On the Minimum Number of Monochromatic Generalized Schur Triples

The solution to the problem of finding the minimum number of monochromatic triples $(x,y,x+ay)$ with $a\geq 2$ being a fixed positive integer over any 2-coloring of $[1,n]$ was conjectured by Butler, Costello, and Graham (2010) and Thanathipanonda (2009). We solve this problem using a method based on Datskovsky's proof (2003) on the minimum number of monochromatic Schur triples $(x,y,x+y)$. We do this by exploiting the combinatorial nature of the original proof and adapting it to the general problem.


Introduction
Ramsey theory has a rich history first popularized in 1935 by Erdös and Szekeres in their seminal paper [3]. We investigate a part of the theory that was orginally developed by Issai Schur. The formulation of Schur's theorem was first derived from Van der Waerden's theorem in 1927 [10]. Van der Waerden proved that any r-coloring of Z + must admit a monochromatic 3-term arithmetic progression {a, a + d, a + 2d} for some a, d > 1. A particular choice of x, y and z in terms of a and d admits a monochromatic solution to x + y = 2z, on a plane whose coordinates are the positive integers. Hence, a similar question regarding the coloring of monochromatic solutions to a simpler equation can be posed; namely, does there exist a least positive integer s = s(r) such that for any r-coloring of [1, s] there is a monochromatic solution to x + y = z? Schur determined that the answer is yes, and we call the solution (x, y, z) to such an equation a Schur triple.
In 1959, Goodman [4] was able to determine the minimum number of monochromatic triangles under a 2-edge coloring of a complete graph on n vertices, which turned out to be the same order as the average, n 3 24 +O(n 2 ). This motivated Graham [5] to pose the problem of finding the minimum number of monochromatic Schur triples over any 2-coloring of [1, n] at a conference and to offer 100 USD for the result. Graham initially conjectured that the average value should be the minimum at n 2 16 + O(n). However, Zeilberger and his student Robertson [7] used discrete calculus to show that the minimum number must be n 2 22 + O(n) and won the cash prize. Around the same time, Schoen [8], followed four years later by Datskovsky [2], furnished different proofs using Fourier analysis to show that indeed, n 2 22 + O(n) is the correct minimum. Ultimately, their idea had reduced to one in combinatorics. More recently in 2009, Thanatiponanda [9] confirmed the result using a new technique with computer algebra and a greedy algorithm. He conjectured a minimum number of monochromatic Schur triples for all r-colorings and a minimum number of monochromatic triples satisfying x + ay = z for a fixed integer a 2 over any 2-coloring of [1, n]. We solve the latter part of the conjecture in this paper using a purely combinatorial approach.

The Minimum Number of Monchromatic Schur Triples
We first show how to find the minimum coloring of x + y = z in an elementary way using the method by Datkovsky [2]. Then, we explore the more general case in the next two sections. We start by employing a 2-coloring on all integers in [1, n] for n < ∞ and count the number of monochromatic Schur triples (x, y, z) where z = x + y. Denote the colors to be red (R) and blue (B).
The number of Schur triples includes the number of monochromatic Schur triples |M(n)| and non-monochromatic Schur triples |N (n)|.
Lemma 2. The number of Schur triples in [1, n] is 1 2 Figure 1: The sets N − and N + Proof.
Each non-monochromatic triple gives two non-monochromatic pairs which gives the first equality. To get the second equality, we observe that the pairs in N − will contribute to two triples but the pairs in N + will only contribute to one. For example, in [1,10], (5,3) gives the triples (3, 2, 5) and (5,3,8). But, in [1,10], (8,9) only gives (1,8,9). Finally, the last equality comes from the fact that By putting together Lemmas 1 and 2 and Proposition 5, we obtain the next lemma.
Lemma 6. The number of monochromatic Schur triples under a 2-coloring on [1, n] is In order to find the minimum value of |M(n)|, we must obtain the lower bound of |N + | in terms of µ B and µ R . To do this more efficiently, we denote D := |N − | − |N + | and find an upper bound on D instead. The proof requires the following notation: the electronic journal of combinatorics 22 (2015), #P00 Definition 7. Let S be the set of pairs of the form {s, n + 1 − s} where 1 s n/2. Denote by µ CC ′ the number of sets S with colorings C and C ′ in this order. γn is the number of non-monochromatic pairs in S.
