*Article* **A Proof of a Conjecture on Bipartite Ramsey Numbers** *B***(2, 2, 3)**

**Yaser Rowshan 1, Mostafa Gholami <sup>1</sup> and Stanford Shateyi 2,\***


**Abstract:** The bipartite Ramsey number *B*(*n*1, *n*2, ... , *nt*) is the least positive integer *b*, such that any coloring of the edges of *Kb*,*<sup>b</sup>* with *t* colors will result in a monochromatic copy of *Kni*,*ni* in the *i*-th color, for some *i*, 1 ≤ *i* ≤ *t*. The values *B*(2, 5) = 17, *B*(2, 2, 2, 2) = 19 and *B*(2, 2, 2) = 11 have been computed in several previously published papers. In this paper, we obtain the exact values of the bipartite Ramsey number *B*(2, 2, 3). In particular, we prove the conjecture on *B*(2, 2, 3) which was proposed in 2015—in fact, we prove that *B*(2, 2, 3) = 17.

**Keywords:** Ramsey numbers; bipartite Ramsey numbers; Zarankiewicz number

#### **1. Introduction**

The bipartite Ramsey number *B*(*n*1, *n*2, ... , *nt*) is the least positive integer *b*, such that any coloring of the edges of *Kb*,*<sup>b</sup>* with *t* colors will result in a monochromatic copy of *Kni*,*ni* in the *i*-th color, for some *i*, 1 ≤ *i* ≤ *t*. The existence of such a positive integer is guaranteed by a result of Erd˝os and Rado [1].

The Zarankiewicz number *z*(*Km*,*n*, *t*) is defined as the maximum number of edges in any subgraph *G* of the complete bipartite graph *Km*,*n*, such that *G* does not contain *Kt*,*t* as a subgraph. Zarankiewicz numbers and related extremal graphs have been studied by many authors, including Kóvari [2], Reiman [3], and Goddard, Henning, and Oellermann in [4].

The study of bipartite Ramsey numbers was initiated by Beineke and Schwenk in 1976 [5], and continued by others, in particular Exoo [6], Hattingh, and Henning [7]. The following exact values have been established: *B*(2, 5) = 17 [8], *B*(2, 2, 2, 2) = 19 [9], *B*(2, 2, 2) = 11 [6]. In the smallest open case for five colors, it is known that 26 ≤ *B*(2, 2, 2, 2, 2) ≤ 28 [9]. One can refer to [2,9–14] and it references for further studies. Collins et al. in [8] showed that 17 ≤ *B*(2, 2, 3) ≤ 18, and in the same source made the following conjecture:

**Conjecture 1** ([8])**.** *B*(2, 2, 3) = 17*.*

We intend to get the exact value of the multicolor bipartite Ramsey numbers *B*(2, 2, 3). We prove the following result:

**Theorem 1.** *B*(2, 2, 3) = 17*.*

In this paper, we are only concerned with undirected, simple, and finite graphs. We follow [15] for terminology and notations not defined here. Let *G* be a graph with vertex set *V*(*G*) and edge set *E*(*G*). The degree of a vertex *v* ∈ *V*(*G*) is denoted by deg*G*(*v*), or simply by deg(*v*). The neighborhood *NG*(*v*) of a vertex *v* is the set of all vertices of *G* adjacent to *v* and satisfies |*NG*(*v*)| = deg*G*(*v*). The minimum and maximum degrees of vertices of *G* are denoted by *δ*(*G*) and Δ(*G*), respectively. Additionally, the complete bipartite graph with bipartition (*X*,*Y*), where |*X*| = *m* and |*Y*| = *n*, is denoted by *Km*,*n*. We use [*X*,*Y*] to denote the set of edges between the bipartition (*X*,*Y*) of *G*. Let *G* = (*X*,*Y*) be a bipartite graph

**Citation:** Rowshan, Y.; Gholami, M.; Shateyi, S. A Proof of a Conjecture on Bipartite Ramsey Numbers *B*(2, 2, 3). *Mathematics* **2022**, *10*, 701. https:// doi.org/10.3390/math10050701

Academic Editors: Janez Žerovnik and Darja Rupnik Poklukar

Received: 7 December 2021 Accepted: 23 December 2021 Published: 23 February 2022

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**Copyright:** © 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

and *<sup>Z</sup>* ⊆ *<sup>X</sup>* or *<sup>Z</sup>* ⊆ *<sup>Y</sup>*, the degree sequence of *<sup>Z</sup>* denoted by *DG*(*Z*)=(*d*1, *<sup>d</sup>*2, ... , *<sup>d</sup>*|*Z*|), is the list of the degrees of all vertices of *Z*. The complement of a graph *G*, denoted by *G*, is a graph with same vertices such that two distinct vertices of *G* are adjacent if and only if they are not adjacent in G. *H* is *n*-colorable to (*H*1, *H*2, ... , *Ht*) if there exists a *t*-coloring of the edges of *H* such that *Hi* - *<sup>H</sup><sup>i</sup>* for each 1 <sup>≤</sup> *<sup>i</sup>* <sup>≤</sup> *<sup>t</sup>*, where *<sup>H</sup><sup>i</sup>* is the spanning subgraph of *H* with edges of the *i*-th color.

#### **2. Some Preliminary Results**

To prove our main result—namely, Theorem 1—we need to establish some preliminary results. We begin with the following proposition:

**Proposition 1** ([8,13])**.** *The following results about the Zarankiewicz number are true:*


**Proof of Proposition 1.** By using the bounds in Table 3 and Table 4 of [8] and Table *C*.3 of [13], the proposition holds.

**Theorem 2** ([8])**.** 17 ≤ *B*(2, 2, 3) ≤ 18*.*

**Proof of Theorem 2.** The lower bound witness is found in Table 2 of [8]. The upper bound is implied by using the bounds in Table 3 and Table 4 of [8]. We know that *z*(*K*18,18, 2) = 81, *z*(*K*18,18, 3) ≤ 156, and 2 × 81 + 156 = 318 < 324 = |*E*(*K*18,18)|.

Suppose that (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K*17,17, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *G<sup>b</sup>* and *K*3,3 - *Gg*; in the following theorem, we specify some properties of the subgraph with color g. The properties are regarding Δ(*Gg*), *δ*(*Gg*), *E*(*Gg*), and degree sequence of vertices *X*, *Y* in the induced graph with color *g*.

**Theorem 3.** *Assume that* (*G<sup>r</sup>* , *Gb*, *Gg*) *is a* 3*-edge coloring of K*17,17*, where K*2,2 - *G<sup>r</sup> , K*2,2 - *Gb, and K*3,3 -*Gg. So:*


**Proof of Theorem 3.** Assume that *X* = {*x*1, *x*2, ... , *x*17}, *Y* = {*y*1, *y*2, ... , *y*17} is a partition set of *K* = *K*17,17 and (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K*, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *Gb*, and *K*3,3 - *<sup>G</sup>g*. Since <sup>|</sup>*E*(*K*)<sup>|</sup> <sup>=</sup> 289, if <sup>|</sup>*E*(*Gg*)| ≤ 140 then <sup>|</sup>*E*(*Gg*)| ≥ 149—that is, either <sup>|</sup>*E*(*Gr*)| ≥ 75 or <sup>|</sup>*E*(*Gb*)| ≥ 75. In any case, by Proposition 1, either *<sup>K</sup>*2,2 <sup>⊆</sup> *<sup>G</sup><sup>r</sup>* or *<sup>K</sup>*2,2 <sup>⊆</sup> *<sup>G</sup>b*, a contradiction. Hence, assume that <sup>|</sup>*E*(*Gg*)| ≥ 141. If <sup>|</sup>*E*(*Gg*)| ≥ 142 then by Proposition 1, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*, a contradiction again; that is, <sup>|</sup>*E*(*Gg*)<sup>|</sup> <sup>=</sup> 141 and part (*a*) is true.

