SOME TWO COLOR, FOUR VARIABLE RADO NUMBERS. Aaron Robertson Department of Mathematics, Colgate University, Hamilton, NY

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1 SOME TWO COLOR, FOUR VARIABLE RADO NUMBERS Aaron Robertson Department of Mathematics, Colgate University, Hamilton, NY 1336 and Kellen Myers 1 Abstract There exists a minimum integer N such that any -coloring of {1,,..., N} admits a monochromatic solution to x+y +z = lw for, l Z +, where N depends on and l. We determine N when l {0, 1,, 3,, 5}, for all, l for which 1 l l + 1 l, as well as for arbitrary when l =. 1. Introduction For r, 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 x 1,..., x n, we say a solution { x i } n i=1 is monochromatic if χ x i = χ x j for every i, j pair. A well-nown theorem of Rado states that a linear homogeneous equation c 1 x c n x n = 0 with each c i Z admits a monochromatic solution in Z + under any r-coloring of Z +, for any r, if and only if some nonempty subset of {c i } n i=1 sums to zero. Such an equation is said to satisfy Rado s regularity condition. The smallest N such that any r-coloring of {1,,..., N} = [1, N] satisfies this condition is called the r-color Rado number for the equation E. Rado also proved the following, much lesser nown, result. Theorem 1 Rado [R] Let E = 0 be a linear homogeneous equation with integer coefficients. Assume that E has at least 3 variables with both positive and negative coefficients. Then any -coloring of Z + admits a monochromatic solution to E = 0. For Rado s original proof in German see [R]; for a proof in English see [MR]. In this article we study the equation x + y + z = lw for positive integers and l. As such, we mae the following notation. Notation For a positive integer and j > an integer, let E, j represent the equation x + y + z = + jw. 1 This wor was done as part of a summer REU, funded by Colgate University, while the second author was an undergraduate at Colgate University, under the directorship of the first author. 1

2 . A General Upper Bound Definition Let E be any equation that satisfies the conditions in Theorem 1. Denote by RRE the minimum integer N such that any -coloring of [1, N] admits a monochromatic solution to E. Part of the following result is essentially a result due to Burr and Loo [BL] who show that, for j, we have RRx + y = jw =. Their result was never published. Below we present our independently derived proof. Theorem Let, j Z + with j. Then RRE, j. Furthermore, for all j, we have RRE, j =. Proof. Let F denote the equation x + y = jw. We will show that RRF. Since any solution to F is also a solution to E, j for any Z +, the first statement will follow. Assume, for a contradiction, that there exists a -coloring of [ 1, ] with no monochromatic solution to F. Using the colors red and blue, we let R be the set of red integers and B be the set of blue integers. We denote solutions of F by x, y, w where x, y, w Z +. Since x, y, w = j 1, 1, 1 solves F, we may assume that 1 R and j 1 B. We separate the proof into two cases. Case 1. j +1 B. Assume i 1 is red. Considering 1, ij 1, i gives ij 1 B. If i, this, in turn, gives us i + 1 R by considering ij 1, j + 1, i + 1. Hence 1,,..., + 1 are all red. But then j,, 1 is a red solution, a contradiction. Case. j + 1 R. This implies that j B. Note also that the solutions 1, j 1, 1 and j j 1, j j 1, j 1 give us j 1 B and j j 1 R. First consider the case when j is even. By considering j, j, 1 we see that j B. Assume, for i 1, that i 1j B. Considering j, i 1j, j + i we have j + i R. This, in turn, implies that i+1j B by considering j j 1, i+1j,. j + i Hence we have j + i is red for 1 i j. This gives us that j 1 R when i = j 1, a contradiction. Now consider the case when j is odd. We consider two subcases. Subcase i. j B. For i 1, assume that ij B. We obtain +i R by considering the solution j, ij, + i. This gives us i+1j B by considering j j 1, i + 1j, + i Hence, we have that j, j,..., j are all blue, contradicting the deduction that j j 1 R. Subcase ii. j R. We easily have B. Next, we conclude that j j 3 R by considering j, j j 3, j 1. Then, the solution j j 3, j j 1, j gives us j B. We use j j 1, j, to see that B. From j,, we have j R. To avoid j j 1 +, j, j being a red solution, we have j j 1 + B. This gives