Lemma 8. Assume, without loss of generality, that µ B µ R . Then Proof. Assuming 1 y < x n 2 , denote the sets X and Y as Ordered pairs in X × Y when colored differently are contained in N + ∪ N − , that is: We outline all possible colorings of the X and Y sets that contribute to the value of D in the table in Figure 2. We see that with the exception of the first four cases, the contribution to D is 0.
We now obtain an upper bound of D: The inequality (2) comes from our assumption that µ BB µ RR . The last equality comes Calculus shows that the maximum of (3) occurs when γ = µ B 2n which simplifies our inequality to Non-monochromatic X set and Monochromatic Y set 9 red blue red red (n + 1 − x, y) 0 (n + 1 − x, n + 1 − y) 10 blue red blue blue (n + 1 − x, y) 0 (n + 1 − x, n + 1 − y) 11 red blue blue blue (x, y) 0 (x, n + 1 − y) 12 blue red red red (x, y) 0 (x, n + 1 − y) Non-monochromatic X and Y sets 13 red blue blue red (x, y) Theorem 9. Over all 2-colorings of [1, n], the minimum number of monochromatic Schur triples is n 2 22 + O(n). Proof. An upper bound of the minimum can be obtained from a coloring on [1, n]. We color R 4n/11 , B 6n/11 , R n/11 as illustrated in Figure 3. This proportion comes from a brute force computer search first proposed by Zeilberger [7]. This coloring will give us n 2 22 + O(n) monochromatic triples. Next, we look for the lower bound of the minimum. By using Lemma 8 and the fact that which, together with Lemma 6, gives The right hand side of (4) achieves a maximum when µ R = 5n 11 and µ B = 6n 11 . As a result, we get the lower bound for the minimum to be: Because the lower and upper bounds match, we have therefore shown the desired result.
Remark. We can be confident that the bounds for equations (1) and (2) is sharp relative to the optimal coloring because we know that cases 3 and 4 from Figure 2 will not occur and that µ BR = 0.
This method of Datkovsky's also gives the optimal coloring for fixed ratios of µ B and µ R .
Corollary 10. For any fixed µ B µ R , the coloring on [1, n] that gives the minimum number of monochromatic Schur triples is R For the case µ B 2n 3 , the maximum of D occurs when γ = µ B 2n . So, For the case µ B > 2n 3 , the maximum of D occurs when γ = µ R n . With similar calculations, The colorings mentioned in the statement of the corollary give us upper bounds for the minimum, which happens to match the lower bounds.
Corollary 11. For any fixed µ B µ R , the coloring on [1, n] that gives the maximum number of monochromatic Schur triples is R 3 . Proof. Using a similar calculation to Lemma 8, we have that For the case µ B > 2n 3 , the minimum of D occurs when γ = µ R n . Therefore, The colorings mentioned in the statement of the corollary give us lower bounds of the maximum, which happens to match the upper bounds.
3 The Minimum Number of Monochromatic Triples (x, y, x + 2y) The technique illustrated in the previous section can be extended to x + ay = z for any fixed integer a 2. However, the nice symmetry we had previously with the equation x + y = z is no longer there. In this section we deal with the specific case a = 2. The general case will be outlined in Section 4.
We parallel the same argument as in Section 2. First, we write the number of nonmonochromatic triples |N (n)| in terms of variables we can optimize.
Definition 12. Denote by µ B 1 and µ R 1 the number of blue and red colorings respectively on 1, n 2 . Furthermore, denote by µ B 2 and µ R 2 the number of blue and red colorings respectively on n 2 , n .
the electronic journal of combinatorics 22 (2015), #P00 Note that µ B 1 + µ R 1 = n 2 and µ R 1 + µ R 2 = µ R . Definition 13. The sets of non-monochromatic pairs in [1, n]×[1, n 2 ] are defined as follows: Proof. We count sets of ordered pairs that come from non-monochromatic triples of type (x, y, z) where z = x + 2y as follows: We then have that: To obtain the second inequality, observe that (refer to Lemma 3 of Thanatipanonda [9]). The last equality comes The next lemma follows immediately from Proposition 14.