To prove part (*b*), since <sup>|</sup>*E*(*Gg*)<sup>|</sup> <sup>=</sup> 141 by part (*a*), we can check that <sup>Δ</sup>(*Gg*) <sup>≥</sup> 9. Assume that there exists a vertex of *<sup>V</sup>*(*K*) say *<sup>x</sup>*, such that <sup>|</sup>*NGg* (*x*)| ≥ 10—that is, <sup>Δ</sup>(*Gg*) <sup>≥</sup> 10. Consider *x* and set *G<sup>g</sup>* <sup>1</sup> <sup>=</sup> *<sup>G</sup><sup>g</sup>* \ {*x*}, hence by part (*a*), <sup>|</sup>*E*(*G<sup>g</sup>* <sup>1</sup> )| ≤ 141 − 10 = 131. Therefore, since <sup>|</sup>*E*(*K*16,17)<sup>|</sup> <sup>=</sup> 272, so <sup>|</sup>*E*(*G<sup>g</sup>* <sup>1</sup> )| ≥ 141—that is, either <sup>|</sup>*E*(*G<sup>r</sup>* <sup>1</sup>)| ≥ 71 or <sup>|</sup>*E*(*G<sup>b</sup>* <sup>1</sup>)| ≥ 71. In any case, by Proposition <sup>1</sup> either *<sup>K</sup>*2,2 <sup>⊆</sup> *<sup>G</sup><sup>r</sup>* <sup>1</sup> <sup>⊆</sup> *<sup>G</sup><sup>r</sup>* or *<sup>K</sup>*2,2 <sup>⊆</sup> *<sup>G</sup><sup>b</sup>* <sup>1</sup> <sup>⊆</sup> *<sup>G</sup>b*, a contradiction. So, <sup>Δ</sup>(*Gg*) = 9. To prove *<sup>δ</sup>*(*Gg*) = 8, assume that *<sup>M</sup>* <sup>=</sup> {*<sup>x</sup>* <sup>∈</sup> *<sup>X</sup>*, <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> 9} and *N* = {*x* ∈ *X*, |*NGg* (*x*)| = 8}; by part (*a*) one can say that |*M*| ≥ 5, if |*M*| = 6, then *<sup>δ</sup>*(*Gg*) <sup>≤</sup> 7—that is, there is a vertex of *<sup>X</sup>* (say *<sup>x</sup>*) such that <sup>|</sup>*NGg* (*x*)| ≤ 7; therefore,

<sup>|</sup>*N*| ≤ 10. If <sup>|</sup>*N*<sup>|</sup> <sup>=</sup> 10, then <sup>|</sup>*E*(*Gg*[*<sup>M</sup>* <sup>∪</sup> *<sup>N</sup>*,*Y*])<sup>|</sup> <sup>=</sup> 134, so by Proposition 1, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*, a contradiction. Now assume that <sup>|</sup>*N*| ≤ 9, thus <sup>|</sup>*E*(*Gg*)| ≤ (<sup>6</sup> <sup>×</sup> <sup>9</sup>)+(<sup>9</sup> <sup>×</sup> <sup>8</sup>)+(<sup>2</sup> <sup>×</sup> <sup>7</sup>) = 140, a contradiction again. For <sup>|</sup>*M*<sup>|</sup> <sup>=</sup> 7 if <sup>|</sup>*N*| ≥ 6, then <sup>|</sup>*E*(*Gg*[*<sup>M</sup>* <sup>∪</sup> *<sup>N</sup>* ,*Y*])| = 111, where *N* ⊆ *N* and |*N* <sup>|</sup> <sup>=</sup> 6, so by Proposition 1, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*, a contradiction. Hence assume that <sup>|</sup>*N*| ≤ 5; therefore, <sup>|</sup>*E*(*Gg*)| ≤ (<sup>7</sup> <sup>×</sup> <sup>9</sup>)+(<sup>5</sup> <sup>×</sup> <sup>8</sup>)+(<sup>5</sup> <sup>×</sup> <sup>7</sup>) = 138, a contradiction again. For <sup>|</sup>*M*<sup>|</sup> <sup>=</sup> 8 if <sup>|</sup>*N*| ≥ 5, then <sup>|</sup>*E*(*Gg*[*<sup>M</sup>* <sup>∪</sup> *<sup>N</sup>* ,*Y*])| = 112, where *N* ⊆ *N* and |*N* | = 5; therefore, by Proposition 1, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*, a contradiction, so assume that <sup>|</sup>*N*| ≤ 4—that is, <sup>|</sup>*E*(*Gg*)| ≤ (<sup>8</sup> <sup>×</sup> <sup>9</sup>)+(<sup>4</sup> <sup>×</sup> <sup>8</sup>)+(<sup>5</sup> <sup>×</sup> <sup>7</sup>) = 139, a contradiction again. For <sup>|</sup>*M*<sup>|</sup> <sup>=</sup> 9 if <sup>|</sup>*N*| ≥ 3, then <sup>|</sup>*E*(*Gg*[*<sup>M</sup>* <sup>∪</sup> *<sup>N</sup>* ,*Y*])| = 105, where *N* ⊆ *N* and |*N* <sup>|</sup> <sup>=</sup> 3, so by Proposition 1, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*, a contradiction. Thus <sup>|</sup>*N*| ≤ 2—that is, <sup>|</sup>*E*(*Gg*)| ≤ (<sup>9</sup> <sup>×</sup> <sup>9</sup>)+(<sup>2</sup> <sup>×</sup> <sup>8</sup>)+(<sup>6</sup> <sup>×</sup> <sup>7</sup>) = 139, which is a contradiction again. For <sup>|</sup>*M*<sup>|</sup> <sup>=</sup> 10, if <sup>|</sup>*N*| ≥ 1, then <sup>|</sup>*E*(*Gg*[*<sup>M</sup>* <sup>∪</sup> *<sup>N</sup>* ,*Y*])| = 98, where *N* ⊆ *N* and |*N* <sup>|</sup> <sup>=</sup> 1; so, by Proposition <sup>1</sup> *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*, a contradiction. Thus, assume that <sup>|</sup>*N*<sup>|</sup> <sup>=</sup> 0, so <sup>|</sup>*E*(*Gg*)| ≤ (<sup>10</sup> <sup>×</sup> <sup>9</sup>)+(<sup>7</sup> <sup>×</sup> <sup>7</sup>) = 139, a contradiction again. Therefore, <sup>|</sup>*M*<sup>|</sup> <sup>=</sup> <sup>5</sup> and <sup>|</sup>*N*<sup>|</sup> <sup>=</sup> 12—that is, *<sup>δ</sup>*(*Gg*) = 8, and part (*b*) is true.

