Prove that there exist three numbers of different colors such that $a+b=c$ We color the numbers from $1$ to $n$ ($n\ge 7$, I think) with three distinct colors, say white, black and red such that every color appears more than $n/4$ times. Prove that there exist three numbers of different colors such that $a+b=c$.
I started by assuming that there doesn't exist such integers then if we assume that the number $1$ is colored black that would mean that $$(C(i),C(i+1))\neq (W,R)\quad \forall i\ge 2$$
Where $C(k)$ is the color of the number $k$. Because otherwise we would have $C(i)=W$ and $C(i+1)=R$ Which contradicts our assumption . And that's where I got stuck
 A: Partial Progress
For convenience, let the colors be $A, B, C$. Assume WLOG that $1$ is colored with $A$, and let the color with the largest minimum integer colored with the respective color be $C$. Now, let $v$ be the minimum integer colored with $C$, note that $v \geqslant 3$ (since $A$ and $B$ must precede it at least once). Moreover, it is not possible to have $v = 3$ since you cannot color $1,2,3$ with $A,B,C$ respectively, because $1 + 2 = 3$. Thus, $v \geqslant 4$.
Now, for each positive integer $k$ with $1 \leqslant k < v$, note that $k + (v-k) = v$. Since $v$ is colored with $C$ and neither of $k$ or $v-k$ are colored with $C$, the colors of $k$ and $v-k$ must be the same. In specific, $v-1$ must be colored with $A$.
Now, what can the color of $v+1$ be? Since $1 + v = (v+1)$, the color of $v+1$ cannot be $B$. Assume for the sake of contradiction that the color of $v+1$ is $C$. Let $u$ be the minimum integer colored with $B$ (note that $1 < u < v$). We know that $v-u$ also has color $B$. By the minimality of $u$, we know that $u-1$ must have color $A$, and so must $v-u+1$. However, we now have $(v-u+1) + u = (v+1)$, and their respective colors are $A, B, C$, which is a contradiction. Thus, $v+1$ must be colored $A$.
Notice how the first $C$ is sandwiched between $A$'s? This is going to be the case in general. We prove this by induction. Our base case has already been established. Take any $C$ that occurs after the first one, say at position $w$ (where $1 < u < v < w$), and assume that the claim holds true for the previous $C$'s. We know that $w-1$ cannot be colored $B$, since $1 + (w-1) = w$. We also know that $w-1$ cannot be colored with $C$ since every previous $C$ is succeeded by $A$. Thus, $w-1$ is colored with $A$. Furthermore, assume $w+1 \leqslant n$ (else, there is no successor, so the $C$ at the end is... half-sandwiched?). We can again see that $w+1$ cannot be colored with $B$ since $1 + w = (w+1)$.
Assume for the sake of contradiction that $w+1$ is colored with $C$. Recollect that $u$ is colored with $B$. Since $u + (w-u+1) = w+1$, we cannot color $w-u+1$ with $A$. As $(u-1) + (w-u+1) = w$, we cannot color $w-u+1$ with $B$. This implies that $w-u+1$ is colored with $C$. The induction hypothesis implies that $w-u$ is colored with $A$. Now, note that $u + (w-u) = w$ yields a contradiction. Thus, the sandwiching claim is proved by induction.
Even if $n$ is colored with $C$, this is followed by $A$ at position $1$ if we think about our numbers being in a circle. Thus, every $C$ corresponds to two $A$s. Now, if no two $C$'s share an $A$ in between them, then each $C$ would correspond to two unique $A$s. However, this would imply that the number of $C$s and $A$s are more than $\frac{n}{4}$ and $\frac{n}{2}$ respectively, which would be a contradiction since there are too few options left for $B$ (at most $\frac{n}{4}$).
Consider the minimal $x < n$ such that $x-1$ and $x+1$ are both colored with $C$. This means that $x-2$ and $x$ are colored with $A$. We can see that $x-u-1$ and $x-u+1$ cannot be colored with $A$ since $u$ is colored with $B$. By the minimality of $x$, not both of $x-u-1$ and $x-u+1$ can be colored with $C$. This forces one of them to be colored with $B$.
Assume $u > 2$, then $(u-2) + (x-u+1) = (x-1)$, where $u-2$ is colored with $A$ and $x-1$ is colored with $C$. This means $x-u+1$ must be colored with $C$, and hence, $x-u-1$ with $B$. However, $2$ is colored with $A$, and $2 + (x-u-1) = (x-u+1)$, yielding a contradiction. This forces $u = 2$.