Lemma 15. The number of monochromatic triples of type (x, y, x+2y) under a 2-coloring on [1, n] is Minimizing |M(n)| can be reduced to finding the upper bound of CC ′ the number of sets S with colorings C and C ′ in this order. The superscript (1) refers to the coloring of pairs on 1, n 2 . Also, denote by γ 1 n the number of nonmonochromatic pairs in S.
With this notation, γ 1 n = µ (1) Definition 17. Define sets X and Y 1 as follows, EE ′ is defined by the number of pairs (X, Y 1 ) where X has the coloring {C, C ′ } and Y 1 has the coloring {E, E ′ } under the condition 2y < x.
Proof. Assuming 1 2y < x n 2 , we observe that the ordered pairs in X × Y 1 when colored differently are contained in N − x ∪ N + y , that is: The table in Figure 5 shows that there are only four cases that contribute any value to D 2 , while the other cases contribute 0, similar to the table in Figure 2.
The result follows immediately.
The next proposition gives the upper bound for D 2 .
the electronic journal of combinatorics 22 (2015), #P00 Monochromatic X set and Non-monochromatic Y 1 set Case Proposition 19. Assume, without loss of generality, that µ B µ R and suppose the number of non-monochromatic pairs in S, γ 1 n, is fixed. Then where A 1 is the largest possible area under the curve in Figure 6, with a base of length γ 1 n for γ 1 Proof. From Lemma 18, We configure X to gain the maximum of µ BB ⊗ µ  condition 2y < x . With this set up for X, we can count the number of pairs gained for every y in Y 1 as shown in Figure 7a.
Similarly, the configuration of X to gain the maximum of µ RR ⊗ µ (1) BR is when we color the far left of the interval 1, n 2 blue and the remainder of the interval red. With this set up for X, we can count the number of pairs gained for every y in Y 1 as shown in Figure 7b.  Proof. Assume, without loss of generality, that µ B µ R . The proof ultimately depends on determining the area under the curve of Figure 6, which we break down into three cases according to the value of γ 1 . The three cases are illustrated in Figure 8. We complete the proof of this proposition as follows. For each case, we write A 1 in terms of the variables µ R , µ B , γ and γ 1 . Then, we optimize γ, γ 1 with respect to µ B 1 , µ R 1 and µ R 2 . Finally, we use the optimal γ and γ 1 values to maximize Denote this maximum to be ∆ max . Propositions 14 and 19 show that ∆ max will be the upper bound for |N (n)|. The optimization of ∆ has been done using Maple and for curious readers, the code can be found at Thanatipanonda's website. We note that ∆ can the electronic journal of combinatorics 22 (2015), #P00 ultimately be written as a function of only two variables µ R and µ R 1 . In our calculations, we use the following lower bound of γn and upper bound of γ 1 n: Maple's current technology does not allow us to optimize with absolute value and minimum functions. Thus, we separate each case into the following pieces. The subcases are summarized in the table in Figure 9. It is important to note that subcase D can be ignored in our calculation because it only produces one pair, namely µ R 1 = n 4 and µ R = n 2 (recall that µ R n 2 ). . This case is illustrated in Figure 8a.
In order to maximize A 1 , we maximize γ 1 n and minimize γn. To be able to determine the values of γ 1 and γ, we consider two further subcases as follows: the electronic journal of combinatorics 22 (2015), #P00 The optimal values of (γ 1 n, γn) is (min (µ R 1 , µ B 1 ) , |µ R 1 − µ R 2 |) as shown in Figure 10a.
The optimal values of (γ 1 n, γn) lies on the line γn = 2γ 1 n as shown in Figure 10b. This gives Thus, A 1 attains its maximum value at γn = µ B 2 .
The calculations for ∆ max for all subcases are summarized in the table in Figure 11 and the admissible regions for subcases A, B, and C are shown in Figure 14a. In this case, ∆ max occurs under subcase A. Figure 10: Finding the optimal values of γ 1 n and γn in Case 1 Case 2: γn 2 γ 1 n < µ R 2 . This case is illustrated in Figure 8b. In this case, we take the area of the rectangle but subtract those pairs that are outside of the region. Thus, we have that: In order to maximize A 1 , we want to make γ 1 n as close to µ B 2 as possible and minimize γn. We break this case down into subcases depending on whether or not the upper bound of γ 1 n is less than µ R 2 .
