Tiling a Square by Rectangles I have to prove that you can't create a square with side length $10$ by arranging $25$ rectangles with side lengths $4$ and $1$, where no pair of rectangles may overlap and the whole square must be filled.
If you have a very good design so you always have got $4$ fields on this square but you can't fill them because the $4$ left fields are not a rectangle with the side lengths $4$ and $1$.
Any hints or suggestions for my problem are appreciated. Thank you.
 A: Assign integer coordinates to the (centres of) the little squares, in more or less the usual way. The bottom left square is $(0,0)$, the one immediately to its right is $(1,0)$, the next one is $(2,0)$, and so on up to $(9,0)$.  The next row up is labelled $(0,1)$, $(1,1)$, and so on.
The sum of the $x$-coordinates of all points is $(10)(45)$, as is the sum of all the $y$-coordinates, for a total of $900$.  
Any $1\times 4$ rectangle covers $4$ points the sum of whose coordinates has remainder $2$ on division by $4$. For suppose for example that the rectangle has long side in the horizontal direction. The four $y$-coordinates are all the same, so their sum is divisible by $4$. The four $x$-coordinates are four consecutive integers, and therefore their sum has remainder $2$ on division by $4$. 
Now we suppose that $25$ such rectangles cover our $10\times 10$ square, and derive a contradiction.  If $25$ rectangles covered, then the sum of the coordinates of all points would have remainder $2$ on division by $4$. However, $900$ has remainder $0$ on division by $4$.  
A: Color the board as follows
$\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\
\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\
\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square\\
\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square\\
\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\
\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\
\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square\\
\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square\\
\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\
\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\$
Note that a $1\times4$ rectangle placed on the board always covers two black squares and two white squares. Therefore if it's possible to cover the board with $25$ $1\times4$ rectangles, there must be $50$ black and $50$ white squares on the board, what's not the case, so it's not possible.
