A $7\times 14$ chessboard is cut along its gridlines into squares and three-square corners. Show that there are at least as many corners as squares... 
The problem: A $7$ by $14$ chessboard is cut along the grid lines into some squares and some three-square corners. Show that there are at least as many corners as squares. Can equality hold?

I do not think I understand the problem statement. Say we have a $2$ by $2$ square. Each $2$ by $2$ square will be able to fit one three-square corner and one square, and we can let the square be placed on the top-right of each $2$ by $2$. If we line the first six rows with these squares, we will have an equal number of squares and corners. However, we cannot fit any more corners in the last row this way, which means that we will have $14$ more squares than corners. This clearly is not right given the question description. Where is the error in my thinking?
How do we approach such a question? I understand that colouring and breaking up the $7$ by $14$ chessboard into smaller segments is a technique, but I have yet to find one that works, probably also due to my lack of understanding of the question at hand. I attempted to colour it like a chessboard, with alternating black and white squares, but it does not seem to give me much information.
 A: As stated the problem is not true.
Cover the chessboard with two $7\times 7$ squares and there will be no corner.
Also, it can be found other configuration's with smaller squares:
\begin{align}11200111222333\\
12200111222333\\
33499111222333\\
34499445566778\\
55111445566788\\
56111990220033\\
66111900220033
\end{align}

Perhaps all squares must have size $2\times 2$. In this case the statement is true. Color every even row with red color. Suppose we have $a$ corners and $b$ $2\times 2$ squares. Let $x$ be the number of corners with $2$ red in $1$ white unit square and let $y$ be the number of corners with $1$ red in $2$ white unit squares. Since each $2x2$ square covers 2 red and 2 white unit squares we have
\begin{align}
2x+y+2b &=42\\
2y+x+2b &= 56
\end{align}
which means that $y=x+14\geq 14$ so $a=x+y\ge14$. Since $$98=3a+4b \geq 42+4b\implies b\leq 14\leq a$$ we are done.
Now it is not hard to find a configuration with $a=b=14$. Fill first 4 rows with 14 $2\times 2$ squares and the rest of the chessboard with cornes.
