I studied a result that says: for $k \ge 2$ colors if we have a coloring of the numbers in the set $N=[n]$ in $k$ colors such that every color class is sum free (no 3 numbers $a,b,c$ of the same color with $a+b=c$), then we can extend that to a sum-free coloring of $N'=[3n+1]$ with $k+1$ colors.
The proof says that if $i \in N$ has color $x$ then we assign that color to both $3i$ and $3i-1$ in $N'$. The ones that are left over are those $\equiv1 (\text{mod}\ 3)$ and those are assigned the new color $k+1$. Its easy to check this is sum free if we consider the sum of two numbers in $N' \setminus N$. But my question is, what if one of the numbers (say $a$) is in $N$ and the other (say $b$) is in $N' \setminus N$? If their sum ($c$) is in $N'$, can we still show that $a, b, c$ cannot be the same color?