue.

*We now have a contradiction, since $\color{red}{3}\cdot m=\color{blue}{2}\cdot(m-1)+\color{blue}{m+2}$, and no matter whether $3m$ is red or blue, we obtain a monochromatic solution to our equation, against the assumption.    


By the way, in Ramsey theory on the integers, by Bruce Landman and Aaron Robertson, you can find this result and generalizations, discussed in section $8.2$. The argument for the upper bound I just showed is an adaptation of a result they prove there (see Theorem $8.21$).
