Two Color Off-diagonal Rado-type Numbers

We show that for any two linear homogeneous equations E0, E1, each with at least three variables and coefficients not all the same sign, any 2-coloring of Z+ admits monochromatic solutions of color 0 to E0 or monochromatic solutions of color 1 to E1. We define the 2-color off-diagonal Rado number RR(E0, E1) to be the smallest N such that [1, N ] must admit such solutions. We determine a lower bound for RR(E0, E1) in certain cases when each Ei is of the form a1x1 + . . .+anxn = z as well as find the exact value of RR(E0, E1) when each is of the form x1 + a2x2 + . . . + anxn = z. We then present a Maple package that determines upper bounds for off-diagonal Rado numbers of a few particular types, and use it to quickly prove two previous results for diagonal Rado numbers. 0 Introduction For r ≥ 2, an r-coloring of the positive integers Z is an assignment χ : Z → {0, 1, . . . , r− 1}. Given a diophantine equation E in the variables x1, . . . , xn, we say a solution {x̄i} n i=1 is monochromatic if χ(x̄i) = χ(x̄j) for every i, j pair. A well-known theorem of Rado states that, for any r ≥ 2, a linear homogeneous equation c1x1 + . . .+ cnxn = 0 with each ci ∈ Z admits a monochromatic solution in Z + under any r-coloring of Z if and only if some nonempty subset of {ci} n i=1 sums to zero. The smallest N such that any r-coloring of {1, 2, . . . , N} = [1, N ] satisfies this condition is called the r-color Rado number for the equation E . However, Rado also proved the following, much lesser known, result. Theorem 0.1 (Rado [6]) Let E be a linear homogeneous equation with integer coefficients. Assume that E has at least 3 variables with both positive and negative coefficients. Then any 2-coloring of Z admits a monochromatic solution to E . ∗This work was done as part of a summer REU, funded by Colgate University, while the first author was an undergraduate at Colgate University, under the directorship of the second author. the electronic journal of combinatorics 13 (2007), #R53 1 Remark. Theorem 0.1 cannot be extended to more than 2 colors, without restriction on the equation. For example, Fox and Radoičić [2] have shown, in particular, that there exists a 3-coloring of Z that admits no monochromatic solution to x + 2y = 4z. For more information about equations that have finite colorings of Z with no monochromatic solution see [1] and [2]. In [4], the 2-color Rado numbers are determined for equations of the form a1x1 + . . .+ anxn = z where one of the ai’s is 1. The case when min(a1, . . . , an) = 2 is done in [5], while the general case is settled in [3]. In this article, we investigate the “off-diagonal” situation. To this end, for r ∈ Z define an off-diagonal Rado number for the equations Ei, 0 ≤ i ≤ r − 1, to be the least integer N (if it exists) for which any r-coloring of [1, N ] must admit a monochromatic solution to Ei of color i for some i ∈ [0, r− 1]. In this paper, when r = 2 we will prove the existence of such numbers and determine particular values and lower bounds in several specific cases when the two equations are of the form a1x1 + . . . + anxn = z. 1 Existence The authors were unable to find an English translation of the proof of Theorem 0.1. For the sake of completeness, we offer a simplified version of Rado’s original proof. Proof of Theorem 0.1 (due to Rado [6]) Let ∑k i=1 αixi = ∑` i=1 βiyi be our equation, where k ≥ 2, ` ≥ 1, αi ∈ Z + for 1 ≤ i ≤ k, and βi ∈ Z + for 1 ≤ i ≤ `. By setting x = x1 = x2 = · · · = xk−1, y = xk, and z = y1 = y2 = · · · = y`, we may consider solutions to ax + by = cz, where a = ∑k−1 i=1 αi, b = ck, and c = ∑` i=1 βi. We will denote ax + by = cz by E . Let m = lcm ( a gcd(a,b) , c gcd(b,c) ) . Let (x0, y0, z0) be the solution to E with max(x, y, z) a minimum, where the maximum is taken over all solutions of positive integers to E . Let A = max(x0, y0, z0). Assume, for a contradiction, that there exists a 2-coloring of Z with no monochromatic solution to E . First, note that for any n ∈ Z, the set {in : i = 1, 2, . . . , A} cannot be monochromatic, for otherwise x = x0n, y = y0n, and z = z0n is a monochromatic solution, a contradiction. Let x = m so that bx a , bx c ∈ Z. Letting red and blue be our two colors, we may assume, without loss of generality, that x is red. Let y be the smallest number in {im : i = 1, 2, . . . , A} that is blue. Say y = `m so that 2 ≤ ` ≤ A. For some n ∈ Z, we have that z = b a (y−x)n is blue, otherwise {i b a (y−x) : i = 1, 2, . . .