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Let \(\varepsilon_{0}\), \(\varepsilon_{1}\) be two linear homogenous equations, each with at least three variables and coefficients not all the same sign. Define the \(2\)-color off-diagonal Rado number \(R_2(\varepsilon_{0},\varepsilon_{1})\) to be the smallest \(N\) such that for any 2-coloring of \([1,N]\), it must admit a monochromatic solution to \(\varepsilon_{0}\) of the first color or a monochromatic solution to \(\varepsilon_{1}\) of the second color. Mayers and Robertson gave the exact \(2\)-color off-diagonal Rado numbers \(R_2(x+qy=z,x+sy=z). \) Xia and Yao established the formulas for \(R_2(3x+3y=z,3x+qy=z) \) and \(R_2(2x+3y=z,2x+2qy=z) \). In this paper, we determine the exact numbers \(R_2(2x+qy=2z,2x+sy=2z)\), where \(q, s\) are odd integers with \(q>s\geq1\).
Let \([a, b]\) denote the set \(\{x\in \mathbb{Z}|a \leq x \leq b\}\). A function \(\Delta\): \([1, n] \rightarrow[0, k-1]\) is called a \(k\)-coloring of the set \([1, n]\). If \(\varepsilon\) is a system of equations in \(m\) variables, then we say that a solution \({x_{1}, x_{2}, \ldots , x_{m}}\) to \(\varepsilon\) is monochromatic if and only if \[\Delta(x_{1}) =\Delta(x_{2})=\cdots =\Delta(x_{m}).\]
In 1916, Schur [1] proved that for every integer \(k\geq 2\), 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 \(\varepsilon\), 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 \(\varepsilon\), is called \(k\)-color Rado number. If such an integer \(n\) does not exist, then the \(k\)-color Rado number for the system \(\varepsilon\) 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 \(k\geq 2\) and equations \(\varepsilon_{i}\), where \(i=0,1,\ldots, k-1\), 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 \(\varepsilon_{i}\) for some \(i\in[0,k-1]\). Note that if \(\varepsilon_i=\varepsilon_j\) for all \(i\), \(j\) \((1\leq i,j\leq k)\), 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 \[R_2(x+qy=z,x+y=z)=2q+2\left\lfloor \frac{q+1}{2}\right\rfloor+1\] and for \(s\geq 2\), \[R_2(x+qy=z,x+sy=z)=qs+q+2s+1.\] Motivated by Myers and Robertson’s work, Yao and Xia [15] proved the following results: \[R_2(3x+3y=z,3x+3qy=z)=54q+57, \qquad q\geq 2\] and \[R_2(2x+3y=z,2x+2qy=z)=20q+26, \qquad q\geq 2.\]
In this paper, we are also interested in precise values of \(2\)-color off-diagonal Rado numbers and let \(R_2(\varepsilon_{0}, \varepsilon_{1})\) 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 \(q>s\geq1\). The main results of this paper can be stated as follows.