The optimal value of (γ 1 n, γn) is (min The optimal value of (γ 1 n, γn) is The calculations for ∆ max for all subcases are summarized in the table in Figure 12 and the admissible regions for subcases A, B, and C are shown in Figure 14b. In this case, ∆ max occurs under subcase A. Case 3: µ R 2 < γ 1 n. This case is illustrated in Figure 8c. In this final case, A 1 indicated by nearly the entire region under the graph. We take the area of the rectangle but subtract those pairs that are outside of the region. Thus, we the electronic journal of combinatorics 22 (2015), #P00 have that In order to maximize A 1 , we want to make γ 1 n as to close to n 4 as possible and minimize γn. The calculations for ∆ max for all subcases are summarized in the table in Figure 13 and the admissible regions for subcases A, B, and C are shown in Figure 14c. Once again, ∆ max occurs under subcase A. In all three cases, we can see that ∆ max occurs in Case 3 when µ R 1 = 3n 11 and µ R = 4n 11 with ∆ max = 5n 2 22 + O(n). Theorem 21. Over all 2-colorings of [1, n], the minimum number of monochromatic triples satisfying x + 2y = z is n 2 44 + O(n). Proof. An upper bound of the minimum can be obtained from a coloring on [1, n]. We color R 3n/11 , B 7n/11 , R n/11 as illustrated in Figure 15. This solution was discovered in Butler, Costello, and Graham [1] and in Thanathipanonda [9].
Because the lower and upper bounds match, we have therefore shown the desired result.
4 The General Case x + ay = z, a 2 We now generalize our result.
Theorem 22. Over all 2-colorings of [1, n], the minimum number of monochromatic triples satisfying x + ay = z, a 2 is n 2 2a(a 2 +2a+3) + O(n). The set up of this proof is similar to the set up in Section 3. We will outline it here. Definition 23. Denote by µ B 1 and µ R 1 the number of blue and red colorings respectively on 1, n a . Furthermore, denote by µ B 2 and µ R 2 the number of blue and red colorings respectively on n a , n . It is now easy to adapt this notation to prove the following analog to Proposition 14.
Proposition 25. |N (n)| 1 2 Proposition 27. Assume, without loss of generality, that µ B µ R and suppose the number of non-monochromatic pairs in S, γ 1 n, is fixed. Then where A 1 is the largest possible area that can be placed under the curve in Figure 17 Proposition 28. Over all 2-colorings of [1, n], the maximum number of non-monochromatic triples satisfying x + ay = z, a 2 is Proof. Without loss of generality, we assume µ B µ R . Suppose Propositions 25 and 27 show that optimizing ∆ will give us the upper bound for |N (n)|. We will call this optimum ∆ max . In order to find ∆ max , we consider three different cases to compute A 1 . Like before, each case will be subjected to the conditions listed in Figure  9.
Here, we show only the details for Case 3A which will give us the best upper bound like in the previous section.
Similar to before, we want γ 1 n to be as close to n 2a as possible and γn should be as small as possible. This is achieved by setting γ 1 n = µ B 1 and γn = µ R 1 − µ R 2 . Then which is attained when µ R 1 = a+1 a 2 +2a+3 and µ R = a+2 a 2 +2a+3 .
Proof of Theorem 22. An upper bound of the minimum can be obtained from a coloring on [1, n]. We color [R, B, R] with the ratio 1, a + 1 a+1 , 1 a+1 as illustrated in Figure 18, which was discovered in [1] and [9].
This coloring gives us Because the lower and upper bounds match, we have therefore shown the desired result.

Conjectures
In this section, we present conjectures on variations of Graham's original problem. Denote by R, B, G the colors red, blue, and green respectively.

Equation:
ax + by = az where a, b are integers. This has also been conjectured in [1, p. 409].

Equation:
x + y + w = z The coloring that gives the minimum number of monochromatic solutions over any 2-coloring of [1, n] is , with the number of rainbow solutions to be n(n + 1) 10 .