3 us a contradiction; the solution j j 1 +, j, is blue. This completes the proof of the first statement of the theorem. For the proof of the second statement of the theorem, we need only provide a lower bound 1. We first show that any solution to x + y + z = + jw with x, y, z, w < of must have z = w when j. Assume, for a contradiction, that z w. If z < w, then +jw +jz +1 > z +. However, x+y < jj +1 while > jj +1 for j 3. Hence, 1. Since we have x + y + z = + jw z w. If z > w, then + jw z 1 + j we now have + z x + y + z z 1 + j contradicting the given bound on. Thus, z = w. 1. Hence, j 1, Now, any solution to x + y + z = + jw with z = w is a solution to x + y = jw. From Burr and Loo s result, there exists a -coloring of [ 1, ] 1 with no monochromatic solution to x + y = jw. This provides us with a -coloring with no monochromatic solution to x + y + z = + jw, thereby finishing the proof of the second statement. 3. Some Specific Numbers In this section we determine the exact values for RRE, j for j {0, 1,, 3,, 5}, most of which are cases not covered by Theorem. Theorem 3 For, RRE, 0 = and RRE 1, 0 = 3 1. Furthermore RRE, 0 = 5 and RRE1, 0 = 11. Proof. The cases RRE, 0 = 5, RRE, 0 =, and RRE3, 0 = 5 are easy calculations, as is RRE1, 0 = 11, which first appeared in [BB]. Hence, we may assume 3 in the following arguments. We start with RRE, 0 =. To show that RRE, 0 consider the -coloring of [1, 1] defined by coloring the odd integers red and the even integers blue. To see that there is no monochromatic solution to x + y + z = w, note that we must have x + y. This implies that x + y = since x, y 1. Thus, w = z + 1. However, no consecutive integers have the same color. Hence, any solution to E, 0 is necessarily bichromatic. Next, we show that RRE, 0. Assume, for a contradiction, that there exists a -coloring of [1, ] with no monochromatic solution to our equation. Using the colors red and blue, we may assume, without loss of generality, that is red. This gives us that 1 and + 1 are blue, by considering x, y, z, w =,, 1, and,,, + 1. Using these in the solution,, 1, + 1 we see that must be red, which implies that is blue using,,,. However, this gives the blue solution 1, + 1,, 1, a contradiction. 3

4 We move on to RRE 1, 0. To show that RRE 1, consider the following -colorings of [1, 3 ], dependent on we use r and b for red and blue, respectively: brrbbrrbb... bbr if 0 mod rrbbrrbb... bbr if 1 mod brrbbrrbb... rrb if mod rrbbrrbb... rrb if 3 mod. Since we need 1 x + y and x, y 3, we have x + y { 1, }. By construction, if x + y = 1, then x and y have different colors. Hence, the only possibility is x + y =. But then w = z + and we see that w and z must have different colors. Next, we show that RRE 1, Assume, for a contradiction, that there exists a -coloring of [1, 3 1] with no monochromatic solution to our equation. Using the colors red and blue, we may assume, without loss of generality, that 1 is red. To avoid 1, 1, z, z + being a red solution, we see that + 1 and 3 are blue using z = 1 and 3, respectively. If is red, then is blue using,,,. From 3 1, 3 1,, +1 we see that 3 1 is red. This, in turn, implies that is blue using 3 1, 3 1,, 1. So that, +,, is not a blue solution, we require + to be red. But then 1, 1,, + is a red solution, a contradiction. If is blue, then must be red. So that 3 1, 3 1, 3, is not a blue solution, we have that 3 1 is red. Also, + must be red by considering 3, + 1,, +. But this implies that 3 1, 3 1, 1, + is a red solution, a contradiction. We proceed with a series of results for the cases j = 1, 3,, 5. When j =, the corresponding number is trivially 1 for all Z +. Below, we will call a coloring of [1, n] valid if it does not contain a monochromatic solution to E, j. Theorem For Z +, RRE, 1 = { for 3 5 for. Proof. Assume, for a contradiction, that there exists a -coloring of [1, 5] with no monochromatic solution to x + y + z = + 1w. We may assume that 1 is red. Considering the solutions 1, 1,,,,,,, 1, 3,,, and, 3, 5, 5, in order, we find that is blue, is red, 3 is blue, and 5 is red. But then 1,, 5, 5 is a red solution, a contradiction. Hence, RRE, 1 5 for all Z +. We see from the above argument that the only valid colorings of [1, 3] assuming, without loss of generality, that 1 is red are rbr and rbb where we use r for red and b for blue. Furthermore, the only valid coloring of [1, ] is rbbr. We use these colorings to finish the proof. First consider the valid coloring rbr. The possible values of x + y + z when x, y, z are all red form the set { +, +, + 6, 3 +, 3 +, 3 + 6}. The possible values when x, y, z are all

