
92.4. Peter has many squares of equal side. Some of the squares are black, some are white. Peter wants to assemble a big square, with side equal to $n$ sides of the small squares, so that the big square has no rectangle formed by the small squares such that all the squares in the vertices of the rectangle are of equal colour. How big a square is Peter able to assemble?


Solution. We show that Peter only can make a $4 \times 4$ square. The construction is possible, if $n=4$ :

![](https://cdn.mathpix.com/cropped/2024_06_06_f01d327b68eb688a4496g-27.jpg?height=263&width=263&top_left_y=577&top_left_x=888)

Now consider the case $n=5$. We may assume that at least 13 of the 25 squares are black. If five black squares are on one horizontal row, the remaining eight ones are distributed on the other four rows. At least one row has two black squres. A rectangle with all corners black is created. Next assume that one row has four black squares. Of the remaing 9 squares, at least three are one row. At least two of these three have to be columns having the assumed four black squares. If no row has more than four black squares, there have to be at least three rows with exactly three black squares. Denote these rows by $A, B$, and $C$. Let us call the columns in which the black squares on row $A$ lie black columns, and the other two columns white columns. If either row $B$ or row $C$ has at least two black squares which are on black columns, a rectancle with black corners arises. If both rows $B$ and $C$ have only one black square on the black columns, then both of them have two black squares on the two white columns, and they make the black corners of a rectangle. So Peter cannot make a $5 \times 5$ square in the way he wishes.
