Every point of a grid is colored in blue, red or green. How to prove there is a monochromatic rectangle? I have a $3$-coloring of $\mathbb{Z}\times\mathbb{Z}$, i.e. a function $f:\mathbb{Z}\times\mathbb{Z}\to\{\color{red}{\text{red}},\color{green}{\text{green}},\color{blue}{\text{blue}}\}$.
I have to prove that there is a monochromatic rectangle with its sides being parallel to the axis, i.e. to prove that for some choice of $a,b,c,d\in\mathbb{Z}$ with $a\neq b$ and $c\neq d$, all the points $(a,c),(a,d),(b,c),(b,d)\,$ have the same color.
I tried to work by contradiction, without achieving much.
Additionally, can we prove some upper bound on $|a-b|$ and $|c-d|$?
 A: For a slightly more quantitative proof, we may show that there is a monochromatic rectangle with its largest dimension being $\color{red}{\leq 82}$ and its smallest dimension being $\color{red}{\leq 4}$.
Step 1. We may give to any integer point $(x,0)$ on the $x$-axis a maxi-color in the set $\{1,2,\ldots,80,81\}$ depending on the actual colors of $(x,0),(x,1),(x,2),(x,3)$.
Step 2. By the Dirichlet box principle, among $82$ consecutive lattice points on the $x$-axis at least two of them, say $(a,0)$ and $(b,0)$, have the same maxi-color.
Step 3. Since we have only three actual colors, at least two points among $(a,0),(a,1),(a,2),(a,3)$ have the same color. Such points, coupled with the corresponding points on $x=b$, give the wanted monochromatic rectangle.

The $82$ above can be replaced by a smaller number, namely $1+\alpha(G)$, i.e. one plus the size of the largest independent set in a quite complicated graph (of the Haggkvist-Hell type) on $81$ vertices. So we may also improve the given bound by using some rather deep results about topological graphs (see Kneser's conjecture, proved by Lovasz in 1978 through the Borsuk-Ulam theorem).
However, an elementary proof that $\alpha(G)=18$ is presented above by Joffan.
A: Such a rectangle will occur in a grid of $19$ columns by $4$ rows.
By the pigeonhole principle, each column must have a repeated colour point. Ignoring any later repeats, classify the columns according to the first two repeat colour positions; there are $6$ options: $(1,2), (1,3), (1,4), (2, 3), (2, 4)$ and $(3,4)$. So since there are also $3$ colour options for the repeated colours there are only $6\times3=18$ different options for repeated colour by position. Therefore, by the pigeonhole principle again, in a $19\times 4$ grid we must have a suitable rectangle with identically coloured corners.

From an observation by Pere in comments: 
For $n$ colours, using the same approach, we would adopt a grid of $(n+1)$ rows and then require $n{n+1 \choose 2}+1 = n(n+1)n/2 +1 = (n^3+n^2)/2 + 1$ columns to find a rectangle with same-coloured corners. 
A: Theorem: Let $X,Y$ be infinite sets. If the points of $X\times Y$ are colored with a finite number of colors in such a way that there exists a set $C$ of $c$ colors such that for every $x\in X$ only colors in $C$ appear an infinite number of times among pairs of the form $(x,y)$, then there exist $x_1,x_2\in X,y_1,y_2\in Y$ so that $(x_1,y_1),(x_1,y_2),(x_2,y_1),(x_2,y_2)$ all have the same color.
Proof:
We proceed by induction on $c$, when $c=1$ it is clear. Take any $x_1,x_2$ and let $\alpha$ be the only color that appears an infinite number of times. Notice that the set of $A_{x_1}=\{y\in Y | (x_1,y) \text{ has color }  \alpha\}$ has finite complement, and so does the set $A_{x_2}=\{y\in Y | (x_2,y) \text{ has color }  \alpha\}$. This implies $A_{x_1}\cap A_{x_2}$ is infinite. Taking $x1,x2$ and any $y_1,y_2\in A_{x_1}\cap A_{x_2}$ does the trick.
Inductive step: Consider any $x\in X$, there must be an $\alpha \in Y$ such that the set $A_{x}=\{y\in Y | (x,y) \text{ has color }  \alpha\}$ is infinite. Now consider the set $X'=X\setminus x$ and $Y'=A_x$. If there is an $x'\in X'$ such that there is an infinite number of $y'\in Y'$ such that $(x',y')$ has color $\alpha$ then we are done. just take $x,x'$ and $y_1,y_2$ such that $(x',y_1)$ and $(x',y_2)$ have color $\alpha$. Otherwise notice that the set $C'=C\setminus\alpha$ has $c-1$ elements, and for every $x\in X'$ the only colors that appear an infinite number of times among the pairs $(x,y)$ are in $C'$. By the inductive hypothesis there exist $x_1,x_2\in X'$ and $y_1,y_2\in Y'$ such that $(x_1,y_1),(x_1,y_2),(x_2,y_1),(x_2,y_2)$ all have the same color.
A: We will decide how to choose rows first. So basically there are three colors so applying the pigeonhole principle if we choose the minimum number of rows to be 4 then it is bound at the worst case one color will repeat.
Now for choosing the minimum number of columns out of 4 rows we will choose any 2 rows and assign its color to be the same, either red blue, or green. This will lead us to the combination of 18 possible options (4C2*3 (for each color)).
So again applying pigeonhole there should be at least 19 (18+1) columns for the possibility of two colors in 4 rows to be the same