Now, by parts (*a*) and (*b*) it is straightforward to say that *DGg* (*X*) = *DGg* (*Y*) = (9, 9, 9, 9, 9, 8, 8, . . . , 8)—that is, part (*c*) is true, and this completes the proof.

#### **3. Proof of the Main Theorem**

neighbors of *Gg*(*x*) and *Gg*(*x*

In this section, by using the results of Section 2, we will prove the main theorem. Suppose that (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K*17,17, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *G<sup>b</sup>* and *K*3,3 -*Gg*. In the following theorem, we discuss the maximum number of common

**Theorem 4.** *Assume that* (*G<sup>r</sup>* , *Gb*, *Gg*) *is a* 3*-edge coloring of K*17,17*, where K*2,2 - *G<sup>r</sup> , K*2,2 - *G<sup>b</sup> and K*3,3 -*<sup>G</sup>g. Let* <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> <sup>9</sup> *and NGg* (*x*) = *<sup>Y</sup>*1*; the following results are true:*

) for *x*, *x* ∈ *X*.


**Proof of Theorem 4.** Assume that *X* = {*x*1, *x*2, ... , *x*17}, *Y* = {*y*1, *y*2, ... , *y*17} is a partition set of *K* = *K*17,17, and (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K*, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *G<sup>b</sup>* and *K*3,3 - *<sup>G</sup>g*. Without loss of generality (W .l.g.) assume that *<sup>x</sup>* <sup>=</sup> *<sup>x</sup>*<sup>1</sup> and *<sup>Y</sup>*<sup>1</sup> <sup>=</sup> {*y*1, ... , *<sup>y</sup>*9}. To prove part (*a*), by contrast assume that there exists a vertex of *X* \ {*x*1} (say *x*) such that |*NGg* (*x*) ∩ *Y*1| ≥ 6. W.l.g., suppose that *x* = *x*<sup>2</sup> and *Y*<sup>2</sup> = {*y*1, *y*2, ... , *y*6} ⊆ *NGg* (*x*2). Since *K*3,3 - *<sup>G</sup>g*, for each *<sup>x</sup>* <sup>∈</sup> *<sup>X</sup>* \ {*x*1, *<sup>x</sup>*2}, so <sup>|</sup>*NGg* (*x*) <sup>∩</sup> *<sup>Y</sup>*2| ≤ 2—that is, *i*=17 ∑ *i*=1 |*NGg* (*xi*) ∩ *Y*2| ≤ <sup>6</sup> <sup>+</sup> <sup>6</sup> + (<sup>15</sup> <sup>×</sup> <sup>2</sup>) <sup>≤</sup> 42. Now, since <sup>|</sup>*E*(*Gg*[*X*,*Y*2])| ≤ 42, one can check that there exists at

least one vertex of *Y*<sup>2</sup> (say *y*), such that |*NGg* (*y*)| ≤ 7, a contradiction to part (*c*) of Theorem 3. Hence, |*NGg* (*x*) ∩ *Y*1| ≤ 5 for each *x* ∈ *X* \ {*x*1}—that is, part (*a*) is true.

To prove part (*b*), if *n* ≤ 71, then by part (*c*) of Theorem 3, it can be checked that there exists at least one vertex of *Y*<sup>1</sup> (say *y*), such that |*NGg* (*y*)| ≤ 7, a contradiction. Therefore, *n* ≥ 72. Assume that *n* ≥ 74 and let *DGg* (*Y*1)=(*d*1, *d*2, ... , *d*9). Since *i*=17 ∑ *i*=1 |*NGg* (*xi*) ∩ *Y*1| ≥ 74, there exist at least two vertices of *Y*<sup>1</sup> (say *y* , *y*), such that |*NGg* (*y* )| = |*NGg* (*y*)| = 9. Since *n* ≥ 74 and |*X* \ {*x*1}| = 16, there exists at least one vertex of *X* \ {*x*1} (say *x* ), such that |*NGg* (*x* ) ∩ *Y*1| = 5. W.l.g., suppose that *x* = *x*<sup>2</sup> and *NGg* (*x*2) ∩ *Y*<sup>1</sup> = *Y*<sup>2</sup> = {*y*1, ... , *y*5}. Now we have the following claims:

**Claim 1.** *For each x* ∈ *X* \ {*x*1, *x*2}*, we have* |*NGg* (*x*) ∩ *Y*2| = 2 *and DGg* (*Y*2)=(8, 8, 8, 8, 8)*.*

**Proof of Claim 1.** Since *K*3,3 - *<sup>G</sup><sup>g</sup>* for each *<sup>x</sup>* <sup>∈</sup> *<sup>X</sup>* \ {*x*1, *<sup>x</sup>*2}, thus <sup>|</sup>*NGg* (*x*) <sup>∩</sup> *<sup>Y</sup>*2| ≤ 2—that is, *i*=17 ∑ *i*=1 <sup>|</sup>*NGg* (*xi*) <sup>∩</sup>*Y*2| ≤ <sup>5</sup> <sup>+</sup> <sup>5</sup> + (<sup>15</sup> <sup>×</sup> <sup>2</sup>) <sup>≤</sup> 40. Now, since <sup>|</sup>*E*(*Gg*[*X*,*Y*2])| ≤ 40 and <sup>|</sup>*Y*2<sup>|</sup> <sup>=</sup> 5, if there exists a vertex of *X*<sup>1</sup> )(say *x* ), such that <sup>|</sup>*NGg* (*x*) <sup>∩</sup> *<sup>Y</sup>*2| ≤ 1, then <sup>|</sup>*E*(*Gg*[*X*,*Y*2])| ≤

39; therefore, there exists at least one vertex of *Y*<sup>2</sup> (say *y*), such that |*NGg* (*y*)| ≤ 7, a contradiction to part (*c*) of Theorem 3. So, |*NGg* (*x*) ∩ *Y*2| = 2 and ∑ *y*∈*Y*<sup>2</sup> |*NGg* (*y*)| = 40, therefore by part (*c*) of Theorem 3 *DGg* (*Y*2)=(8, 8, 8, 8, 8), and the proof of the claim is

**Claim 2.** *DGg* (*X*1)=(5, 4, 4, ... , 4) *where X*<sup>1</sup> = *X* \ {*x*1}*, in other word* |*NGg* (*xi*) ∩ *Y*1| = 4 *for each i* ∈ {3, 4, . . . , 17}*.*

**Proof of Claim 2.** By contradiction, assume that there exists a vertex of *X* \ {*x*1, *x*2} (say *x*), such that |*NGg* (*x*) ∩ *Y*1| = 5. W.l.g suppose that *x* = *x*<sup>3</sup> and *NGg* (*x*3) ∩ *Y*<sup>1</sup> = *Y*3, now by Claim 1, |*NGg* (*x*3) ∩ *Y*2| = 2. W.l.g., assume that *Y*<sup>3</sup> = {*y*1, *y*2, *y*6, *y*7, *y*8}, thus by Claim 1, *DGg* (*Y*3)=(8, 8, 8, 8, 8)—that is, <sup>|</sup>*NGg* (*y*)<sup>|</sup> <sup>=</sup> 8 for each *<sup>y</sup>* <sup>∈</sup> *<sup>Y</sup>*<sup>1</sup> \ {*y*9}. Since <sup>Δ</sup>(*Gg*) = 9, we can check that *<sup>n</sup>* <sup>=</sup> *<sup>i</sup>*=<sup>17</sup> ∑ *i*=1 <sup>|</sup>*NGg* (*xi*) <sup>∩</sup> *<sup>Y</sup>*1<sup>|</sup> <sup>=</sup> *<sup>i</sup>*=<sup>9</sup> ∑ *i*=1 |*NGg* (*yi*)| ≤ (8 × 8) + 9 = 73, a contradiction. So, *DGg* (*X*1)=(5, 4, 4, . . . , 4), and the proof of the claim is complete.