Theorem 1. For odd integers \(q,s\) and \(q>s>2\), we have \[R_2(2x+qy=2z,2x+sy=2z)= \min\{qs+q+2s+1,qs+2q+1\}.\]
Theorem 2. For odd integer \(q \geq 3\), we have \[R_2(2x+qy=2z,2x+y=2z)= 2q+4.\]
Lemma 1. For odd integers \(q,s\) and \(q>s>2\), we have \[R_2(2x+qy=2z,2x+sy=2z)\geq \min\{qs+q+2s+1,qs+2q+1\}.\]
Proof. Let \(N=\min\{qs+q+2s+1,qs+2q+1\}\), consider the 2-coloring of \([1,N-1]\) defined by coloring \[\begin{aligned} R=\{2i|i=1,2,\ldots, s\}\cup\{j|qs+q+1\leq j\leq N-1 \ {\rm and} \ j\ {\rm is \ an \ odd \ integer}\} \end{aligned}\] red and its complement blue. Now, we show that there is no suitable red solution to \(2x+sy=2z\) and no suitable blue solution to \(2x+qy=2z\). Assume that \((x_{0},y_{0},z_{0})\) is a red solution to \(2x+sy=2z\). Note that \[qs+q+1>s(s+2)\] since \(q>s\). If \(z_{0}<qs+q+1\), then there exist integers \(a\) and \(b\) such that \(x_0=2a\), \(z_0=2b\) and \(1\leq a <b< s\). Thus, \(sy_0=4(b-a)\). The fact that \(s\) is an odd integer implies that \(4|y_0\) and \(y_0\geq 4\). However, \[z_0=\frac{2x_0+sy_0}{2}\geq \frac{4+4s}{2}=2s+2,\] which is a contradiction and therefore \(z_{0}\geq qs+q+1\). If \(y_0\geq qs+q+1\), then \[z_0=\frac{2x_0+sy_0}{2} \geq \frac{4+s(qs+q+1)}{2}> N-1,\] which is a contradiction and hence, \(y_{0}\in\{2i|i=1,2,\ldots, s\}\). If \(x_0\in\{2i|i=1,2,\ldots, s\}\), we see that \[z_0=\frac{2x_0+sy_0}{2}\leq 2s+s^2<qs+q+1,\] which is a contradiction and thus, \(x_{0}\geq qs+q+1\). Note that \(z_{0}\) and \(x_{0}\) are all odd, therefore, \(4|(2z_{0}-2x_{0})\), i.e., \(4|sy_{0}\). Recall that \(s\) is odd, therefore, \(4|y_{0}\) and \(y_{0}\geq 4\). However, \[z_{0}=\frac{2x_{0}+sy_{0}}{2}\geq\frac{2(qs+q+1)+4s}{2}=qs+q+2s+1>N-1,\] which is a contradiction, and there is no suitable red solution to \(2x+sy=2z\).
Assume that \((x_{0},y_{0},z_{0})\) is a blue solution to \(2x+qy=2z\). Note that \(y_{0}\) must be even this is because \(q\) is an odd integer. Therefore \(y_{0}\geq 2s+2\) and \[z_0=\frac{2x_0+qy_0}{2}\geq \frac{2+q(2s+2)}{2} =qs+q+1,\] which implies that \(z_{0}\) must be even. If \(y_{0}=2s+2\), then \(2x_{0}=2z_{0}-q(2s+2)\), that is, \(x_{0}=z_{0}-q(s+1)\). Therefore, \(2|x_{0}\) and \(x_{0}\geq2s+2\). However, \[z_{0}=\frac{2x_{0}+qy_{0}}{2}\geq\frac{2(2s+2)+q(2s+2)}{2}=qs+q+2s+2>N-1,\] which is a contradiction. If \(y_{0}\geq 2s+4\), then \[z_{0}=\frac{2x_{0}+qy_{0}}{2}\geq\frac{2\times 1+q(2s+4)}{2}=qs+2q+1>N-1,\] which is a contradiction. Thus, there is no suitable blue solution to \(2x+qy=2z\). The proof is complete. ◻
Lemma 2. For odd integer \(q>2\), we have \[\begin{aligned} \label{I-5} R_2(2x+qy=2z,2x+y=2z)\geq 2q+4. \end{aligned}\]
Proof. Let \(\{1\}\bigcup[3,2q]\bigcup\{2q+2\}\) be colored blue and \(\{2,2q+1,2q+3\}\) be colored red. It is easy to verify that there is no suitable blue solutions to \(2x+qy=2z\) and no suitable red solution to \(2x+y=2z\). ◻
