On Some Exact Formulas for $2$-Color Off-diagonal Rado Numbers

1. Introduction

Let $[a,b]$ denote the set $\{x\in \mathbb{Z}|a\le x\le b\}$. A function $\mathrm{\Delta }$: $[1,n]\to [0,k-1]$ is called a $k$-coloring of the set $[1,n]$. If $\epsilon$ is a system of equations in $m$ variables, then we say that a solution $x_1,x_2,\dots ,x_m$ to $\epsilon$ is monochromatic if and only if $$\mathrm{\Delta }(x_1)=\mathrm{\Delta }(x_2)=\cdots =\mathrm{\Delta }(x_m).$$

In 1916, Schur [1] proved that for every integer $k\ge 2$, there exists a least integer $n=S(k)$ such that for every $k$-coloring of the set $[1,n]$, there exists a monochromatic solution to $x+y=z$, the integer $S_k$ is called Schur number. Rado [2, 3] generalized the work of Schur to arbitrary system of linear equations and found necessary and sufficient conditions to determine if an arbitrary system of linear equations admits a monochromatic solution for every coloring of the natural numbers with a finite number of colors. For a given system of linear equations $\epsilon$, the least integer $n$, provided that it exists, such that for every coloring of the set $[1,n]$ with $k$ colors, there exists a monochromatic solution to $\epsilon$, is called $k$-color Rado number. If such an integer $n$ does not exist, then the $k$-color Rado number for the system $\epsilon$ is infinite. In recent years there has been considerable interest in finding the exact Rado numbers for particular linear equations and in several other closely related problems, see for example [4, 5, 6, 7, 8, 9, 10, 11, 12].

For positive integer $k\ge 2$ and equations ${\epsilon }_i$, where $i=0,1,\dots ,k-1$, the $k$-color off-diagonal Rado number is the least integer $N$ provided that it exists for which any $k$-coloring of $[1,N]$ must admit a monochromatic solution of color $i$ to ${\epsilon }_i$ for some $i\in [0,k-1]$. Note that if ${\epsilon }_i={\epsilon }_j$ for all $i$, $j$ $(1\le i,j\le k)$, then the $k$-color off-diagonal Rado number is the $k$-color Rado number. Robertson and Schaal [13] gave some 2-color off-diagonal Rado numbers for particular linear equations. In [14], Mayers and Robertson gave the exact $2$-color off-diagonal Rado numbers when the two equations are of the form $x+qy=z$ and $x+sy=z$. They showed that $$R_2(x+qy=z,x+y=z)=2q+2\lfloor \frac{q+1}{2}\rfloor +1$$ and for $s\ge 2$, $$R_2(x+qy=z,x+sy=z)=qs+q+2s+1.$$ Motivated by Myers and Robertson’s work, Yao and Xia [15] proved the following results: $$R_2(3x+3y=z,3x+3qy=z)=54q+57,q\ge 2$$ and $$R_2(2x+3y=z,2x+2qy=z)=20q+26,q\ge 2.$$

In this paper, we are also interested in precise values of $2$-color off-diagonal Rado numbers and let $R_2({\epsilon }_0,{\epsilon }_1)$ denote it. Let blue and red be the two colors and denoted by 0 and 1, respectively. We determine the exact numbers $R_2(2x+qy=2z,2x+sy=2z)$, where $q,s$ are odd integers and $q>s\ge 1$. The main results of this paper can be stated as follows.

2. Lower Bounds

3. Proof of Theorem 1

It follows from Lemma 1 that $$\begin{array}{r}{R}_2(2x+qy=2z,2x+sy=2z)\ge min\{qs+q+2s+1,qs+2q+1\},\end{array}$$ where $q,s$ are odd integers and $q>s\ge 3$. In order to prove this theorem, it suffices to show that $$R_2(2x+qy=2z,2x+sy=2z)\le min\{qs+q+2s+1,qs+2q+1\},$$ where $q,s$ are odd integers and $q>s\ge 3$. Let $\mathrm{\Delta }$ be a 2-coloring of $[1,min\{qs+q+2s+1,qs+2q+1\}]$ using the colors red and blue. Without loss of generality, we assume, for contradiction, that there is no blue solution to $2x+qy=2z$ and no red solution to $2x+sy=2z$. We break the argument into 4 cases;