5 blue is +. The possible values of + 1w when w is red form the set { + 1, 3 + 3}; when w is blue, + is the only possible value. We denote these results by: R x,y,z = { +, +, + 6, 3 +, 3 +, 3 + 6} B x,y,z = { + } R w = { + 1, 3 + 3} B w = { + }. Next, we determine those values of, if any, for which R x,y,z R w or B x,y,z B w. Clearly, there is no such for these sets. Hence, we conclude that RRE, 1 for all. We need not consider the valid coloring rbb since we now now that RRE, 1 for all. We move on to the valid coloring of [1, ], which is rbbr. We find that R x,y,z = { +, + 5, + 8, +, + 5, + 8} B x,y,z = { +, + 5, + 6, 3 +, 3 + 5, 3 + 6} R w = { + 1, + } B w = { +, 3 + 3}. We see that B x,y,z B w when = 1 + = 3 + 3, = + 5 = 3 + 3, and = = For all other values of, B x,y,z B w = and R x,y,z R w =. Hence, we conclude that RRE, 1 5 for 3, while, since rbbr is the only valid coloring of [1, ], RRE, 1 for = 1,, 3. This completes the proof of the theorem. The proofs below refer to the small Maple pacage FVR. The description of FVR is given in Section 3.1, which follows the next 3 theorems. Theorem 5 For Z +, for 5 and = 7 RRE, 3 = 6 for = 8, 11 9 for = 6, 9, 10 and 1 Proof. The method of proof is the same as that for Theorem, but we will wor it out in some detail commenting on the use of the Maple pacage FVR. It is easy to chec that the only valid -colorings using r for red, b for blue, and assuming that 1 is red of [1, n] for n =, 5,..., 8 are as in the following table. The determinations of 5

6 R x,y,z, R w, B x,y,z, and B w are equally easy. n coloring sets rbrr R x,y,z = {i + j : i = 1, 3, ; j =,, 5, 6, 7, 8}; R w = {i + 3 : i = 1, 3, } B x,y,z = { + }; B w = { + 6} 5 rbrrb R x,y,z = {i + j : i = 1, 3, ; j =,, 5, 6, 7, 8}; R w = {i + 3 : i = 1, 3, } B x,y,z = {i + j : i =, 5; j =,, 7, 10}; B w = { + 6, } 6 rbrrbb R x,y,z = {i + j : i = 1, 3, ; j =, 5, 6, 7, 8}; R w = {i + 3 : i = 1, 3, } B x,y,z = {i + j : i =, 5, 6; j =, 7, 8, 10, 11, 1}; B w = { + 6, , } 7 rbrrbbr R x,y,z = {i + j : i = 1, 3,, 7; j =,..., 8, 10, 11, 1}; R w = {i + 3 : i = 1, 3,, 7} B x,y,z = {i + j : i =, 5, 6; j =, 7, 8, 10, 11, 1}; B w = { + 6, , } 8 rbrrbbrb R x,y,z = {i + j : i = 1, 3,, 7; j =,..., 8, 10, 11, 1}; R w = {i + 3 : i = 1, 3,, 7} B x,y,z = {i + j : i =, 5, 6; j =, 7, 8, 10,..., 1, 16}; B w = {i + 3 : i =, 5, 6, 8} The sets R x,y,z, R w, B x,y,z, and B w are automatically found by FVR, which then gives us the values of that induce a nonempty intersection of either R x,y,z R w or B x,y,z B w. For completeness, we give the details. For the coloring rbrr, we have R x,y,z R w when = = 3 + 5, = = + 7, = = + 6, = = + 5, = = +, and = = +. Since rbrr is the only valid coloring of [1, ], we have RRE, 3 = for = 1,, 3,, 5, 7. For the coloring rbrrb, we have no new additional elements in R x,y,z R w. Hence, any possible additional intersection point comes from B x,y,z B w. However, B x,y,z B w = for all Z +. Hence, RRE, 3 6 for Z + \ {1,, 3,, 5, 7}. For rbrrbb, we again have no new additional elements in the red intersection. We do, however, have additional elements in B x,y,z B w. When = = and = = 6 + we have a blue intersection. We conclude that RRE, 3 = 6 for = 8, 11 and RRE, 3 7 for Z + \ {1,, 3,, 5, 7, 8, 11}. Considering rbrrbbr, we have no new additional elements in B x,y,z B w. Furthermore, we have no new additional intersection points in R x,y,z R w. Hence, RRE, 3 8 for Z + \ {1,, 3,, 5, 7, 8, 11}. Lastly, we consider rbrrbbrb, which gives no new additional elements in R x,y,z R w. Furthermore, we have no new additional intersection points in B x,y,z B w. Thus, RRE, 3 9 for Z + \ {1,, 3,, 5, 7, 8, 11}. Analyzing the valid coloring of [1, 8] we see that we cannot extend it to a valid coloring of [1, 9]. Hence, RRE, 3 9 for all so that RRE, 3 = 9 for Z + \ {1,, 3,, 5, 7, 8, 11}. 6

