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Problem: Show that a $8 \times 8$ chessboard cannot be perfectly covered by $1$ square tetramino, and 15 other tetraminoes chosen from straight tetraminoes and Z-tetraminoes.

My attempt: I tried to color the board with black and white in several different ways but none of them yielded a contradiction. I tried coloring in such that a straight tetramino and a Z-tetramino would cover different number (and different parity) of blacks and different number of whites.

So, someone please provide some help. Also, it would be great if someone could provide the motivation behind there proof or provide some link to some general methods to deal with these problems as i frequently encounter these kinds of problems but I lack an organised approach to tackle them.

A Z-tetramino would look like the picture above. enter image description here

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  • $\begingroup$ Ideally, include a picture in the question. At minimum, provide a link to a page with the tetraminos pictured. $\endgroup$
    – Thomas Andrews
    Commented Feb 4, 2016 at 19:55
  • $\begingroup$ What is a perfect covering of a chessboard? $\endgroup$
    – null
    Commented Feb 4, 2016 at 19:59
  • $\begingroup$ @null , a perfect covering of a chessboard with tiles means, each cell is covered with some tiles and no two tiles overlap and each cell of each tile is on some cell of the board. $\endgroup$
    – Subham Jaiswal
    Commented Feb 4, 2016 at 20:03

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Hint: There is a four-coloring of the board so that

  1. Any straight tetranimo covers one of each color, and
  2. Any zig-zag covers either two of two colors or one each of all four coors.
  3. The square must color two colors exactly once, and another twice.
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  • $\begingroup$ Thank you, sir. I should have tried to use four-coloring. $\endgroup$
    – Subham Jaiswal
    Commented Feb 4, 2016 at 20:14

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