1. Case 1: $\mathrm{\Delta }(2)=0$ and $\mathrm{\Delta }(s)=0$.

   $\mathrm{\Delta }(2)=0$ and $\mathrm{\Delta }(s)=0$ imply that $\mathrm{\Delta }(q+s)=1$, otherwise $(s,2,q+s)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(q+s)=1$, we must have $\mathrm{\Delta }(q+s+\frac{(q+s)s}{2})=0$ or else $(q+s,q+s,q+s+\frac{(q+s)s}{2})$ is a red solution to $2x+sy=2z$. Now, $\mathrm{\Delta }(q+s+\frac{(q+s)s}{2})=0$ and $\mathrm{\Delta }(2)=0$, so $\mathrm{\Delta }(2q+s+\frac{(q+s)s}{2})=1$, otherwise $(q+s+\frac{(q+s)s}{2},2,2q+s+\frac{(q+s)s}{2})$ is a blue solution to $2x+qy=2z$. Note that $$2q+s+\frac{(q+s)s}{2}\le min\{qs+q+2s+1,qs+2q+1\}$$ since $q-s\ge 2$. $\mathrm{\Delta }(2q+s+\frac{(q+s)s}{2})=1$ and $\mathrm{\Delta }(q+s)=1$ imply that $\mathrm{\Delta }(2q+s)=0$, otherwise $(2q+s,s+q,2q+s+\frac{(q+s)s}{2})$ is a red solution to $2x+sy=2z$. Since $\mathrm{\Delta }(2q+s)=0$ and $\mathrm{\Delta }(s)=0$, we must have $\mathrm{\Delta }(4)=1$ or else $(s,4,2q+s)$ is a blue solution to $2x+qy=2z$. If $\mathrm{\Delta }(q+3s)=1$, then $(q+s,4,q+3s)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(q+3s)=0$. If $\mathrm{\Delta }(3s)=0$, then $(3s,2,q+3s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(3s)=1$. Now, $\mathrm{\Delta }(3s)=1$ and $\mathrm{\Delta }(4)=1$, we must have $\mathrm{\Delta }(5s)=0$ or else $(3s,4,5s)$ is a red solution to $2x+sy=2z$. $\mathrm{\Delta }(5s)=0$ and $\mathrm{\Delta }(2)=0$, imply that $\mathrm{\Delta }(q+5s)=1$, otherwise $(5s,2,q+5s)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(q+5s)=1$ and $\mathrm{\Delta }(q+s)=1$, we must have $\mathrm{\Delta }(8)=0$ or else $(q+s,8,q+5s)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(4q+s)=0$, then $(s,8,4q+s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(4q+s)=1$. If $\mathrm{\Delta }(4q-s)=1$, then $(4q-s,4,4q+s)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(4q-s)=0$. Since $\mathrm{\Delta }(4q-s)=0$ and $\mathrm{\Delta }(2)=0$, we must have $\mathrm{\Delta }(3q-s)=1$ or else $(3q-s,2,4q-s)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(3q-s)=1$ and $\mathrm{\Delta }(4)=1$ imply that $\mathrm{\Delta }(3q+s)=0$, otherwise $(3q-s,4,3q+s)$ is a red solution to $2x+sy=2z$. Now we have $\mathrm{\Delta }(3q+s)=0$, $\mathrm{\Delta }(2)=0$ and $\mathrm{\Delta }(2q+s)=0$, and then $(2q+s,2,3q+s)$ is a blue solution to $2x+qy=2z$, which is a contradiction.

2. Case 2: $\mathrm{\Delta }(2)=0$ and $\mathrm{\Delta }(s)=1$.

   If $\mathrm{\Delta }(2)=\mathrm{\Delta }(4)=\cdots =\mathrm{\Delta }(2q+2)=0$ and then $(2,4,2q+2)$ is a blue solution to $2x+qy=2z$, which is a contradiction. So we can assume that $k$ $(1\le k<q+1)$ is the least number such that $\mathrm{\Delta }(2)=\cdots =\mathrm{\Delta }(2k)=0$ and $\mathrm{\Delta }(2k+2)=1$.