7 Theorem 6 For Z +, RRE, = 3 for =, 3, 5 for = 6, 7, 8, 10, 11, 1 6 for = 5, 9, 1, 13, 15, 18 8 for = 17, 19, 9 for = 1, 3, 10 for = 16, 0, 1 and 5. Proof. Use the Maple pacage FVR with the following valid colorings which are easily obtained: n valid colorings 3 rbb rbbr 5 rbbrr 6 rbbrrr 7 rbbrrrr, rbbrrrb 8 rbbrrrrb, rbbrrrbb 9 rbbrrrrbr, rbbrrrbbr 10 none Note that if [1, n] has more than one valid coloring, we can conclude that RRE, n for = ˆ only if ˆ is an intersection point for all valid colorings. Otherwise, there exists a coloring of [1, n] that avoids monochromatic solutions to E, when = ˆ. Theorem 7 For Z +, RRE, 5 = for = 1,, 3 6 for =, 13, 1 7 for = 16, 17, 18, 3 8 for 5 1 and = 1 10 for = 19,, 6, 7, 8, 9, for =, 30, 31, 3, 3, 36, 37, 38, 39, 1,, 3, 8 1 for = 15, 35,, 6, 7, for = 51, 5 15 for = 0, 5, 0, 5, 9, 50 and 5. Proof. Use the Maple pacage FVR with the following valid colorings which are easily obtained: 7

8 n valid colorings rrrb, rrbb, rbrb, rbbb 5 rrbbb, rbrbr, rbbbr 6 rrbbbr, rbrbrr, rbbbrr 7 rrbbbrr, rrbbbrb, rbrbrrr, rbrbrrb, rbbbrrr 8 rrbbbrrb, rbrbrrbr, rbbbrrrr 9 rrbbbrrbb, rbrbrrbrb, rbbbrrrrr.rbbbrrrrb 10 rrbbbrrbbr, rbbbrrrrbr, rbbbrrrrrr 11 rrbbbrrbbrr, rbbbrrrrrrr, rbbbrrrrbrr 1 rrbbbrrbbrrr, rbbbrrrrrrrr, rbbbrrrrbrrr 13 rrbbbrrbbrrrr, rrbbbrrbbrrrb 1 rrbbbrrbbrrrrr, rrbbbrrbbrrrrb, rrbbbrrbbrrrbr, rrbbbrrbbrrrbb 15 none. 3.1 About FVR In the above theorems, we find our lower bounds by considering all valid colorings of [1, n] for some n Z + and deducing the possible elements that x + y + z can be when x, y, and z are monochromatic and the possible elements that +jw can be, i.e., determining R x,y,z, B x,y,z, R w, and B w. We then looed for intersections that would mae x, y, z, w a monochromatic solution. The intersections are specific values of which show that the given coloring has monochromatic solutions for these values of. This process has been automated in the Maple pacage FVR, which is available from the first author s website. The input is a list of all valid colorings of [1, n]. The output is a list of values of for which we have monochromatic solutions. By increasing n we are able to determine the exact Rado numbers for all Z +. An example of this is explained in detail in the proof of the next theorem.. A Formula for x + y + z = w In [HS] and [GS] a formula for, in particular, x + y + z = w is given: RRx + y + z = w = In this section we provide a formula for the next important equation of this form, namely the one in this section s title. To the best of our nowledge this is the first formula given for a linear homogeneous equation E = 0 of more than three variables with a negative coefficient not equal to 1 assuming, without loss of generality, at least as many positive coefficients as negative ones that does not satisfy Rado s regularity condition. aaron/programs.html 8

