The Problem about 2-coloring finite plane Suppose we color  a   $X \times X$ finite plane by red and blue arbitrarily. How large does X need to be to guarantee a  monochromatic combinatorial square $k \times k$ 
1 0 1 0 1 1
1 1 1 1 1 1 
1 0 1 0 1 1
1 1 1 1 1 1
The above figure shows a combinatorial $2 \times 2$ filled  square filled by zeros.
 A: The exact answer, $15$, to this question is the content of my paper with Shalom Eliahou:
Here a copy of the corresponding entry of Math-Review:
Bacher, Roland; Eliahou, Shalom
Extremal binary matrices without constant 2-squares
J. Comb. 1 (2010), no. 1, [ISSN 1097-959X on cover], 77–100.
05D10 (11B75)
Summary: "In this paper we solve, by computational means, an open problem of Erickson: Let $[n]=\{1,…,n\}$; what is the smallest integer $n_0$ such that, for every $n\ge n_0$ and every 2-coloring of the grid $[n]\times[n]$, there is a constant 2-square, i.e. a $2\times2$ subgrid $S=\{i,i+t\}\times\{j,j+t\}$ whose four points are colored the same? It has been shown recently that $13\le n_0\le\min(W(2,8),5\cdot2240)$, where $W(2,8)$ is the still unknown eighth classical van der Waerden number. We obtain here the exact value $n_0=15$. In the process, we display 2-colorings of $[13]\times{\bf Z}$ and $[14]\times[14]$ without constant 2-squares, and show that this is best possible.'' 
A: From http://csce.uark.edu/~dapon/thesis.pdf, 13 isn't large enough to guarantee a monochromatic square.
0000001001111
0101100101010
0011001111001
1110100010011
1011111001000
0110010011110
1101001010100
1000011110010
1011000100111
0001110010101
0101011000011
1100010101001
0110111100100