Assume that *NGg* (*x*2) ∩ *Y*<sup>1</sup> = *Y*<sup>2</sup> = {*y*1. ... , *y*5}, by Claim 1 *DGg* (*Y*2)=(8, 8, 8, 8, 8). Since there exist at lest two vertices of *Y*<sup>1</sup> (say *y* , *y*), such that |*NGg* (*y* )| = |*NGg* (*y*)| = 9, thus *y* , *y* ∈ {*y*6, *y*7, *y*8, *y*9}. W.l.g., we can suppose that *y* = *y*<sup>6</sup> and *NGg* (*y*6) = *X*<sup>2</sup> = {*x*1, *x*3, ... , *x*10}. By Claim 2, |*NGg* (*x*) ∩ *Y*1| = 4 and |*NGg* (*x*) ∩ *Y*2| = 2 for each *x* ∈ *X*<sup>2</sup> \ {*x*1}—that is, |*NGg* (*x*) ∩ {*y*7, *y*8, *y*9}| = 1 for each *x* ∈ *X*<sup>2</sup> \ {*x*1}. Since |*X*<sup>2</sup> \ {*x*1}| = 8 and |*NGg* (*x*) ∩ {*y*7, *y*8, *y*9}| = 1, by the pigeon-hole principle, there exists a vertex of {*y*7, *y*8, *y*9} (say *y*), such that |*NGg* (*y*) ∩ *X*<sup>2</sup> \ {*x*1}| ≥ 3. W.l.g., we can suppose that *y* = *y*<sup>7</sup> and {*x*3, *x*4, *x*5} ⊆ *NGg* (*y*7) ∩ *X*<sup>2</sup> \ {*x*1}. As |*Y*2| = 5 and |*NGg* (*xi*) ∩ *Y*2| = 2 for *i* = 3, 4, 5, there exist *i*, *i* ∈ {3, 4, 5}, such that |*NGg* (*xi*) ∩ *NGg* (*xi*) ∩ *Y*2| = 0. W.l.g., suppose that *i* = 3, *i* <sup>=</sup> 4 and *<sup>y</sup>*<sup>1</sup> <sup>∈</sup> *NGg* (*x*3) <sup>∩</sup> *NGg* (*x*4) <sup>∩</sup> *<sup>Y</sup>*2. Therefore, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*[{*x*1, *<sup>x</sup>*3, *<sup>x</sup>*4}, {*y*1, *<sup>y</sup>*6, *<sup>y</sup>*7}], a contradiction. So, *n* ≤ 73 and the proof of the theorem is complete.

In part (b) of Theorem 4, we showed that 72 <sup>≤</sup> *<sup>n</sup>* <sup>=</sup> *<sup>i</sup>*=<sup>17</sup> ∑ *i*=1 |*NGg* (*xi*) ∩ *Y*1| ≤ 73. Now we consider these two cases independently.

#### *3.1. The Case That n = 73*

complete.

In the following theorem, we prove that in any 3-edge coloring of *K*17,17 (say (*G<sup>r</sup>* , *Gb*, *Gg*), where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *<sup>G</sup>b*), if there exists a vertex of *<sup>V</sup>*(*K*) (say *<sup>x</sup>*), such that <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> <sup>9</sup> and ∑ *xi*∈*X*\{*x*} <sup>|</sup>*NGg* (*xi*) <sup>∩</sup> *NGg* (*x*)<sup>|</sup> <sup>=</sup> 64, then *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*.

**Theorem 5.** *Assume that* (*G<sup>r</sup>* , *Gb*, *Gg*) *is a* 3*-edge coloring of K* = *K*17,17*, such that K*2,2 - *G<sup>r</sup> , K*2,2 - *<sup>G</sup>b. Assume that there exists a vertex of <sup>V</sup>*(*K*) *(say x), such that* <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> <sup>9</sup>*. If i*=17 ∑ *i*=1 <sup>|</sup>*NGg* (*xi*) <sup>∩</sup> *<sup>Y</sup>*1<sup>|</sup> <sup>=</sup> <sup>73</sup> *where Y*<sup>1</sup> <sup>=</sup> *NGg* (*x*)*, then K*3,3 <sup>⊆</sup> *<sup>G</sup>g.*

**Proof of Theorem 5.** By contradiction, assume that *K*3,3 - *Gg*. Therefore, by Theorem 3 and Theorem 4, we have the following results:


each *x* ∈ *A*.

Assume that *X* = {*x*1, *x*2, ... , *x*17}, *Y* = {*y*1, *y*2, ... , *y*17} is the partition set of *K* = *K*17,17, and (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K*, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *G<sup>b</sup>* and *K*3,3 -*Gg*. W.l.g., assume that *<sup>x</sup>* <sup>=</sup> *<sup>x</sup>*1, *<sup>Y</sup>*<sup>1</sup> <sup>=</sup> {*y*1, *<sup>y</sup>*2, ... , *<sup>y</sup>*9}, and *<sup>n</sup>* <sup>=</sup> *<sup>i</sup>*=<sup>17</sup> ∑ *i*=1 |*NGg* (*xi*) ∩ *Y*1| = 73. Since *n* = 73, by (*c*) we can say that *DGg* (*Y*1)=(*d*1, *d*2, ... , *d*9)=(9, 8, 8, ... , 8)—that is, there exists a vertex of *Y*<sup>1</sup> (say *y*), such that |*NGg* (*y*)| = 9. By (d), |*NGg* (*x*1) ∩ *NGg* (*x*)| ≤ 5 for each *x* ∈ *X*{\*x*1}. Set *C* = {*x* ∈ *X*, |*NGg* (*x*) ∩ *NGg* (*x*1)| = 5}. Now by argument similar to the proof of Claim 1, we have the following claim:

**Claim 3.** *Assume that x* ∈ *C and NGg* (*x*) ∩ *Y*<sup>1</sup> = *Y , then for each x* ∈ *X* \ {*x*1, *x*}*, we have* |*NGg* (*x* ) ∩ *Y* | = 2 *and DGg* (*Y* )=(8, 8, 8, 8, 8)*.*

Here there exists a claim about |*C*| as follows:

**Claim 4.** |*C*| ≤ 2*.*

**Proof of Claim 4.** By contradiction, assume that |*C*| ≥ 3. W.l.g., suppose that {*x*2, *x*3, *x*4} ⊆ *C* and *NGg* (*x*2) ∩ *Y*<sup>1</sup> = *Y*<sup>2</sup> = {*y*1, ... , *y*5}. By Claim 3, |*NGg* (*x*) ∩ *Y*2| = 2 for each *x* ∈ *X* \ {*x*1, *x*2}. W.l.g., suppose that *NGg* (*x*3) ∩ *Y*<sup>1</sup> = *Y*<sup>3</sup> = {*y*1, *y*2, *y*6, *y*7, *y*8}. Since *x*<sup>4</sup> ∈ *C* and |*NGg* (*x*4) ∩ *Yi*| = 2 for *i* = 2, 3, *y*<sup>9</sup> ∈ *NGg* (*x*4) ∩ *Y*1. Hence, for each *y* ∈ *Y*1, there is at least one *i* ∈ {2, 3, 4} such that *y* ∈ *NGg* (*xi*); therefore, by Claim 3, *DGg* (*Y*1) = (8, 8, 8, 8, 8, 8, 8, 8, 8), which is in contrast to *i*=17 ∑ *i*=1 <sup>|</sup>*NGg* (*xi*) <sup>∩</sup> *<sup>Y</sup>*1<sup>|</sup> <sup>=</sup> *<sup>i</sup>*=<sup>9</sup> ∑ *i*=1 |*NGg* (*yi*)| = 73, so |*C*| ≤ 2.

