(a) Assume the opposite. Orient the board such that there are 4 rows and 7 columns. Suppose that a colour appear three times in some column, say white squares appear in the first three squares of the first column. In each of the first three squares of each column, there are at least two black squares. But there are only $ {3\choose2}=3$ possible positions for the two black squares in each column, hence there must be four black squares which form a rectangle, contradiction. Now assume that each colour appear exactly twice in each column. But there are only $ {4\choose2}=6$ possible positions for each column, hence two columns are identical which means they contain a rectangle, contradiction. (b) [asy][asy]/* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(300); real lsf = 0.5; /* changes label-to-point distance */ pen xdxdff = rgb(0.49,0.49,1); pen zzttqq = rgb(0.6,0.2,0); /* segments and figures */ draw((0,0)--(1,0)--(1,2)--(0,2)--cycle); draw((1,0)--(2,0)--(2,1)--(1,1)--cycle); draw((1,2)--(1,3)--(2,3)--(2,2)--cycle); draw((2,0)--(3,0)--(3,1)--(2,1)--cycle); draw((2,3)--(3,3)--(3,4)--(2,4)--cycle); draw((3,1)--(4,1)--(4,3)--(3,3)--cycle); draw((4,1)--(5,1)--(5,2)--(4,2)--cycle); draw((4,3)--(5,3)--(5,4)--(4,4)--cycle); draw((5,2)--(6,2)--(6,4)--(5,4)--cycle); draw((0,0)--(1,0),zzttqq); draw((1,0)--(1,2),zzttqq); draw((1,2)--(0,2),zzttqq); draw((0,2)--(0,0),zzttqq); draw((1,0)--(2,0),zzttqq); draw((2,0)--(2,1),zzttqq); draw((2,1)--(1,1),zzttqq); draw((1,1)--(1,0),zzttqq); draw((1,2)--(1,3),zzttqq); draw((1,3)--(2,3),zzttqq); draw((2,3)--(2,2),zzttqq); draw((2,2)--(1,2),zzttqq); draw((2,0)--(3,0),zzttqq); draw((3,0)--(3,1),zzttqq); draw((3,1)--(2,1),zzttqq); draw((2,1)--(2,0),zzttqq); draw((2,3)--(3,3),zzttqq); draw((3,3)--(3,4),zzttqq); draw((3,4)--(2,4),zzttqq); draw((2,4)--(2,3),zzttqq); draw((3,1)--(4,1),zzttqq); draw((4,1)--(4,3),zzttqq); draw((4,3)--(3,3),zzttqq); draw((3,3)--(3,1),zzttqq); draw((4,1)--(5,1),zzttqq); draw((5,1)--(5,2),zzttqq); draw((5,2)--(4,2),zzttqq); draw((4,2)--(4,1),zzttqq); draw((4,3)--(5,3),zzttqq); draw((5,3)--(5,4),zzttqq); draw((5,4)--(4,4),zzttqq); draw((4,4)--(4,3),zzttqq); draw((5,2)--(6,2),zzttqq); draw((6,2)--(6,4),zzttqq); draw((6,4)--(5,4),zzttqq); draw((5,4)--(5,2),zzttqq); draw((0,4)--(6,4)); draw((6,4)--(6,0)); draw((0,0)--(6,0)); draw((6,0)--(0,0)); draw((0,4)--(0,0)); draw((0,1)--(6,1)); draw((0,2)--(6,2)); draw((0,3)--(6,3)); draw((1,4)--(1,0)); draw((2,4)--(2,0)); draw((3,4)--(3,0)); draw((4,4)--(4,0)); draw((5,4)--(5,0)); draw((0,1)--(3,4)); draw((0,0)--(1,1)); draw((1,0)--(2,1)); draw((2,0)--(4,2)); draw((3,1.99)--(4,3)); draw((4,1)--(6,3)); draw((4,3)--(5,4)); draw((5,3)--(6,4)); /* points and labels */ dot((0,0)); dot((6,0)); dot((6,4)); dot((0,4)); dot((1,0)); dot((1,2)); dot((0,2)); dot((2,0)); dot((2,1)); dot((1,1)); dot((1,3)); dot((2,3)); dot((2,2)); dot((3,0)); dot((3,1)); dot((3,3)); dot((3,4)); dot((2,4)); dot((4,1)); dot((4,3)); dot((5,1)); dot((5,2)); dot((4,2)); dot((5,3)); dot((5,4)); dot((4,4)); dot((6,2)); dot((6,4)); dot((6,4)); dot((0,1)); dot((6,1)); dot((0,3)); dot((6,3)); dot((1,4)); dot((4,0)); dot((5,0)); dot((3,1.99)); clip((-3.43,-4.16)--(-3.43,6.66)--(11.98,6.66)--(11.98,-4.16)--cycle);[/asy][/asy]

