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.
On
Color the board as follows
$\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\ \blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\ \square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square\\ \square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square\\ \blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\ \blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\ \square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square\\ \square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square\\ \blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\ \blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare~\square~\square~\blacksquare~\blacksquare\\$
Notice 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, which is not the case.
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 $4$.