I need to proof something

Question

The checker with dimensions $2018 \times 2018$ was covered with one a square tile with dimensions of $2 \times 2$ and $\frac{2018^2-4}5$ rectangular tiles with dimensions $1 \times 5$ in such a way that each chessboard is covered through exactly one tile (tiles can be rotated). Prove that the tile $2 \times 2$ does not cover any field with an edge contained in the edge of the chessboard.

Answer 1

Assume that the $2\times2$ tile is on the edge, and WLOG assume that it is the top edge.

The proof will assign a letter in $\{a,b,c,d,e\}$ to every square of the $2018\times2018$ grid in two different ways, and show that the $2\times2$ tile on the top edge cannot cover the correct letters.

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Much like the checker board pattern of black or white, assign $\{a,b,c,d,e\}$ diagonally:

$$\begin{array}{|cccccccccc|} \hline a&b&c&d&e&a&b&\cdots&b&c\\ b&c&d&e&a&b&c&\cdots&c&d\\ c&d&e&a&b&c&d&\cdots&d&e\\ d&e&a&b&c&d&e&\cdots&e&a\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ b&c&d&e&a&b&c&\cdots&c&d\\ c&d&e&a&b&c&d&\cdots&d&e\\ \hline \end{array}\tag1$$

Each $1\times5$ tile should cover a set of $\{a,b,c,d,e\}$, either vertically or horizontally.

The numbers of occurences $n_x$ for letter $x$ are

$$\begin{align*} n_a = n_e &= (1+6+11+\cdots + 2016) + (2015 + 2010 + \cdots + 5)\\ &= 814464\\ n_b = n_d &=(2+7+12+\cdots + 2017 )+ (2014+ 2009 + \cdots + 4 )\\ &= 814465\\ n_c &=(3+8+13+\cdots + 2018)+ (2013+ 2008 + \cdots + 3)\\ &= 814466\\ \end{align*}$$

So relative to the number of $a$'s and $e$'s, there are $1$ extra $b$, $2$ extra $c$'s and $1$ extra $d$.

The $2\times 2$ tile must cover those extra letters, in order to accommodate the $1\times5$ rectangular tiles, e.g.

$$\begin{array}{|cccccccccc|} \hline a&(b)&(c)&d&e&a&b&\cdots&b&c\\ b&(c)&(d)&e&a&b&c&\cdots&c&d\\ c&d&e&a&b&c&d&\cdots&d&e\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ \end{array}$$

In particular, the two letters covered on the top row must be $b$ and $c$.

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Similarly, the set of letters can be assigned along the other diagonal:

$$\begin{array}{|cccccccccc|} \hline a&b&c&d&e&a&b&\cdots&b&c\\ e&a&b&c&d&e&a&\cdots&a&b\\ d&e&a&b&c&d&e&\cdots&e&a\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ \end{array}\tag2$$

In arrangement $(2)$, there are similarly $2$ extra $a$'s, $1$ extra $b$ and $1$ extra $e$ to be covered by the $2\times2$ tile. In particular, the letters covered on the top row must be $a$ and $b$.

But this is a contradiction, because the top rows of arrangements $(1)$ and $(2)$ are unchanged, and the $2\times 2$ tile cannot simultaneously cover $a$, $b$ and $c$ on the top row.