It follows from Lemma 1 that \[\begin{aligned} \label{In-1} R_2(2x+qy=2z,2x+sy=2z)\geq \min\{qs+q+2s+1,qs+2q+1\}, \end{aligned}\] where \(q, s\) are odd integers and \(q>s\geq 3\). In order to prove this theorem, it suffices to show that \[R_2(2x+qy=2z,2x+sy=2z)\leq \min\{qs+q+2s+1,qs+2q+1\},\] where \(q, s\) are odd integers and \(q>s\geq 3\). Let \(\Delta\) 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;
\(\Delta(2)=0\) and \(\Delta(s)=0\) imply that \(\Delta(q+s)=1\), otherwise \((s,2,q+s)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(q+s)=1\), we must have \(\Delta(q+s+\frac{(q+s)s}{2})=0\) or else \((q+s,q+s,q+s+\frac{(q+s)s}{2})\) is a red solution to \(2x+sy=2z\). Now, \(\Delta(q+s+\frac{(q+s)s}{2})=0\) and \(\Delta(2)=0\), so \(\Delta(2q+s+\frac{(q+s)s}{2})=1\), 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 \[2q+s+\frac{(q+s)s}{2}\leq \min\{qs+q+2s+1,qs+2q+1\}\] since \(q-s\geq 2\). \(\Delta(2q+s+\frac{(q+s)s}{2})=1\) and \(\Delta(q+s)=1\) imply that \(\Delta(2q+s)=0\), otherwise \((2q+s,s+q,2q+s+\frac{(q+s)s}{2})\) is a red solution to \(2x+sy=2z\). Since \(\Delta(2q+s)=0\) and \(\Delta(s)=0\), we must have \(\Delta(4)=1\) or else \((s,4,2q+s)\) is a blue solution to \(2x+qy=2z\). If \(\Delta(q+3s)=1\), then \((q+s,4,q+3s)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(q+3s)=0\). If \(\Delta(3s)=0\), then \((3s,2,q+3s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(3s)=1\). Now, \(\Delta(3s)=1\) and \(\Delta(4)=1\), we must have \(\Delta(5s)=0\) or else \((3s,4,5s)\) is a red solution to \(2x+sy=2z\). \(\Delta(5s)=0\) and \(\Delta(2)=0\), imply that \(\Delta(q+5s)=1\), otherwise \((5s,2,q+5s)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(q+5s)=1\) and \(\Delta(q+s)=1\), we must have \(\Delta(8)=0\) or else \((q+s,8,q+5s)\) is a red solution to \(2x+sy=2z\). If \(\Delta(4q+s)=0\), then \((s,8,4q+s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(4q+s)=1\). If \(\Delta(4q-s)=1\), then \((4q-s,4,4q+s)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(4q-s)=0\). Since \(\Delta(4q-s)=0\) and \(\Delta(2)=0\), we must have \(\Delta(3q-s)=1\) or else \((3q-s,2,4q-s)\) is a blue solution to \(2x+qy=2z\). \(\Delta(3q-s)=1\) and \(\Delta(4)=1\) imply that \(\Delta(3q+s)=0\), otherwise \((3q-s, 4,3q+s)\) is a red solution to \(2x+sy=2z\). Now we have \(\Delta(3q+s)=0\), \(\Delta(2)=0\) and \(\Delta(2q+s)=0\), and then \((2q+s,2,3q+s)\) is a blue solution to \(2x+qy=2z\), which is a contradiction.
If \(\Delta(2)=\Delta(4)=\cdots=\Delta(2q+2)=0\) and then \((2,4,2q+2)\) is a blue solution to \(2x+qy=2z\), which is a contradiction. So we can assume that \(k\) \((1\leq k<q+1)\) is the least number such that \(\Delta(2)=\cdots=\Delta(2k)=0\) and \(\Delta(2k+2)=1\).