1. Subcase 1: $k\le min\{\frac{(s-1)q+1}{s},q-2\}$.

   $\mathrm{\Delta }(2k+2)=1$ and $\mathrm{\Delta }(s)=1$ implies that $\mathrm{\Delta }(ks+2s)=0$, otherwise $(s,2k+2,ks+2s)$ is a red solution to $2x+sy=2z$. Since $\mathrm{\Delta }(ks+2s)=0$ and $\mathrm{\Delta }(2)=0$, we must have $\mathrm{\Delta }(q+ks+2s)=1$ or else $(ks+2s,2,q+ks+2s)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(q+ks+2s)=1$ and $\mathrm{\Delta }(2k+2)=1$ imply that $\mathrm{\Delta }(q+s)=0$, otherwise $(q+s,2k+2,q+ks+2s)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(2q+s)=0$, then $(q+s,2,2q+s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(2q+s)=1$. If $\mathrm{\Delta }(2q+ks+2s)=1$ then $(2q+s,2k+2,2q+ks+2s)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(2q+ks+2s)=0$. Note that $$2q+ks+2s\le min\{qs+q+2s+1,qs+2q+1\}$$ since $k\le min\{\frac{(s-1)q+1}{s},q-2\}$. Since $\mathrm{\Delta }(2q+ks+2s)=0$ and $\mathrm{\Delta }(ks+2s)=0$, we must have $\mathrm{\Delta }(4)=1$, otherwise $(ks+2s,4,ks+2s+2q)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(4)=1$ and $\mathrm{\Delta }(2q+s)=1$ imply that $\mathrm{\Delta }(2q-s)=0$ or else $(2q-s,4,2q+s)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(3q-s)=0$, then $(2q-s,2,3q-s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(3q-s)=1$. Since $\mathrm{\Delta }(3q-s)=1$ and $\mathrm{\Delta }(4)=1$, we must have $\mathrm{\Delta }(3q+s)=0$ or else $(3q-s,4,3q+s)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(4q+s)=0$, then $(3q+s,2,4q+s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(4q+s)=1$. If $\mathrm{\Delta }(4q-s)=1$, then $(4q-s,4,4q+s)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(4q-s)=0$. $\mathrm{\Delta }(s)=1$ and $\mathrm{\Delta }(4)=1$ imply that $\mathrm{\Delta }(3s)=0$ or else $(s,4,3s)$ is a red solution to $2x+sy=2z$. Since $\mathrm{\Delta }(3s)=0$ and $\mathrm{\Delta }(2)=0$, we must have $\mathrm{\Delta }(q+3s)=1$ or else $(3s,2,q+3s)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(4)=1$ and $\mathrm{\Delta }(q+3s)=1$ imply that $\mathrm{\Delta }(q+5s)=0$, otherwise $(q+3s,4,q+5s)$ is a red solution to $2x+sy=2z$. $\mathrm{\Delta }(q+5s)=0$ and $\mathrm{\Delta }(2)=0$ imply that $\mathrm{\Delta }(5s)=1$, otherwise $(5s,2,q+5s)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(5s)=1$ and $\mathrm{\Delta }(s)=1$, we must have $\mathrm{\Delta }(8)=0$ or else $(s,8,5s)$ is a red solution to $2x+sy=2z$. When $s=3$, by the above proof, we have $\mathrm{\Delta }(4q+s)=\mathrm{\Delta }(4q+3)=1$ and $\mathrm{\Delta }(q+3s)=\mathrm{\Delta }(q+9)=1$, then $\mathrm{\Delta }(2q-4)=0$, otherwise $(q+9,2q-4,4q+3)$ is a red solution to $2x+3y=2z$. $\mathrm{\Delta }(2)=0$ implies that $\mathrm{\Delta }(q+2)=1$ or else $(2,2,q+2)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(q+2)=1$ and $\mathrm{\Delta }(4)=1$, we must have $\mathrm{\Delta }(q-4)=0$, otherwise $(q-4,4,q+2)$ is a red solution to $2x+3y=2z$. Now we have $\mathrm{\Delta }(2q-4)=0$, $\mathrm{\Delta }(q-4)=0$ and $\mathrm{\Delta }(2)=0$, and then $(q-4,2,2q-4)$ is a blue solution to $2x+qy=2z$, which is a contradiction. When $s\ge 5$, since $\mathrm{\Delta }(8)=0$ and $\mathrm{\Delta }(q+s)=0$, we must have $\mathrm{\Delta }(5q+s)=1$, otherwise $(q+s,8,5q+s)$ is a blue solution to $2x+qy=2z$. Note that $$5q+s\le min\{qs+q+2s+1,qs+2q+1\},$$ since $s\ge 5$. $\mathrm{\Delta }(5q+s)=1$ and $\mathrm{\Delta }(4)=1$ imply that $\mathrm{\Delta }(5q-s)=0$ or else $(5q-s,4,5q+s)$ is a red solution to $2x+sy=2z$. Now we have $\mathrm{\Delta }(5q-s)=0$, $\mathrm{\Delta }(2)=0$ and $\mathrm{\Delta }(4q-s)=0$, and then $(4q-s,2,5q-s)$ is a blue solution to $2z+qy=2z$, which is a contradiction.