Now by considering |*C*| there are three cases as follows:

Case 1: |*C*| = 0. Since *n* = 73, |*Y*1| = 9 and |*C*| = 0, *DGg* (*X* \ {*x*1})=(4, 4, ... , 4, 4), *DGg* (*Y*1)=(9, 8, 8, 8, 8, , 8, 8, 8, 8), *i*=17 ∑ *i*=1 |*NGg* (*xi*)∩*Y* | = 68, and *DGg* (*Y* )=(9, 9, 9, 9, 8, 8, 8, 8), where *Y* = *Y* \ *Y*1. Set *B* = {*y* ∈ *Y* , |*NGg* (*y*)| = 9}, so |*B*| = 4. Now we are ready to prove the following claim:

**Claim 5.** *There exists a vertex of A* \ {*x*1} *(say x), such that:*

$$\sum\_{y \in N\_{G\mathcal{S}}(\ge)} |N\_{G^{\mathcal{S}}}(y)| \ge 74,$$

*in which A* = {*x* ∈ *X*, |*NGg* (*x*)| = 9}*.*

**Proof of Claim 5.** *DGg* (*X*1)=(4, 4, ... , 4, 4) and *DGg* (*Y*1)=(9, 8, 8, 8, 8, , 8, 8, 8, 8) for each *x* ∈ *A* \ {*x*1}; thus:

$$\sum\_{y \in N\_{G\mathcal{S}}(\mathfrak{x}) \cap Y\_1} |N\_{G\mathcal{S}}(y)| \ge 32$$

As |*Y* | = 8, |*B*| = 4, and |*NGg* (*xi*) ∩ *Y* | = 5 for each *x* ∈ *A* \ {*x*1}, there exists at least one vertex of *<sup>A</sup>* \ {*x*1} (say *<sup>x</sup>*), such that <sup>|</sup>*NGg* (*x*) <sup>∩</sup> *<sup>B</sup>*| ≥ 2, otherwise *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*[*A*,*Y* \ *<sup>B</sup>*], a contradiction. Hence, w.l.g., suppose that *x*<sup>2</sup> ∈ *A*, where |*NGg* (*x*2) ∩ *B*| ≥ 2. So:

$$\sum\_{y \in N\_{G\mathcal{S}}(\alpha\_2) \cap Y'} |N\_{G\mathcal{S}}(y)| \ge 42.$$

That is,

$$\sum\_{\mathcal{Y}\in\mathcal{N}\_{\mathcal{G}\mathcal{F}}(\mathbf{x}\_{2})} |\mathcal{N}\_{\mathcal{G}\mathcal{F}}(\mathcal{y})| = \sum\_{\mathcal{Y}\in\mathcal{N}\_{\mathcal{G}\mathcal{F}}(\mathbf{x}\_{2})\cap Y'} |\mathcal{N}\_{\mathcal{G}\mathcal{F}}(\mathcal{y})| + \sum\_{\mathcal{Y}\in\mathcal{N}\_{\mathcal{G}\mathcal{F}}(\mathbf{x})\cap Y\_{1}} |\mathcal{N}\_{\mathcal{G}\mathcal{F}}(\mathcal{y})| \ge 42 + 32 = 74.7$$

Now by considering *<sup>x</sup>*<sup>2</sup> and *NGg* (*x*2) and by (e) (or part (b) of Theorem 4) *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*, a contradiction again.

Case 2: |*C*| = 1. W.l.g., suppose that *C* = {*x*2}, *NGg* (*x*2) ∩ *Y*<sup>1</sup> = *Y*<sup>2</sup> = {*y*1, ... , *y*5}. By Claim 3, |*NGg* (*x*2) ∩ *NGg* (*x*) ∩ *Y*1| = 2 for each *x* ∈ *X* \ {*x*1, *x*2} and |*NGg* (*yi*)| = 8 for each *i* ∈ {1, 2, ... , 5}. Since there exists a vertex of *Y*<sup>1</sup> named *y*, such that |*NGg* (*y*)| = 9, w.l.g. we can suppose that *y* = *y*<sup>6</sup> and *NGg* (*y*6) = {*x*1, *x*3, *x*<sup>4</sup> ... , *x*10}. Since *n* = 73 and |*C*| = 1, *DGg* (*X*1)=(5, 4, 4, ... , 4, 3)—that is, there exist at least seven vertices of *NGg* (*y*6) \ {*x*1} (say *X*<sup>3</sup> = {*x*3, *x*<sup>4</sup> ... , *x*9}), such that |*NGg* (*x*) ∩ *Y*1| = 4 for each *x* ∈ *X*3. Since |*X*3| = 7, |*Y*2| = 5, |*NGg* (*x*) ∩ *Y*1| = 4 and |*NGg* (*xi*) ∩ *Y*2| = 2 for each *x* ∈ *X*3, |*NGg* (*x*) ∩ {*y*7, *y*8, *y*9}| = 1 for each *x* ∈ *X*3. Therefore, by the pigeon-hole principle there exists a vertex of {*y*7, *y*8, *y*9} (say *y* ), such that |*NGg* (*y* ) ∩ *X*3| ≥ 3. W.l.g., suppose that *y* = *y*<sup>7</sup> and {*x*3, *x*4, *x*5} ⊆ *NGg* (*y*7). Therefore, since |*Y*2| = 5, there exists *i*, *i* ∈ {3, 4, 5} such that |*NGg* (*xi*) ∩ *NGg* (*xi*) ∩ *Y*2| = 0. W.l.g., suppose that *i* = 3, *i* = 4 and *<sup>y</sup>*<sup>1</sup> <sup>∈</sup> *NGg* (*x*3) <sup>∩</sup> *NGg* (*x*4) <sup>∩</sup> *<sup>Y</sup>*2. Therefore, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*[{*x*1, *<sup>x</sup>*3, *<sup>x</sup>*4}, {*y*1, *<sup>y</sup>*6, *<sup>y</sup>*7}], which is a contradiction.