9 Theorem 8 For Z +, RRx + y + z = w = if 0 mod if 1 mod if mod if 3 mod. Proof. We begin with the lower bounds. Let N i be one less than the stated formula for i mod, with i {0, 1,, 3}. We will provide -colorings of [1, N i ], for i = 0, 1,, 3, that admit no monochromatic solution to x + y + z = w. For i = 0, color all elements in [ 1, ] red and all remaining elements blue. If we assume x, y, and z are all red, then x + y + z + so that for any solution we have w + 1. Thus, there is no red solution. If we assume x, y, and z are all blue, then x + y + z + + = > N 0, showing that there is no blue solution. For i = 1, color N 1 and all elements in [ ] 1, +1 red. Color the remaining elements blue. Similarly to the last case, we have no blue solution since x + y + z > N 1 1. If we assume x, y, z and w are all red, then we cannot have all of x, y, z in [ ] 1, +1. If we do, since is odd so that we must have x + y odd, then x + y + z = + 3 > +1. Thus, w must be blue. Now we assume, without loss of generality, that x = N 1. In this situation, we must have w = N 1. Hence, since N 1 + y + z = N 1 we see that y + z = N 1. Hence, y, z +1. But then y + z ++1 < N 1, a contradiction. Hence, there is no red solution under this coloring. The cases i = and i = 3 are similar to the above cases. As such, we provide the colorings and leave the details to the reader. For i =, we color N and all elements in [ 1, ] red, while the remaining elements are colored blue. For i = 3, color all elements in [ ] 1, +1 red and all remaining elements blue. We now turn to the upper bounds. We let M i be equal to the stated formula for i mod, with i {0, 1,, 3}. We employ a forcing argument to determine the color of certain elements. We let R denote the set of red elements and B the set of blue elements. We denote by a - tuple x, y, z, w a solution to x + y + z = w. In each of the following cases assume, for a contradiction, that there exists a -coloring of [1, M i ] with no monochromatic solution to the equation. In each case we assume 1 R. Case 1. 0 mod. We will first show that R. Assume, for a contradiction, that B. Then + R by considering, +,, +. Also, + 1 B comes from the similar solution 1, + 1, 1, + 1. Now, from 3 + 3, 1, 1, + we have B. As a consequence, we see that 3 R by considering 3, 3 + 3, 3, From here we use 3 + 1, 3, 1, + to see that B. But then 3 + 1, + 1,, is a blue solution, a contradiction. Hence, R. 9