\(\Delta(2k+2)=1\) and \(\Delta(s)=1\) implies that \(\Delta(ks+2s)=0\), otherwise \((s,2k+2,ks+2s)\) is a red solution to \(2x+sy=2z\). Since \(\Delta(ks+2s)=0\) and \(\Delta(2)=0\), we must have \(\Delta(q+ks+2s)=1\) or else \((ks+2s,2,q+ks+2s)\) is a blue solution to \(2x+qy=2z\). \(\Delta(q+ks+2s)=1\) and \(\Delta(2k+2)=1\) imply that \(\Delta(q+s)=0\), otherwise \((q+s,2k+2,q+ks+2s)\) is a red solution to \(2x+sy=2z\). If \(\Delta(2q+s)=0\), then \((q+s,2,2q+s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(2q+s)=1\). If \(\Delta(2q+ks+2s)=1\) then \((2q+s,2k+2,2q+ks+2s)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(2q+ks+2s)=0\). Note that \[2q+ks+2s\leq \min\{qs+q+2s+1,qs+2q+1\}\] since \(k\leq \min\{\frac{(s-1)q+1}{s},q-2\}\). Since \(\Delta(2q+ks+2s)=0\) and \(\Delta(ks+2s)=0\), we must have \(\Delta(4)=1\), otherwise \((ks+2s,4,ks+2s+2q)\) is a blue solution to \(2x+qy=2z\). \(\Delta(4)=1\) and \(\Delta(2q+s)=1\) imply that \(\Delta(2q-s)=0\) or else \((2q-s,4,2q+s)\) is a red solution to \(2x+sy=2z\). If \(\Delta(3q-s)=0\), then \((2q-s,2,3q-s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(3q-s)=1\). Since \(\Delta(3q-s)=1\) and \(\Delta(4)=1\), we must have \(\Delta(3q+s)=0\) or else \((3q-s,4,3q+s)\) is a red solution to \(2x+sy=2z\). If \(\Delta(4q+s)=0\), then \((3q+s,2,4q+s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(4q+s)=1\). If \(\Delta(4q-s)=1\), then \((4q-s,4,4q+s)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(4q-s)=0\). \(\Delta(s)=1\) and \(\Delta(4)=1\) imply that \(\Delta(3s)=0\) or else \((s,4,3s)\) is a red solution to \(2x+sy=2z\). Since \(\Delta(3s)=0\) and \(\Delta(2)=0\), we must have \(\Delta(q+3s)=1\) or else \((3s,2,q+3s)\) is a blue solution to \(2x+qy=2z\). \(\Delta(4)=1\) and \(\Delta(q+3s)=1\) imply that \(\Delta(q+5s)=0\), otherwise \((q+3s,4,q+5s)\) is a red solution to \(2x+sy=2z\). \(\Delta(q+5s)=0\) and \(\Delta(2)=0\) imply that \(\Delta(5s)=1\), otherwise \((5s,2,q+5s)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(5s)=1\) and \(\Delta(s)=1\), we must have \(\Delta(8)=0\) or else \((s,8,5s)\) is a red solution to \(2x+sy=2z\). When \(s=3\), by the above proof, we have \(\Delta(4q+s)=\Delta(4q+3)=1\) and \(\Delta(q+3s)=\Delta(q+9)=1\), then \(\Delta(2q-4)=0\), otherwise \((q+9,2q-4,4q+3)\) is a red solution to \(2x+3y=2z\). \(\Delta(2)=0\) implies that \(\Delta(q+2)=1\) or else \((2,2,q+2)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(q+2)=1\) and \(\Delta(4)=1\), we must have \(\Delta(q-4)=0\), otherwise \((q-4,4,q+2)\) is a red solution to \(2x+3y=2z\). Now we have \(\Delta(2q-4)=0\), \(\Delta(q-4)=0\) and \(\Delta(2)=0\), and then \((q-4,2,2q-4)\) is a blue solution to \(2x+qy=2z\), which is a contradiction. When \(s\geq 5\), since \(\Delta(8)=0\) and \(\Delta(q+s)=0\), we must have \(\Delta(5q+s)=1\), otherwise \((q+s,8,5q+s)\) is a blue solution to \(2x+qy=2z\). Note that \[5q+s\leq \min\{qs+q+2s+1,qs+2q+1\},\] since \(s\geq 5\). \(\Delta(5q+s)=1\) and \(\Delta(4)=1\) imply that \(\Delta(5q-s)=0\) or else \((5q-s,4,5q+s)\) is a red solution to \(2x+sy=2z\). Now we have \(\Delta(5q-s)=0\), \(\Delta(2)=0\) and \(\Delta(4q-s)=0\), and then \((4q-s,2,5q-s)\) is a blue solution to \(2z+qy=2z\), which is a contradiction.