Case 3: |*C*| = 2. W.l.g., suppose that *C* = {*x*2, *x*3}, *NGg* (*x*2) ∩ *Y*<sup>1</sup> = *Y*<sup>2</sup> = {*y*1, ... , *y*5}. By Claim 3, |*NGg* (*x*2) ∩ *NGg* (*x*3) ∩ *Y*1| = 2. So, w.l.g. we can suppose that *NGg* (*x*3) ∩ *Y*<sup>1</sup> = *Y*<sup>3</sup> = {*y*1, *y*2, *y*6, *y*7, *y*8}. Now, by Claim 3, |*NGg* (*yi*)| = 8 for each *i* ∈ {1, 2, ... , 8}. Since there is a vertex of *Y*<sup>1</sup> named *y*, such that |*NGg* (*y*)| = 9, *y* = *y*9. W.l.g., we can assume that *NGg* (*y*9) = *X*<sup>2</sup> = {*x*1, *x*4, *x*<sup>5</sup> ... , *x*11}. Since *n* = 73 and |*C*| = 2, *DGg* (*X*1) = (5, 5, 4, 4, ... , 4, 3, 3)—that is, there exist two vertices of *X* (say *x*, *x* ), such that |*NGg* (*x*) ∩ *Y*1| = 3. If |*NGg* (*y*9) ∩ {*x*, *x* }| ≤ 1, then there exist at least seven vertices of *NGg* (*y*9) \ {*x*1}, such that |*NGg* (*x*) ∩ *Y*1| = 4; in this case, the proof is the same as Case 1. Hence, assume that *x*, *x* ∈ *NGg* (*y*9). Since |*NGg* (*x*) ∩ *Y*2| = |*NGg* (*x* ) ∩ *Y*2| = 2, one can check that |*NGg* (*x*) ∩ {*y*6, *y*7, *y*8}| = |*NGg* (*x* ) ∩ {*y*6, *y*7, *y*8}| = 0. Assume that *Xi* = *NGg* (*yi*) for *i* = 6, 7, 8. Since |*Xi*| = 8 and *x*, *x* ∈/ *Xi*, then for each *x* ∈ *Xi* \ {*x*1} we have |*NGg* (*x*) ∩ *Y*1| = 4. Therefore, by considering *Xi* \ {*x*1} and *yi* for each *i* ∈ {6, 7, 8}, the proof is the same as Case 1 and *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*, a contradiction again.

Therefore, by Cases 1, 2, and 3 the assumption does not hold—that is, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup><sup>g</sup>* and this completes the proof of the theorem.

#### *3.2. The Case That n = 72*

In the following theorem, we prove that in any 3-edge coloring of *K*17,17 (say (*G<sup>r</sup>* , *Gb*, *Gg*), where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *<sup>G</sup>b*), if there exists a vertex of *<sup>V</sup>*(*K*) (say *<sup>x</sup>*), such that <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> <sup>9</sup> and ∑ *xi*∈*X*\{*x*} <sup>|</sup>*NGg* (*xi*) <sup>∩</sup> *NGg* (*x*)<sup>|</sup> <sup>=</sup> 63, then *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*.

**Theorem 6.** *Assume that* (*G<sup>r</sup>* , *Gb*, *Gg*) *is a* 3*-edge coloring of K* = *K*17,17*, where K*2,2 - *G<sup>r</sup> , K*2,2 - *<sup>G</sup>b. Suppose that there exists a vertex of <sup>V</sup>*(*K*) *(say x), such that* <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> <sup>9</sup>*. If i*=17 ∑ *i*=1 <sup>|</sup>*NGg* (*xi*) <sup>∩</sup> *<sup>Y</sup>*1<sup>|</sup> <sup>=</sup> <sup>72</sup>*, where Y*<sup>1</sup> <sup>=</sup> *NGg* (*x*)*, then K*3,3 <sup>⊆</sup> *<sup>G</sup>g.*

**Proof of Theorem 6.** By contradiction, assume that *K*3,3 - *Gg*. Therefore, by Theorems 3 and 4, we have the following results:


each *x* ∈ *A*. Assume that *X* = {*x*1, *x*2, ... , *x*17}, *Y* = {*y*1, *y*2, ... , *y*17} is a partition set of *K* = *K*17,17, and (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K*, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *G<sup>b</sup>* and *K*3,3 - *Gg*. W.l.g., assume that *<sup>x</sup>* <sup>=</sup> *<sup>x</sup>*1, *<sup>Y</sup>*<sup>1</sup> <sup>=</sup> {*y*1, *<sup>y</sup>*2, ... , *<sup>y</sup>*9}, and *<sup>n</sup>* <sup>=</sup> *<sup>i</sup>*=<sup>17</sup> ∑ *i*=1 |*NGg* (*xi*) ∩ *Y*1| = 72. Since

*n* = 73, by (*c*) we can say that *DGg* (*Y*1)=(*d*1, *d*2, ... , *d*9)=(8, 8, 8, ... , 8). Set *C* = {*x* ∈ *X*, |*NGg* (*x*) ∩ *NGg* (*x*1)| = 5}. Define *D* and *E* as follows:

$$D = \{ \mathbf{x} \in X \mid \{ \mathbf{x}\_1 \} \text{ } such \text{ that } |\mathbf{N}\_{\mathbf{G}^\mathbb{Z}}(\mathbf{x}) \cap Y\_1| = \mathbf{5} \}$$

*E* = {*x* ∈ *X* \ {*x*1}, *such that* |*NGg* (*x*) ∩ *Y*1| = 3}.

Here we have a claim about |*D*| and |*E*| as follows:

**Claim 6.** |*D*| ≤ 3 *and* |*E*| ≤ 4*.*

**Proof of Claim 6.** By contradiction, suppose that |*D*| ≥ 4. W.l.g., assume that {*x*2, *x*3, *x*4, *x*5} ⊆ *D*, *NGg* (*x*2) ∩ *Y*<sup>1</sup> = *Y*<sup>2</sup> = {*y*1, ... , *y*5}. Now, by Claim 3, |*NGg* (*x*) ∩ *Y*2| = 2 for each *x* ∈ *X* \ {*x*1, *x*2}. W.l.g., we can suppose that *NGg* (*x*3) ∩ *Y*<sup>1</sup> = *Y*<sup>3</sup> = {*y*1, *y*2, *y*6, *y*7, *y*8}. Consider *NGg* (*xi*) ∩ *Y*1(*i* = 4, 5). Since |*NGg* (*xi*) ∩ *Yj*| = 2 (*i* = 4, 5, *j* = 2, 3) and *xi* ∈ *A*, |*NGg* (*xi*) ∩ {*y*3, *y*4, *y*5}| = 2, |*NGg* (*xi*) ∩ {*y*6, *y*7, *y*8}| = 2, and *y*<sup>9</sup> ∈ *NGg* (*xi*) for *i* = 4, 5; otherwise, if there exists a vertex of {*x*4, *x*5} (say *x*), such that |*NGg* (*xi*) ∩ {*y*1, *<sup>y</sup>*2}| <sup>=</sup> 2, then *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*[{*x*1, *xi*, *<sup>x</sup>*},*Y*1] for some *<sup>i</sup>* ∈ {1, 2}, a contradiction. Therefore, since |{*y*3, *y*4, *y*5}| = |{*y*6, *y*7, *y*8}| = 3 and *x*4, *x*<sup>5</sup> ∈ *A*, by the pigeon-hole principle |*NGg* (*x*4) ∩ *NGg* (*x*5) ∩ {*y*3, *y*4, *y*5}| ≥ 1 and |*NGg* (*x*4) ∩ *NGg* (*x*5) ∩ {*y*6, *y*7, *y*8}| ≥ 1. W.l.g., we can suppose that *y*3, *y*<sup>6</sup> ∈ *NGg* (*x*4) ∩ *NGg* (*x*5), since *y*<sup>9</sup> ∈ *NGg* (*x*4) ∩ *NGg* (*x*5), so *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*[{*x*1, *<sup>x</sup>*4, *<sup>x</sup>*5}, {*y*3, *<sup>y</sup>*6, *<sup>y</sup>*9}], a contradiction. Therefore, <sup>|</sup>*D*| ≤ 3. Now, as *i*=17 ∑ *i*=2 |*NGg* (*xi*) ∩ *Y*1| = 63 and |*D*| ≤ 3, we can say that |*E*| ≤ 4 and the proof of the

claim is complete.