10 Now, since 1, R, in order for 1, 1, 1, + 1 not to be monochromatic, we have + 1 B. Similarly,,, 1, + gives + B. Consequently, so that + 1, + 1, + 1, is not monochromatic, we have R. Our next goal is to show that R. So that + + 1, + + 1, 1, is not red, we have B. In turn, to avoid + 1, + 1,, being blue, we have R, as desired. So that,, 1, and,,, + are not red, we have, + B. Using these in,, 1, + gives 1 R. Since and 1 are both red, 1, 1, 1, 1 gives us 1 B while,, 1, gives us B. This, in turn, gives us R by considering,,, +. Now, from + + 1, 1, + 1, we have +1 B. We use this in the two solutions 1, + 1, + 1, +3 and,, + 1, + to find that +3, + R. But this gives us the red solution +,,, +3, a contradiction. Case. 1 mod. The argument at the beginning of Case 1 holds for this case, so we have R. We consider two subcases. Subcase i. +3 R. From ++3, +3, +3, ++3 we have ++3 B. This gives us +1 R by considering + 1, +1, +1, ++3 where + 1 B comes from 1, 1,, + 1. We also have, from 1,, 1, +3 +3, that B. Consequently, +5+6 R so that +3, +3, +3, +5+6 is not monochromatic. We next have that +5 B so that 3,, 1, +5 is not monochromatic we may assume that 9. Hence, R by considering, +5, +3, But this gives us the monochromatic solution +5+10, +1, +3, +5+6, a contradiction. Subcase i. +3 B. Via arguments similar to those in Subcase i, we have ++3, +5+6 R. From ++3, ++9, 1, +5+6 we have ++9 B. This gives us +15 R by considering +15, +3, +3, ++9. We now show that 1 R by showing that for any i 1 we must have i R. To this end, assume, for a contradiction, that i 1 R but i B where i 3. From i, i + i, i, i + i we have i + i R. In turn we have i i B by considering i + i, i + i,, i i. We next see from i i, i, i + 1, i i that i + 1 R. Using our assumption that i 1 R in i 1, i + 1,, + i we have + i B. But then i i, + i, i, i i is 10

11 a blue solution, provided i i M 1, which by the bound given on i is valid. By applying this argument to i = 3,,..., 1, in order, we see that all positive integers less than or equal to 1 must be red. In particular, 1 R. Using 1 R in +15, +15, 1, +15 we have +15 B. This, in turn, gives us For our contradiction, we see now that R by considering ++15, +15, 1, ++15, +15, +3, ++15 is a red solution. Case 3. mod. From Case 1 we have, ++1 R and +1, 1, + B. From 1, 1, +, we see that + B. This gives us + + R by considering +, +, + 1, + +. Using this fact in + +,, +, we have + B. But then 1, + + 1,, + is a blue solution, a contradiction. Case. 3 mod. Let i R. From Case we may assume i so that 1,,..., i are all red, By considering i, i + i, i, i + i we have i + i B so that we may assume + 1, +,..., i 1 + i 1, i + i are all blue. Since i + 1, i 1 + i 1, i + 1, i + i is a solution, we have i + 1 R. Hence, i R for 1 i In particular, R. From Case we also have +3 then, ,, +5, B. By considering, +5, +3 is a red solution, a contradiction., +5+8 we have +5+8 R. But. Conclusion The next important numbers to determine are in the first row of Tables 1. As such, it would be nice to have a formula for RRx + y + z = lw. We have been unable to discover one. By analyzing Table 1, given below, we have noticed, to some extent, certain patterns that emerge. In particular, we mae the following conjecture. Conjecture For l fixed and l +, we have + l + 1 RRx + y + z = lw = + O, l l where the O l part depends on the residue class of modulo l. For a concrete conjecture, we believe the following holds. 11

12 Conjecture Let 5. Then + RRx + y + z = 3w = + 3 if 0 mod if 1, 6 mod if mod if 3, 8 mod 9 1 if mod 9 + if 5 mod if 7 mod 9 9. Acnowledgment We than Dan Saracino for a very careful reading that caught some errors in a previous draft. We end with a table of calculated values of RRE, j for small values of and j. These were calculated by a standard bactrac algorithm. We than Joey Parrish for helping with implementation efficiency of the algorithm. The program can be downloaded as RADONUMBERS at the second author s homepage aaron/programs.html. l = =

13 = Table 1: Some Values of RRx + y + z = lw References [BB] A. Beutelspacher and W. Brestovansy, Generalized Schur numbers, Combinatorial theory Schloss Rauischholzhausen, 198, pp , Lecture Notes in Math., 969, Springer, Berlin-New Yor, 198. [BL] S. Burr and S. Loo, On Rado Numbers II, unpublished. [GS] S. Guo and Z-W. Sun, Determination of the two-color Rado number for a 1 x a m x m = x 0, to appear in JCTA, preprint available at arxiv://math.co/ [HS] B. Hopins and D. Schaal, On Rado Numbers for m 1 i=1 a ix i = x m, Adv. Applied Math , [MR] K. Myers and A. Robertson, Two Color Off-diagonal Rado-type Numbers, preprint available at arxiv://math.co/ [R] R. Rado, Studien zur Kombinatori, Mathematische Zeitschrift ,