It is easy to verify that \[q+1\leq 2 \min\left\{\frac{(s-1)q+1}{s}, q-2\right\}\] and \[s+1<2s\leq 2\min\left\{ \frac{(s-1)q+1}{s}, q-2\right\}.\] Thus, \(q+1\leq 2k\) and \(s+1<2s\leq 2k\), and we have \(\Delta(s+1)=\Delta(2s)=\Delta(q+1)=0\). Obviously, \(k\geq2\) and \(\Delta(2)=\Delta(4)=0\), then \(\Delta(2q+2)=1\), otherwise \((2,4, 2q+2)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(q+1)=0\) and \(\Delta(2)=0\), we must have \(\Delta(1)=1\) or else \((1,2,q+1)\) is a blue solution to \(2x+qy=2z\). \(\Delta(1)=1\) and \(\Delta(2q+2)=1\) imply that \(\Delta(qs+s+1)=0\), otherwise \((1,2q+2,qs+s+1)\) is a red solution to \(2x+sy=2z\). Now we have \(\Delta(qs+s+1)=0\), \(\Delta(2s)=0\) and \(\Delta(s+1)=0\), and then \((s+1,2s,qs+s+1)\) is a blue solution to \(2x+qy=2z\), which is a contradiction.
If \(\Delta(2)=\Delta(4)=\cdots=\Delta(2s+2)=1\), then \((2,4,2s+2)\) is a red solution to \(2x+sy=2z\), which is a contradiction. So we can assume that \(l\) (\(1\leq l<s+1\)) is the least integer such that \(\Delta(2)=\cdots =\Delta(2l)=1\) and \(\Delta(2l+2)=0\).
\(\Delta(2)=1\) implies that \(\Delta(s+2)=0\), otherwise \((2,2,s+2)\) is a red solution to \(2x+sy=2z\). If \(\Delta((l+1)q+s+2)=0\), then \((s+2,2l+2,(l+1)q+s+2)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta((l+1)q+s+2)=1\). If \(\Delta((l+1)q+2s+2)=1\), 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 \(\Delta((l+1)q+2s+2)=0\). Note that \[(l+1)q+2s+2\leq \min\{qs+q+2s+1,qs+2q+1\}\] since \(l\leq s-1\). \(\Delta((l+1)q+2s+2)=0\) and \(\Delta(2l+2)=0\) imply that \(\Delta(2s+2)=1\), otherwise \((2s+2,2l+2,(l+1)q+2s+2)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(2s+2)=1\) and \(\Delta(2)=1\), we must have \(\Delta(4)=0\) or else \((2,4,2s+2)\) is a red solution to \(2x+sy=2z\). \(\Delta(s)=0\) and \(\Delta(4)=0\) imply that \(\Delta(2q+s)=1\), otherwise \((s,4,2q+s)\) is a blue solution to \(2x+qy=2z\). If \(\Delta(2q+2s)=1\), then \((2q+s,2,2q+2s)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(2q+2s)=0\). If \(\Delta(2s)=0\), then \((2s,4,2q+2s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(2s)=1\). Now, \(\Delta(2s)=1\) and \(\Delta(2)=1\), we must have \(\Delta(3s)=0\) or else \((2s,2,3s)\) is a red solution to \(2x+sy=2z\). If \(\Delta(3s+2q)=0\), then \((3s,4,3s+2q)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(3s+2q)=1\). Since \(\Delta(3s+2q)=1\) and \(\Delta(2)=1\), we must have \(\Delta(2q+4s)=0\) or else \((3s+2q,2,4s+2q)\) is a red solution to \(2x+sy=2z\). Since \(\Delta(2q+4s)=0\) and \(\Delta(4)=0\), we must have \(\Delta(4s)=1\), otherwise \((4s,4,2q+4s)\) is a blue solution to \(2x+qy=2z\). \(\Delta(4s)=1\) and \(\Delta(2)=1\) imply that \(\Delta(5s)=0\), otherwise \((4s,2,5s)\) is a red solution to \(2x+sy=2z\). If \(\Delta(5s+2q)=0\), then \((5s,4,5s+2q)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(5s+2q)=1\). When \(q=5\), then \(s=3\), by the above proof, we have \(\Delta(2s+2)=\Delta(8)=1\) and \(\Delta(2)=1\), then \(\Delta(14)=0\), otherwise \((2,8,14)\) is a red solution to \(2x+3y=2z\). Now we have \(\Delta(14)=0\) and \(\Delta(4)=0\), so \((4,4,14)\) is a blue solution to \(2x+5y=2z\), which is a contradiction. When \(q>5\) and \(s=3\), \(\Delta(4)=0\) implies that \(\Delta(2q+4)=1\), otherwise \((4,4,2q+4)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(2q+4)=1\) and \(\Delta(2)=1\), we must have \(\Delta(2q+1)=0\) or else \((2q+1,2,2q+4)\) is a red solution to \(2x+3y=2z\). If \(\Delta(4q+1)=0\), then \((2q+1,4,4q+1)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(4q+1)=1\). If \(\Delta(4q+4)=1\), then \((4q+1,2,4q+4)\) is a red solution to \(2x+3y=2z\), so we may assume that \(\Delta(4q+4)=0\). Now we have \(\Delta(8)=0\), \(\Delta(4)=0\) and \(\Delta(4q+4)=0\), so \((4,8,4q+4)\) is a blue solution to \(2x+qy=2z\), which is a contradiction. When \(q>s\geq 5\), we see that \[\max\{2q+6s,6q+s\}\leq \min\{qs+q+2s+1,qs+2q+1\}.\] Since \(\Delta(5s+2q)=1\) and \(\Delta(2)=1\), then \(\Delta(2q+6s)=0\) or else \((2q+5s,2,2q+6s)\) is a red solution to \(2x+sy=2z\). If \(\Delta(6s)=1\), then \((6s,4,2q+6s)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(6s)=1\) and \(\Delta(2s)=1\), we must have \(\Delta(8)=0\), otherwise \((2s,8,6s)\) is a red solution to \(2x+sy=2z\). \(\Delta(2q+s)=1\) and \(\Delta(2)=1\) imply that \(\Delta(2q)=0\) or else \((2q,2,2q+s)\) is a red solution to \(2x+sy=2z\). Since \(\Delta(2q)=0\) and \(\Delta(4)=0\), then \(\Delta(4q)=1\), otherwise \((2q,4,4q)\) is a blue solution to \(2x+qy=2z\). If \(\Delta(4q+s)=1\), then \((4q,2,4q+s)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(4q+s)=0\). If \(\Delta(6q+s)=0\), then \((4q+s,4,6q+s)\) is a blue solution to \(2x+qy=2z\). Since \(\Delta(6q+s)=1\) and \(\Delta(2)=1\), we must have \(\Delta(6q)=0\), otherwise \((6q,2,6q+s)\) is a red solution to \(2x+sy=2z\). Now we have \(\Delta(2q)=0\), \(\Delta(8)=0\) and \(\Delta(6q)=0\), so \((2q,8,6q)\) is a blue solution to \(2x+qy=2z\), which is a contradiction.