} would be red, admitting a monochromatic solution to E . Then w = a c z + b c y must be red, for otherwise az+by = cw and z, y, and w are all blue, a contradiction. Since x and w are both red, we have that q = c a w− b a x = b a (y−x)(n+1) must be blue, for otherwise x, w, and q give a red solution to E . As a consequence, we see that { i b a (y − x) : i = n, n + 1, . . . } the electronic journal of combinatorics 13 (2007), #R53 2 is monochromatic. This gives us that { i b a (y − x)n : i = 1, 2, . . . , A } is monochromatic, a contradiction. Using the above result, we offer an “off-diagonal” consequence. Theorem 1.1 Let E0 and E1 be linear homogeneous equations with integer coefficients. Assume that E0 and E1 each have at least 3 variables with both positive and negative coefficients. Then any 2-coloring of Z admits either a solution to E0 of the first color or a solution to E1 of the second color. Proof. Let a0, a1, b0, b1, c ∈ Z + and denote by Gi the equation aix + biy = cz for i = 0, 1. Via the same argument given in the proof to Theorem 0.1, we may consider solutions to G0 and G1. (The coefficients of z may be taken to be the same in both equations by finding the lcm of the original coefficients of z and adjusting the other coefficients accordingly.) Let the colors be red and blue. We want to show that any 2-coloring admits either a red solution to G0 or a blue solution to G1. From Theorem 0.1, we have monochromatic solutions to each of these equations. Hence, we assume, for a contradiction, that any monochromatic solution to G0 is blue and that any monochromatic solution to G1 is red. This gives us that for any i ∈ Z, if ci is blue, then (a1 + b1)i is red (else we have a blue solution to G1). Now consider monochromatic solutions in cZ. Via the obvious bijection between colorings of cZ and Z and the fact that linear homogeneous equations are unaffected by dilation, Theorem 0.1 gives us the existence of monochromatic solutions in cZ. If cx, cy, cz solve G0 and are the same color, then they must be blue. Hence, x̂ = (a1 + b1)x, ŷ = (a1 + b1)y, and ẑ = (a1 + b1)z are all red. But, x̂, ŷ, ẑ solve G0. Thus, we have a red solution to G0, a contradiction. 2 Two Lower Bounds Given the results in the previous section, we make a definition, which uses the following notation. Notation For n ∈ Z and ~a = (a1, a2, . . . , an) ∈ Z , denote by En(~a) the linear homogeneous equation ∑n i=1 aixi = 0. Definition For k, ` ≥ 3,~b ∈ Z, and ~c ∈ Z, we let RR(Ek(~b), E`(~c)) be the minimum integer N , if it exists, such that any 2-coloring of [1, N ] admits either a solution to Ek(~b) of the first color or a solution to E`(~c) of the second color. We now develop a general lower bound for certain types of those numbers guaranteed to exist by Theorem 1.1. Theorem 2.1 For k, ` ≥ 2, let b1, b2, . . . , bk−1, c1, c2, . . . , c`−1 ∈ Z . Consider Ek = Ek(b1, b2, . . . , bk−1,−1) and E` = E`(c1, c2, . . . , c`−1,−1), written so that b1 = min(b1, b2, . . . , bk−1) and c1 = min(c1, c2, . . . , c`−1). Assume that t = b1 = c1. Let q = ∑k−1 i=2 bi and the electronic journal of combinatorics 13 (2007), #R53 3 s = ∑`−1 i=2 ci. Let (without loss of generality) q ≥ s. Then RR(Ek, E`) ≥ t(t + q)(t + s) + s. Proof. Let N = t(t + q)(t + s) + s and consider the 2-coloring of [1, N − 1] defined by coloring [s + t, (q + t)(s + t) − 1] red and its complement blue. We will show that this coloring avoids red solutions to Ek and blue solutions to E`. We first consider any possible red solution to Ek. The value of xk would have to be at least t(s + t) + q(s + t) = (q + t)(s + t). Thus, there is no suitable red solution. Next, we consider E`. If {x1, x2, . . . , x`−1} ⊆ [1, s + t − 1], then x` < (q + t)(s + t). Hence, the smallest possible blue solution to E` has xi ∈ [(q + t)(s + t), N − 1] for some i ∈ [1, `− 1]. However, this gives x` ≥ t(q + t)(s + t) + s > N − 1. Thus, there is no suitable blue solution. The case when k = ` = 2 in Theorem 2.1 can be improved somewhat in certain cases, depending upon the relationship between t, q, and s. This result is presented below. Theorem 2.2 Let t, j ∈ Z. Let F t j represent the equation tx + jy = z. Let q, s ∈ Z + with q ≥ s ≥ t. Define m = gcd(t,q) gcd(t,q,s) . Then RR(F t q,F t s) ≥ t(t + q)(t + s) + ms. Proof. Let N = t(t + q)(t + s) + ms and consider the 2-coloring χ of [1, N − 1] defined by coloring R = [s + t, (q + t)(s + t) − 1] ∪ {t(t + q)(t + s) + is : 1 ≤ i ≤ m − 1} red and B = [1, N − 1] \ R blue. We will show that this coloring avoids red solutions to F t q and blue solutions to F t s. We first consider any possible red solution to F t q. The value of z would have to be at least t(s + t) + q(s + t) = (q + t)(s + t) and congruent to 0 modulo m. Since t(t+ q)(t+ s) ≡ 0 (mod m) but is 6≡ 0 (mod m) for 1 ≤ i ≤ m−1, there is no suitable red solution. Next, we consider F t s. If {x, y} ⊆ [1, s + t − 1], then s + t ≤ z < (q + t)(s + t). Hence, the smallest possible blue solution to F t s has x or y in [(q + t)(s + t), N − 1]. However, this gives z ≥ t(q + t)(s + t) + s > N − 1. By the definition of the coloring, z must be red. Thus, there is no suitable blue solution to F t s. 3 Some Exact Numbers In this section, we will determine some of the values of RR1(q, s) = RR(x+qy = z, x+sy = z), where 1 ≤ s ≤ q. The subscript 1 is present to emphasize the fact that we are using t = 1 as defined in Theorem 2.1. In this section we will let RRt(q, s) = RR(tx + qy = z, tx + sy = z) and we will denote the equation tx + jy = z by F t j . the electronic journal of combinatorics 13 (2007), #R53 4 Theorem 3.1 Let 1 ≤ s ≤ q. Then