2. Subcase 2: $min\{\frac{(s-1)q+1}{s},q-2\}<k\le q$.

   It is easy to verify that $$q+1\le 2min\{\frac{(s-1)q+1}{s},q-2\}$$ and $$s+1<2s\le 2min\{\frac{(s-1)q+1}{s},q-2\}.$$ Thus, $q+1\le 2k$ and $s+1<2s\le 2k$, and we have $\mathrm{\Delta }(s+1)=\mathrm{\Delta }(2s)=\mathrm{\Delta }(q+1)=0$. Obviously, $k\ge 2$ and $\mathrm{\Delta }(2)=\mathrm{\Delta }(4)=0$, then $\mathrm{\Delta }(2q+2)=1$, otherwise $(2,4,2q+2)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(q+1)=0$ and $\mathrm{\Delta }(2)=0$, we must have $\mathrm{\Delta }(1)=1$ or else $(1,2,q+1)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(1)=1$ and $\mathrm{\Delta }(2q+2)=1$ imply that $\mathrm{\Delta }(qs+s+1)=0$, otherwise $(1,2q+2,qs+s+1)$ is a red solution to $2x+sy=2z$. Now we have $\mathrm{\Delta }(qs+s+1)=0$, $\mathrm{\Delta }(2s)=0$ and $\mathrm{\Delta }(s+1)=0$, and then $(s+1,2s,qs+s+1)$ is a blue solution to $2x+qy=2z$, which is a contradiction.

3. Case 3: $\mathrm{\Delta }(2)=1$ and $\mathrm{\Delta }(s)=0$.

   If $\mathrm{\Delta }(2)=\mathrm{\Delta }(4)=\cdots =\mathrm{\Delta }(2s+2)=1$, then $(2,4,2s+2)$ is a red solution to $2x+sy=2z$, which is a contradiction. So we can assume that $l$ ($1\le l<s+1$) is the least integer such that $\mathrm{\Delta }(2)=\cdots =\mathrm{\Delta }(2l)=1$ and $\mathrm{\Delta }(2l+2)=0$.