We have \(\Delta(2)=\Delta(4)=\cdots =\Delta(2s)=1\) and \(\Delta(2s+2)=0\). Note that \(2|(s+1)\) and \(s+1<2s\), thus, \(\Delta(s+1)=1\). Since \(\Delta(2)=1\) and \(\Delta(s+1)=1\), we must have \(\Delta(2s+1)=0\) and \(\Delta(1)=0\) or else \((s+1,2,2s+1)\) and \((1,2,s+1)\) are red solutions to \(2x+sy=2z\). When \(2s\leq q\), since \(\Delta(2s+1)=0\) and \(\Delta(2s+2)=0\), then \(\Delta(qs+q+2s+1)=1\), otherwise \((2s+1,2s+2,qs+q+2s+1)\) is a blue solution to \(2x+qy=2z\). If \(\Delta(qs+q+1)=0\), then \((1,2s+2,qs+q+1)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(qs+q+1)=1\). Now we have \(\Delta(qs+q+1)=1\), \(\Delta(4)=1\) and \(\Delta(qs+q+2s+1)=1\), thus \((qs+q+1,4,qs+q+2s+1)\) is a red solution to \(2x+sy=2z\), which is a contradiction. When \(2s\geq q+1\), note that \(2|(q+1)\), so \(\Delta(q+1)=1\). If \(\Delta(q+1-s)=1\), then \((q+1-s,2,q+1)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(q+1-s)=0\). Since \(\Delta(q+1-s)=0\) and \(\Delta(2s+2)=0\), we must have \(\Delta(qs+2q+1-s)=1\) or else \((q+1-s,2s+2,qs+2q+1-s)\) is a blue solution to \(2x+qy=2z\). \(\Delta(4)=1\) implies that \(\Delta(2s+4)=0\), otherwise \((4,4,2s+4)\) is a red solution to \(2x+sy=2z\). If \(\Delta(qs+2q+1)=0\), then \((1,2s+4,qs+2q+1)\) is a blue solution to \(2x+qy=qz\), so we may assume that \(\Delta(qs+2q+1)=1\). Now we have \(\Delta(qs+2q+1)=1\), \(\Delta(2)=1\) and \(\Delta(qs+2q+1-s)=1\), so \((qs+2q+1-s,2,qs+2q+1)\) is a red solution to \(2x+sy=2z\), which is a contradiction.
Since \(\Delta(2)=\Delta(s)=1\), we must have \(\Delta(s+2)=0\) and \(\Delta(2s)=0\) or else \((2,2,s+2)\) and \((s,2,2s)\) are red solutions to \(2x+sy=2z\). If \(\Delta(qs+s+2)=0\), then \((s+2,2s, qs+s+2)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(qs+s+2)=1\). If \(\Delta(qs+2s+2)=1\), then \((qs+s+2,2,qs+2s+2)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(qs+2s+2)=0\). Since \(\Delta(qs+2s+2)=0\) and \(\Delta(2s)=0\), we must have \(\Delta(2s+2)=1\), otherwise \((2s+2,2s,qs+2s+2)\) is a blue solution to \(2x+qy=2z\). \(\Delta(2s+2)=1\) and \(\Delta(2)=1\) imply that \(\Delta(4)=0\), otherwise \((2,4,2s+2)\) is a red solution to \(2x+sy=2z\). \(\Delta(4)=0\) implies that \(\Delta(2q+4)=1\), otherwise \((4,4,2q+4)\) is a blue solution to \(2x+qy=2z\). If \(\Delta(2q+4-s)=1\), then \((2q+4-s,2,2q+4)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(2q+4-s)=0\). If \(\Delta(4q+4-s)=0\), then \((2q+4-s,4,4q+4-s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(4q+4-s)=1\). If \(\Delta(4q+4)=1\), then \((4q+4-s,2,4q+4)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(4q+4)=0\). Since \(\Delta(4q+4)=0\) and \(\Delta(4)=0\), we must have \(\Delta(8)=1\), otherwise \((4,8,4q+4)\) is a blue solution to \(2x+qy=2z\). \(\Delta(8)=1\) and \(\Delta(s)=1\) imply that \(\Delta(5s)=0\) or else \((s,8,5s)\) is a red solution to \(2x+sy=2z\). Since \(\Delta(4)=0\) and \(\Delta(5s)=0\), we must have \(\Delta(2q+5s)=1\) or else \((5s,4,2q+5s)\) is a blue solution to \(2x+qy=2z\). Note that \[\min\{qs+q+2s+1,qs+2q+1\}\geq 2q+5s\] since \(s\geq 3\) and \(q\geq 5\). \(\Delta(2q+5s)=1\) and \(\Delta(2)=1\) imply that \(\Delta(2q+4s)=0\), otherwise \((2q+4s,2,2q+5s)\) is a red solution to \(2x+sy=2z\). If \(\Delta(4s)=0\), then \((4s,4,2q+4s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(4s)=1\). If \(\Delta(3s)=1\), then \((3s,2,4s)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(3s)=0\). If \(\Delta(2q+3s)=0\), then \((3s,4,2q+3s)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(2q+3s)=1\). If \(\Delta(2q+2s)=1\), then \((2q+2s,2,2q+3s)\) is a red solution to \(2x+sy=2z\), so we may assume that \(\Delta(2q+2s)=0\). Now we have \(\Delta(2s)=0\), \(\Delta(4)=0\) and \(\Delta(2q+2s)=0\), so \((2s,4,2q+2s)\) is a blue solution to \(2x+qy=2z\), which is a contradiction and the proof is complete.