Interestingly enough, the condition is true for a 3 x 7 chessboard, and the proof is slightly shorter! In each of the seven columns, there are three squares, so there are a pair of the same color in that column. With three possible ways to select the pair out of the three squares and two colors, there are six pair/color possibilities in all. Then in two of the columns, the pairs are in the same position and of the same color, hence our rectangle.

polya78 wrote: Interestingly enough, the condition is true for a 3 x 7 chessboard, and the proof is slightly shorter! In each of the seven columns, there are three squares, so there are a pair of the same color in that column. With three possible ways to select the pair out of the three squares and two colors, there are six pair/color possibilities in all. Then in two of the columns, the pairs are in the same position and of the same color, hence our rectangle. Doesn´t this happen to be USAMO 1976 Problem 1? The solution on that page does mention 3 x 7 chessboards. Also this thread is quite old having reached 1.5 years of age and had been dormant for a long period of time so it is likely considered a revive? though I am a bit of a newbie so... XD

Bad solution (a) By the pigeonhole principle, each row must have at least $4$ of one color. Notice by pigeonhole that we must also have at least two rows with the same color having at least $4$ of one color. WLOG let the color be black and the rows be the first and second rows. FTSOC suppose there are no rectangles with all four corners colored the same. Let there be $x\ge 4$ black squares in the first row. Then, if two black squares in the second row are directly under black squares in the first row, we have a rectangle with all four corners black. Hence, we must place all but one of the black squares in the second row below white squares in the first row. Therefore, we have at most $7-x+1$ black squares in the second row so $4\le 7-x+1$ and $x\le 4$ so $x=4.$ Next, we consider the third row. Notice each black square must go under a black square in the first row or under a black square in the second row (or both). Hence, if there are at least $3$ black squares in the third row, two must go under black squares in the first row or two must go under black squares in the second row by the pigeonhole principle. This means we have a rectangle with all four corners black, so we have at most $2$ black squares in the third row. Similarly, we have at most $2$ black squares in the fourth row. Then, we place the white squares in the third row, noting that we have at least $5.$ When we place white squares in the fourth row, at most $2$ of them can go directly under black squares, so at least $3$ of them must go directly under white squares. But this forms rectangles with all corners white, a contradiction. (b)

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Here is a shorter one. First, we know that there are at least two black (WLOG) squares in a column of 4 squares. Then there are 4 choose 2 ways to arrange 2 squares in four slots for each column, meaning 6, but there are 7 columns so by pigeonhole at least two of them must have the same configuration of black squares. Also, if there are 3 black squares in a column, 4 choose 3 is 4 ways to arrange, 7>4, and if there are 4 black squares, 4 choose 4 is 1, 7>1. $\blacksquare$ @post#3 Yes, 3x7 is easier since 3 choose 2 or 3 choose 3 is way less than 7. #post#4 No, this is the USAMO 1976 Problem 1. Note for part b): Here is a super easy way, but you just use the general pattern of arranging every 4 choose 2 black squares in a column in a different order for all 6 columns so that there is no rectangle, then out of those configurations, which really narrows it down, you can find one that works.

(a) Note that there are $\binom 5 2 \binom 8 2 = 280$ total possible rectangles in the chessboard. However, there are only $2^7 \cdot 2 = 2^8$ possibilities for the rectangle itself since the choosing of the first row determines the second row, so thus by Pigeonhole we are done. $\blacksquare$ (b) insert counterexample