1. Subcase 1: $l\le s-1$.

   $\mathrm{\Delta }(2)=1$ implies that $\mathrm{\Delta }(s+2)=0$, otherwise $(2,2,s+2)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }((l+1)q+s+2)=0$, then $(s+2,2l+2,(l+1)q+s+2)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }((l+1)q+s+2)=1$. If $\mathrm{\Delta }((l+1)q+2s+2)=1$, then $((l+1)q+s+2,2,(l+1)q+2s+2)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }((l+1)q+2s+2)=0$. Note that $$(l+1)q+2s+2\le min\{qs+q+2s+1,qs+2q+1\}$$ since $l\le s-1$. $\mathrm{\Delta }((l+1)q+2s+2)=0$ and $\mathrm{\Delta }(2l+2)=0$ imply that $\mathrm{\Delta }(2s+2)=1$, otherwise $(2s+2,2l+2,(l+1)q+2s+2)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(2s+2)=1$ and $\mathrm{\Delta }(2)=1$, we must have $\mathrm{\Delta }(4)=0$ or else $(2,4,2s+2)$ is a red solution to $2x+sy=2z$. $\mathrm{\Delta }(s)=0$ and $\mathrm{\Delta }(4)=0$ imply that $\mathrm{\Delta }(2q+s)=1$, otherwise $(s,4,2q+s)$ is a blue solution to $2x+qy=2z$. If $\mathrm{\Delta }(2q+2s)=1$, then $(2q+s,2,2q+2s)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(2q+2s)=0$. If $\mathrm{\Delta }(2s)=0$, then $(2s,4,2q+2s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(2s)=1$. Now, $\mathrm{\Delta }(2s)=1$ and $\mathrm{\Delta }(2)=1$, we must have $\mathrm{\Delta }(3s)=0$ or else $(2s,2,3s)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(3s+2q)=0$, then $(3s,4,3s+2q)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(3s+2q)=1$. Since $\mathrm{\Delta }(3s+2q)=1$ and $\mathrm{\Delta }(2)=1$, we must have $\mathrm{\Delta }(2q+4s)=0$ or else $(3s+2q,2,4s+2q)$ is a red solution to $2x+sy=2z$. Since $\mathrm{\Delta }(2q+4s)=0$ and $\mathrm{\Delta }(4)=0$, we must have $\mathrm{\Delta }(4s)=1$, otherwise $(4s,4,2q+4s)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(4s)=1$ and $\mathrm{\Delta }(2)=1$ imply that $\mathrm{\Delta }(5s)=0$, otherwise $(4s,2,5s)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(5s+2q)=0$, then $(5s,4,5s+2q)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(5s+2q)=1$. When $q=5$, then $s=3$, by the above proof, we have $\mathrm{\Delta }(2s+2)=\mathrm{\Delta }(8)=1$ and $\mathrm{\Delta }(2)=1$, then $\mathrm{\Delta }(14)=0$, otherwise $(2,8,14)$ is a red solution to $2x+3y=2z$. Now we have $\mathrm{\Delta }(14)=0$ and $\mathrm{\Delta }(4)=0$, so $(4,4,14)$ is a blue solution to $2x+5y=2z$, which is a contradiction. When $q>5$ and $s=3$, $\mathrm{\Delta }(4)=0$ implies that $\mathrm{\Delta }(2q+4)=1$, otherwise $(4,4,2q+4)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(2q+4)=1$ and $\mathrm{\Delta }(2)=1$, we must have $\mathrm{\Delta }(2q+1)=0$ or else $(2q+1,2,2q+4)$ is a red solution to $2x+3y=2z$. If $\mathrm{\Delta }(4q+1)=0$, then $(2q+1,4,4q+1)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(4q+1)=1$. If $\mathrm{\Delta }(4q+4)=1$, then $(4q+1,2,4q+4)$ is a red solution to $2x+3y=2z$, so we may assume that $\mathrm{\Delta }(4q+4)=0$. Now we have $\mathrm{\Delta }(8)=0$, $\mathrm{\Delta }(4)=0$ and $\mathrm{\Delta }(4q+4)=0$, so $(4,8,4q+4)$ is a blue solution to $2x+qy=2z$, which is a contradiction. When $q>s\ge 5$, we see that $$max\{2q+6s,6q+s\}\le min\{qs+q+2s+1,qs+2q+1\}.$$ Since $\mathrm{\Delta }(5s+2q)=1$ and $\mathrm{\Delta }(2)=1$, then $\mathrm{\Delta }(2q+6s)=0$ or else $(2q+5s,2,2q+6s)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(6s)=1$, then $(6s,4,2q+6s)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(6s)=1$ and $\mathrm{\Delta }(2s)=1$, we must have $\mathrm{\Delta }(8)=0$, otherwise $(2s,8,6s)$ is a red solution to $2x+sy=2z$. $\mathrm{\Delta }(2q+s)=1$ and $\mathrm{\Delta }(2)=1$ imply that $\mathrm{\Delta }(2q)=0$ or else $(2q,2,2q+s)$ is a red solution to $2x+sy=2z$. Since $\mathrm{\Delta }(2q)=0$ and $\mathrm{\Delta }(4)=0$, then $\mathrm{\Delta }(4q)=1$, otherwise $(2q,4,4q)$ is a blue solution to $2x+qy=2z$. If $\mathrm{\Delta }(4q+s)=1$, then $(4q,2,4q+s)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(4q+s)=0$. If $\mathrm{\Delta }(6q+s)=0$, then $(4q+s,4,6q+s)$ is a blue solution to $2x+qy=2z$. Since $\mathrm{\Delta }(6q+s)=1$ and $\mathrm{\Delta }(2)=1$, we must have $\mathrm{\Delta }(6q)=0$, otherwise $(6q,2,6q+s)$ is a red solution to $2x+sy=2z$. Now we have $\mathrm{\Delta }(2q)=0$, $\mathrm{\Delta }(8)=0$ and $\mathrm{\Delta }(6q)=0$, so $(2q,8,6q)$ is a blue solution to $2x+qy=2z$, which is a contradiction.

