This is not a square. Part 2

Consider the figure below.

Each $2\times 2$ square covers exactly $1$ ${\color{#3D99F6}{\text{blue}}}$ square. Each $1\times 4$ tetromino covers an even number of ${\color{#3D99F6}{\text{blue}}}$ squares. There are $64$ red squares. If we used $x$ number of green tiles and the red tiles covered $y$ number of ${\color{#3D99F6}{\text{blue}}}$ squares , then $x+y=64$ where $y$ is an even, so $x$ is even too. But if we use one more green tile, then $x$ will be odd, so then the covering is not possible.