Now, by considering |*D*|, there are three cases as follows:

Case 1: |*D*| = 0. Since *n* = 72 and |*D*| = 0, *DGg* (*X* \ {*x*1})=(4, 4, ... , 4, 3), *DGg* (*Y*1) = (8, 8, 8, 8, 8, 8, 8, 8, 8), *i*=17 ∑ *i*=1 |*NGg* (*xi*) ∩ *Y* | = 69 and *DGg* (*Y* )=(9, 9, 9, 9, 9, 8, 8, 8), where *Y* = *Y* \ *Y*1. Set *B* = {*y* ∈ *Y* , |*NGg* (*y*)| = 9}, hence |*B*| = 5. Now, we have the following claim:

**Claim 7.** *There exists a vertex of A* \ {*x*1} *(say x), such that:*

$$\sum\_{\mathcal{y}\in N\_{G\mathcal{S}}(\mathbf{x})} |N\_{G\mathcal{S}}(\mathcal{y})| \ge 75\mu$$

*in which A* = {*x* ∈ *X*, |*NGg* (*x*)| = 9}*.*

**Proof of Claim 7.** Since *DGg* (*X*1)=(4, 4, ... , 4, 3) and *DGg* (*Y*1)=(8, 8, 8, 8, 8, , 8, 8, 8, 8), so for at least three vertices of *A* \ {*x*1},

$$\sum\_{y \in N\_{G\mathcal{S}}(x) \cap Y\_1} |N\_{G\mathcal{S}}(y)| \ge 32.1$$

Therefore, since |*NGg* (*xi*) ∩ *Y* | = 5 for each *x* ∈ *A* \ {*x*1} and *DGg* (*Y* ) = (9, 9, 9, 9, 9, 8, 8, 8), there exists at least one vertex of *A* \ {*x*1} (say *x*), such that |*NGg* (*x*) ∩ *B*| ≥ 3; otherwise, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*[*A*,*Y* \ *<sup>B</sup>*], a contradiction. Hence, w.l.g., suppose that *<sup>x</sup>*<sup>2</sup> <sup>∈</sup> *<sup>A</sup>* and |*NGg* (*x*2) ∩ *B*| ≥ 3; therefore:

$$\sum\_{y \in N\_{G\overline{\mathbb{K}}}(\mathbf{x}) \cap Y'} |N\_{G\overline{\mathbb{K}}}(y)| \ge 3 \times 9 + 2 \times 8 = 43.$$

That is, we have:

$$\sum\_{\mathcal{Y}\in N\_{\mathcal{G}\mathcal{Y}}(\mathbf{x}\_{2})} |N\_{\mathcal{G}\mathcal{X}}(\mathcal{y})| = \sum\_{\mathcal{Y}\in N\_{\mathcal{G}\mathcal{Y}}(\mathbf{x}\_{2})\cap Y'} |N\_{\mathcal{G}\mathcal{X}}(\mathcal{y})| + \sum\_{\mathcal{Y}\in N\_{\mathcal{G}\mathcal{Y}}(\mathbf{x})\cap Y\_{1}} |N\_{\mathcal{G}\mathcal{Y}}(\mathcal{y})| \ge 43 + 32 = 75.$$

Now, by considering *x*<sup>2</sup> and *NGg* (*x*2) and by (*e*) (or by part (*b*) of Theorem 4), *K*3,3 ⊆ *Gg*, a contradiction again.

Case 2: |*D*| = 1 (for the case that |*D*| = 2, the proof is same). W.l.g., assume that *D* = {*x*2}, *NGg* (*x*2) ∩ *Y*<sup>1</sup> = *Y*<sup>2</sup> = {*y*1, ... , *y*5}. Since *n* = 72, |*D*| = 1 and |*NGg* (*x*) ∩ *Y*1| ≤ 5, |*E*| = 2. As |*NGg* (*x*) ∩ *Y*2| = 2 for each *x* ∈ *X* \ {*x*1, *x*2} and |*E*| = 2, there exists a vertex of {*y*6, *y*7, *y*8, *y*9} (say *y*), such that for each vertex of *NGg* (*y*) ∩ *X* \ {*x*1} (say *x*), |*NGg* (*x*) ∩ *Y*1| = 4. W.l.g., we can suppose that *y* = *y*6, *NGg* (*y*6) ∩ *X* \ {*x*1} = {*x*3, *x*4, ... , *x*9}. Since |*NGg* (*y*6) ∩ *X* \ {*x*1}| = 7 and |*NGg* (*x*) ∩ *Y*2| = 2 for each *x* ∈ *NGg* (*y*6) ∩ *X* \ {*x*1}, |*NGg* (*x*) ∩ {*y*7, *y*8, *y*9}| = 1. Therefore, by the pigeon-hole principle there exists a vertex of {*y*7, *y*8, *y*9} (say *y* ), such that |*NGg* (*y*6) ∩ *NGg* (*y* ) ∩ *X* \ {*x*1}| ≥ 3. W.l.g., suppose that *y* = *y*<sup>7</sup> and {*x*3, *x*4, *x*5} ⊆ *NGg* (*y*6) ∩ *NGg* (*y*7) ∩ *X* \ {*x*1}. Therefore, since |*Y*2| = 5 and |*NGg* (*x*) ∩ *Y*2| = 2, there exist at least two vertices of {*x*3, *x*4, *x*5} (say *x* , *x*), such that |*NGg* (*x* ) ∩ *NGg* (*x*) ∩ *Y*2| = 0. W.l.g., suppose that *x* = *x*3, *x* = *x*<sup>4</sup> and *y*<sup>1</sup> ∈ *NGg* (*x*3) <sup>∩</sup> *NGg* (*x*4). Therefore, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*[{*x*1, *<sup>x</sup>*3, *<sup>x</sup>*4}, {*y*1, *<sup>y</sup>*6, *<sup>y</sup>*7}], a contradiction.

