how many white squares does a $2022 \times 2021$ block have? 
An infinite checkerboard is coloured black and white so that every $2\times 3$ block has exactly two white squares. Prove (or disprove) that every $2022\times 2021$ block has the same number of white squares and if so, find this number. Does the result hold if we additionally assume every $3\times 2$ block has exactly 2 white squares and if so, what is the number of white squares in a $2022\times 2021$ block?

My initial thought was to consider various patterns. Let W be a white square and B a black square. Then we have the following possible patterns for a 2 by 3 block:
$1)\begin{pmatrix}WWB \\ BBB\end{pmatrix},2)\begin{pmatrix}WBB \\ BBW\end{pmatrix},3)\begin{pmatrix}WBB \\ BWB\end{pmatrix},4)\begin{pmatrix}WBB \\ WBB\end{pmatrix},5)\begin{pmatrix}BWW \\ BBB\end{pmatrix},6)\begin{pmatrix}WBW \\ BBB\end{pmatrix},7)\begin{pmatrix}BWB \\ WBB\end{pmatrix},8)\begin{pmatrix}BWB \\ BWB\end{pmatrix},9)\begin{pmatrix}BWB \\ BBW\end{pmatrix},10)\begin{pmatrix}BBW \\ WBB\end{pmatrix},11)\begin{pmatrix}BBW \\ BWB\end{pmatrix},12)\begin{pmatrix}BBW \\ BBW\end{pmatrix}.$
Let the $2022\times 2021$ block have coordinates where the top left entry has coordinates $(1,1)$ and coordinates increase from left to right and top to bottom (coordinates are defined for all squares on the infinite checkerboard). Specify a rectangle with top left corner $(a,b)$ and bottom right corner $(c,d)$ as $[(a,b)-(c,d)].$ Suppose that each case represents the squares with coordinates $(1,1),(1,2),(1,3),(2,1),(2,2),(2,3).$ I'm not sure how to analyze the cases and it seems like it should be possible to take advantage of symmetry.

The answer below assumes that an $x\times y$ block has height y and width x. For the other interpretation, $1\leq x \leq 2021, 1\leq y\leq 2022$ (x increases right, y increases down) $x\equiv 1\mod 3$ (giving $2022\times 674$ white squares) or $x\equiv 0\mod 3$ (giving $2022\times 673$ white squares). Both work because if $[(a,b)-(a+2,b+1)]$ is a $2\times 3$ block, then exactly one of $a,a+1, a+2$ is congruent to $0$ or $1$ modulo 3, say c, and in each case both squares in the 2 by 3 block with x-coordinate c are white.

 A: The answer to the first part is no:
Associate the board's squares with coordinates $(x,y)$, where $1 \leq x \leq 2022$ and $1 \leq y \leq 2021$. Color the board by solid columns. Board A has $(x,y)$ colored white if and only if $y \equiv 1 \pmod 3$. That is, $y = 3k+1$ where $0 \leq k \leq 673$, so Board A has $2022 \times 674$ white squares. Board B has $(x,y)$ colored white if and only if $y \equiv 0 \pmod 3$. Then $1 \leq \frac{y}{3} \leq 673$, and Board B has $2022 \times 673$ white squares.
For the second part, every $2022 \times 2021$ checkerboard has exactly $\frac{2022 \times 2021}{3} = 674 \times 2021$ white squares.
There can't be two adjacent whites:
WWB
BBB
BB

If there were, then at least one orientation relative to the adjacent whites will have spaces within the board in the $3 \times 3$ pattern shown, and all $6$ of the shown squares must be black. But then the bottom $6$ squares already have $5$ of them colored black, and can't contain two white squares.
There can't be two whites in a row or column separated by exactly one black:
WBW
BBB
WBW
BBB

If there were, then at least one orientation relative to the topmost WBW row will have spaces within the board in the $4 \times 3$ pattern shown. Since that row contains two white squares, the adjacent row must have three black squares. The third row must then have two white squares again; since they can't be adjacent, the row must be WBW. Since the third row contains two whites, the fourth row must be BBB. But then this would create two $3 \times 2$ blocks with only one white square each.
There can't be three consecutive black squares in any row or column. If there were, then the three adjacent squares in an adjacent row/column would need to contain two white squares, violating one of the two conclusions above.
Therefore every row and every column contains some rotation of the periodic pattern WBBWBBWBB.... The $2022 \times 2021$ checkboard can be divided into $674 \times 2021$ disjoint rectangles of size $3 \times 1$. Each rectangle contains exactly one white square, so the total number of white squares is also $674 \times 2021$.
(The valid boards for the second part are repeating diagonal stripes, which can be described as either $(x,y)$ is white if $x+y \equiv c \pmod{3}$, or $(x,y)$ is white if $x-y \equiv c \pmod{3}$.)