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It follows from Lemma 2 that when \(q \geq 3\) is an odd integer, \[R_2(2x+qy=2z,2x+y=2z)\geq 2q+4.\] In order to prove this theorem, it suffices to show that when \(q \geq 3\) is an odd integer, \[R_2(2x+qy=2z,2x+y=2z)\leq 2q+4 .\] Let \(\Delta\) be a 2-coloring of \([1,2q+4]\) 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 \(2x+y=2z\). We break our proof into 3 cases.
Since \(\Delta(2)=1\), then the solution \((2,4,4)\) to \(2x+y=2z\) forces \(\Delta(4)=0\) and the solution \((2,2,3)\) to \(2x+y=2z\) forces \(\Delta(3)=0\). Since \(\Delta(4)=0\), then the solution \((4,4,2q+4)\) to \(2x+qy=2z\) forces \(\Delta(2q+4)=1\). Since \(\Delta(2)=1\) and \(\Delta(2q+4)=1\), then the solution \((2q+3,2,2q+4)\) to \(2x+y=2z\) forces \(\Delta(2q+3)=0\). Now we have \(\Delta(3)=\Delta(4)=\Delta(2q+3)=0\), so \((3,4,2q+3)\) is a blue solution to \(2x+qy=2z\), which is a contradiction.
\(\Delta(4)=1\) implies that \(\Delta(6)=0\) or else \((4,4,6)\) is a red solution to \(2x+y=2z\). Since \(\Delta(6)=0\) and \(\Delta(2)=0\), we must have \(\Delta(q+6)=1\), otherwise \((6,2,q+6)\) is a blue solution to \(2x+qy=2z\). If \(\Delta(2q+4)=1\), then \((4,2q+4,q+6)\) is a red solution to \(2x+y=2z\), so we may assume that \(\Delta(2q+4)=0\). If \(\Delta(q+4)=0\), then \((q+4,2,2q+4)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(q+4)=1\). If \(\Delta(q+2)=1\), then \((q+2,4,q+4)\) is a red solution to \(2x+y=2z\), so we have \(\Delta(q+2)=0\). Now we have \(\Delta(2)=0\) and \(\Delta(q+2)=0\), so \((2,2,q+2)\) is a blue solution to \(2x+qy=2z\), which is a contradiction.
\(\Delta(2)=0\) implies that \(\Delta(q+2)=1\), otherwise \((2,2,q+2)\) is a blue solution to \(2x+qy=2z\). If \(\Delta(2q+2)=0\), then \((2,4,2q+2)\) is a blue solution to \(2x+qy=2z\), so we may assume that \(\Delta(2q+2)=1\). Since \(\Delta(q+2)=1\) and \(\Delta(2q+2)=1\), we must have \(\Delta(1)=0\), otherwise \((1,2q+2,q+2)\) is a red solution to \(2x+y=2z\). \(\Delta(1)=0\) and \(\Delta(2)=0\) imply that \(\Delta(q+1)=1\), otherwise \((1,2,q+1)\) is a blue solution to \(2x+qy=2z\). Now we have \(\Delta(2q+2)=1\) and \(\Delta(q+1)=1\), so \((q+1,2q+2,2q+2)\) is a red solution to \(2x+y=2z\), which is a contradiction and this completes the proof.
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We thank the anonymous referees for their helpful comments.