Case 3: |*D*| = 3. W.l.g., suppose that *D* = {*x*2, *x*3, *x*4}, *NGg* (*x*2) ∩ *Y*<sup>1</sup> = *Y*<sup>2</sup> = {*y*1, ... , *y*5}. By Claim 3, |*NGg* (*x*2) ∩ *NGg* (*x*3) ∩ *Y*1| = 2. W.l.g., we can assume that *NGg* (*x*3) ∩ *Y*<sup>1</sup> = *Y*<sup>3</sup> = {*y*1, *y*2, *y*6, *y*7, *y*8}. Since *x*<sup>4</sup> ∈ *D* and |*NGg* (*x*4) ∩ *Yi*| = 2 for *i* = 2, 3, *y*<sup>9</sup> ∈ *NGg* (*x*4). If |*NGg* (*x*4) ∩ {*y*1, *y*2}| = 0, as |*NGg* (*x*2) ∩ *NGg* (*x*4) ∩ *Y*1| = 2 and *x*<sup>4</sup> ∈ *D*, one can check that <sup>|</sup>*NGg* (*x*4) ∩ {*y*6, *<sup>y</sup>*7, *<sup>y</sup>*8}| <sup>=</sup> 2—that is, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*[{*x*1, *<sup>x</sup>*3, *<sup>x</sup>*4},*Y*1], a contradiction. Hence, |*NGg* (*x*4) ∩ {*y*1, *y*2}| = 0. Therefore, |*NGg* (*x*4) ∩ {*y*3, *y*4, *y*5}| = 2 and |*NGg* (*x*4) ∩ {*y*6, *y*7, *y*8}| = 2. W.l.g., we can suppose that *NGg* (*x*4) ∩ *Y*<sup>1</sup> = *Y*<sup>4</sup> = {*y*3, *y*4, *y*6, *y*7, *y*9}. Since |*D*| = 3, so |*E*| = 4. W.l.g., suppose that *E* = {*x*5, *x*6, *x*7, *x*8}. Here, we have a claim as follows:

**Claim 8.** |*NGg* (*y*9) ∩ *E*| = 0*.*

**Proof of Claim 8.** By contradiction, suppose that |*NGg* (*y*9) ∩ *E*| = 0. Assume that *x*<sup>5</sup> ∈ *NGg* (*y*9) <sup>∩</sup> *<sup>E</sup>*—that is, *<sup>x</sup>*5*y*<sup>9</sup> <sup>∈</sup> *<sup>E</sup>*(*Gg*). Since *<sup>x</sup>*<sup>5</sup> <sup>∈</sup> *<sup>E</sup>* and {*x*2, *<sup>x</sup>*3, *<sup>x</sup>*4} <sup>=</sup> *<sup>D</sup>*, by Claim 3, |*NGg* (*x*5) ∩ *NGg* (*xi*)| = |*NGg* (*x*5) ∩ *Yi*| = 2 for *i* = 2, 3, 4. Consider *NGg* (*x*5) ∩ *Y*2, assume that *NGg* (*x*5) ∩ *Y*<sup>2</sup> = {*y* , *y*}, if {*y* , *y*} = {*y*1, *y*2}, then |*NGg* (*x*5) ∩ *Y*4| = 1, a contradiction. Therefore, we can assume that |{*y* , *y*}∩{*y*1, *y*2}| ≤ 1. If |{*y* , *y*}∩{*y*1, *y*2}| = 0, then |*NGg* (*x*5) ∩ *Y*3| = 0, and if |{*y* , *y*}∩{*y*1, *y*2}| = 1, then |*NGg* (*x*5) ∩ *Y*3| ≤ 1. In any case there exists a vertex of *D* (say *x* ), such that |*NGg* (*x*5) ∩ *NGg* (*x* )| = 1, a contradiction. So, the assumption does not hold and the claim is true.

Therefore, by Claim 8, since |*NGg* (*y*9) ∩ *D*| = 0, we can say that for any vertex of *NGg* (*y*9) ∩ *X* \ {*x*1} (say *x*), |*NGg* (*x*) ∩ *Y*1| ≥ 4; therefore, by considering *Y*<sup>2</sup> and *y*9, as |*NGg* (*y*9) ∩ *X* \ {*x*1}| = 7 and |*NGg* (*x*) ∩ *Y*1| ≥ 4 for each *x* ∈ *NGg* (*y*9) ∩ *X* \ {*x*1}, the proof is similar to Case 1, a contradiction.

Therefore, by Cases 1, 2, and 3 the assumption does not hold—that is, *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup><sup>g</sup>* and the proof of the theorem is complete.

Now, combining Theorems 3–6 yields the proof of Theorem 1.

#### **4. Discussion**

There are several papers in which the bipartite Ramsey numbers have been studied. In this paper, we proved the conjecture on *B*(2, 2, 3), which was proposed in 2015 and states that *B*(2, 2, 3) = 17. We proved this conjecture by a combinatorial argument with no computer calculations. This is significant because computing the exact value of Ramsey numbers is a challenge. To approach the proof of this conjecture, we proved four theorems as follows:

1. Assume that (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K*17,17, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *G<sup>b</sup>* and *K*3,3 -*Gg*. Hence, we have:

(a) <sup>|</sup>*E*(*Gg*)<sup>|</sup> <sup>=</sup> 141.


2. Assume that (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K*17,17, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *G<sup>b</sup>* and *K*3,3 -*<sup>G</sup>g*. Let <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> 9 and *NGg* (*x*) = *<sup>Y</sup>*1, the following results are true:


3. Assume that (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K* = *K*17,17, such that *K*2,2 - *G<sup>r</sup>* , *K*2,2 -*<sup>G</sup>b*. Assume that there exists a vertex of *<sup>V</sup>*(*K*) (say *<sup>x</sup>*), such that <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> 9. If

$$\sum\_{i=1}^{i=17} \left| N\_{G^{\mathbb{Z}}}(\boldsymbol{\chi}\_{i}) \cap Y\_{1} \right| = 73 \text{, where } Y\_{1} = N\_{G^{\mathbb{Z}}}(\boldsymbol{\chi}) \text{, then } K\_{3,3} \subseteq G^{\mathbb{Z}}.$$

4. Assume that (*G<sup>r</sup>* , *Gb*, *Gg*) is a 3-edge coloring of *K* = *K*17,17, where *K*2,2 - *G<sup>r</sup>* , *K*2,2 - *<sup>G</sup>b*. Assume that there exists a vertex of *<sup>V</sup>*(*K*) (say *<sup>x</sup>*), such that <sup>|</sup>*NGg* (*x*)<sup>|</sup> <sup>=</sup> 9. If *i*=17 ∑ *i*=1 <sup>|</sup>*NGg* (*xi*) <sup>∩</sup> *<sup>Y</sup>*1<sup>|</sup> <sup>=</sup> 72, where *<sup>Y</sup>*<sup>1</sup> <sup>=</sup> *NGg* (*x*), then *<sup>K</sup>*3,3 <sup>⊆</sup> *<sup>G</sup>g*.

One might also be able to compute *B*(*n*1, ... , *nm*) for small *i*, *ni* like *B*(2, 3, 3, 3) or *B*(3, 3, 3, 3) in the future, using the idea of proofs laid out in this paper.

**Author Contributions:** Conceptualization, Y.R. and M.G.; Formal analysis, Y.R., M.G. and S.S.; Funding acquisition, S.S.; Investigation, Y.R.; Methodology, Y.R., M.G. and S.S.; Resources, Y.R.; Supervision, S.S.; Validation, Y.R.; Writing(original draft), Y.R. and M.G.; Writing(review and editing), M.G. All authors have read and agreed to the published version of the manuscript.

**Funding:** There was no funding for this work.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** This paper focuses on pure graph theory, not involving experiments and data.

**Acknowledgments:** The authors would like to thank the editors and reviewers.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**