2. Subcase 2: $l=s$.

   We have $\mathrm{\Delta }(2)=\mathrm{\Delta }(4)=\cdots =\mathrm{\Delta }(2s)=1$ and $\mathrm{\Delta }(2s+2)=0$. Note that $2|(s+1)$ and $s+1<2s$, thus, $\mathrm{\Delta }(s+1)=1$. Since $\mathrm{\Delta }(2)=1$ and $\mathrm{\Delta }(s+1)=1$, we must have $\mathrm{\Delta }(2s+1)=0$ and $\mathrm{\Delta }(1)=0$ or else $(s+1,2,2s+1)$ and $(1,2,s+1)$ are red solutions to $2x+sy=2z$. When $2s\le q$, since $\mathrm{\Delta }(2s+1)=0$ and $\mathrm{\Delta }(2s+2)=0$, then $\mathrm{\Delta }(qs+q+2s+1)=1$, otherwise $(2s+1,2s+2,qs+q+2s+1)$ is a blue solution to $2x+qy=2z$. If $\mathrm{\Delta }(qs+q+1)=0$, then $(1,2s+2,qs+q+1)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(qs+q+1)=1$. Now we have $\mathrm{\Delta }(qs+q+1)=1$, $\mathrm{\Delta }(4)=1$ and $\mathrm{\Delta }(qs+q+2s+1)=1$, thus $(qs+q+1,4,qs+q+2s+1)$ is a red solution to $2x+sy=2z$, which is a contradiction. When $2s\ge q+1$, note that $2|(q+1)$, so $\mathrm{\Delta }(q+1)=1$. If $\mathrm{\Delta }(q+1-s)=1$, then $(q+1-s,2,q+1)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(q+1-s)=0$. Since $\mathrm{\Delta }(q+1-s)=0$ and $\mathrm{\Delta }(2s+2)=0$, we must have $\mathrm{\Delta }(qs+2q+1-s)=1$ or else $(q+1-s,2s+2,qs+2q+1-s)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(4)=1$ implies that $\mathrm{\Delta }(2s+4)=0$, otherwise $(4,4,2s+4)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(qs+2q+1)=0$, then $(1,2s+4,qs+2q+1)$ is a blue solution to $2x+qy=qz$, so we may assume that $\mathrm{\Delta }(qs+2q+1)=1$. Now we have $\mathrm{\Delta }(qs+2q+1)=1$, $\mathrm{\Delta }(2)=1$ and $\mathrm{\Delta }(qs+2q+1-s)=1$, so $(qs+2q+1-s,2,qs+2q+1)$ is a red solution to $2x+sy=2z$, which is a contradiction.

4. Case 4: $\mathrm{\Delta }(2)=1$ and $\mathrm{\Delta }(s)=1$.

   Since $\mathrm{\Delta }(2)=\mathrm{\Delta }(s)=1$, we must have $\mathrm{\Delta }(s+2)=0$ and $\mathrm{\Delta }(2s)=0$ or else $(2,2,s+2)$ and $(s,2,2s)$ are red solutions to $2x+sy=2z$. If $\mathrm{\Delta }(qs+s+2)=0$, then $(s+2,2s,qs+s+2)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(qs+s+2)=1$. If $\mathrm{\Delta }(qs+2s+2)=1$, then $(qs+s+2,2,qs+2s+2)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(qs+2s+2)=0$. Since $\mathrm{\Delta }(qs+2s+2)=0$ and $\mathrm{\Delta }(2s)=0$, we must have $\mathrm{\Delta }(2s+2)=1$, otherwise $(2s+2,2s,qs+2s+2)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(2s+2)=1$ and $\mathrm{\Delta }(2)=1$ imply that $\mathrm{\Delta }(4)=0$, otherwise $(2,4,2s+2)$ is a red solution to $2x+sy=2z$. $\mathrm{\Delta }(4)=0$ implies that $\mathrm{\Delta }(2q+4)=1$, otherwise $(4,4,2q+4)$ is a blue solution to $2x+qy=2z$. If $\mathrm{\Delta }(2q+4-s)=1$, then $(2q+4-s,2,2q+4)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(2q+4-s)=0$. If $\mathrm{\Delta }(4q+4-s)=0$, then $(2q+4-s,4,4q+4-s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(4q+4-s)=1$. If $\mathrm{\Delta }(4q+4)=1$, then $(4q+4-s,2,4q+4)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(4q+4)=0$. Since $\mathrm{\Delta }(4q+4)=0$ and $\mathrm{\Delta }(4)=0$, we must have $\mathrm{\Delta }(8)=1$, otherwise $(4,8,4q+4)$ is a blue solution to $2x+qy=2z$. $\mathrm{\Delta }(8)=1$ and $\mathrm{\Delta }(s)=1$ imply that $\mathrm{\Delta }(5s)=0$ or else $(s,8,5s)$ is a red solution to $2x+sy=2z$. Since $\mathrm{\Delta }(4)=0$ and $\mathrm{\Delta }(5s)=0$, we must have $\mathrm{\Delta }(2q+5s)=1$ or else $(5s,4,2q+5s)$ is a blue solution to $2x+qy=2z$. Note that $$min\{qs+q+2s+1,qs+2q+1\}\ge 2q+5s$$ since $s\ge 3$ and $q\ge 5$. $\mathrm{\Delta }(2q+5s)=1$ and $\mathrm{\Delta }(2)=1$ imply that $\mathrm{\Delta }(2q+4s)=0$, otherwise $(2q+4s,2,2q+5s)$ is a red solution to $2x+sy=2z$. If $\mathrm{\Delta }(4s)=0$, then $(4s,4,2q+4s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(4s)=1$. If $\mathrm{\Delta }(3s)=1$, then $(3s,2,4s)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(3s)=0$. If $\mathrm{\Delta }(2q+3s)=0$, then $(3s,4,2q+3s)$ is a blue solution to $2x+qy=2z$, so we may assume that $\mathrm{\Delta }(2q+3s)=1$. If $\mathrm{\Delta }(2q+2s)=1$, then $(2q+2s,2,2q+3s)$ is a red solution to $2x+sy=2z$, so we may assume that $\mathrm{\Delta }(2q+2s)=0$. Now we have $\mathrm{\Delta }(2s)=0$, $\mathrm{\Delta }(4)=0$ and $\mathrm{\Delta }(2q+2s)=0$, so $(2s,4,2q+2s)$ is a blue solution to $2x+qy=2z$, which is a contradiction and the proof is complete.