Can a 10x10 square be tiled with 25 1x4 rectangles (straight tetrominoes)?

geometry/tiling/straight.tetromino New

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Problem

Can a 10x10 square be tiled with 25 1x4 rectangles (straight tetrominoes)?
More generally, for which odd m and n can a (2m)x(2n) rectangle be tiled
with 1x4 rectangles?

[ Editor: the original poster, dshemano@desire.wright.edu, noted that the
6x6 cannot be tiled, that the area must be divisible by four, and that
the problem is trivial if either side is divisible by four. ]
[ Editor: problem statement reconstructed from the postings. ]

Show solution

[ Editor: No; two proofs follow. ]

Picture the 10x10 as a 5x5 array of 2x2 squares.  Color the 5x5
array in checkerboard fashion (red and black, say, so there are
13 black 2x2's and 12 red 2x2's).  Now each 1x4 covers exactly
2 red and 2 black 1x1's, so any partial tiling must cover exactly
the same number of red and black 1x1's.  Unfortunately, there are
[Editor: "are" missing in the original] 4 extra black 1x1's, so you can't finish the tiling.

I think this argument could be generalized to prove that
1x4's won't tile 2m x 2n when m and n are both odd.  If so,
this should finish the answer to your general question.

-- Bart Goddard <goddard@nextwork.rose-hulman.edu>

Try to number the 10x10 square like this

        0123012301
        1230123012
        2301230123
        3012301230
        0123012301
        1230123012
        2301230123
        3012301230
        0123012301
        1230123012

It is easy to see that any 1x4 tetromino will cover one each of the
numbers 0-3, regardless of where it is placed, horizontally or
vertically. So any tiling using 25 tetrominoes will cover 25 each of
the numbers 0-3. Since the square contains 25 0's, 26 1's, 25 2's and
24 3's, it can't be tiled. QED.

> For which positive, odd integers m and n is a (2m)x(2n) rectangle
> tileable with 1x4 tetrominoes?

None. If n and m are both odd, there will be an uneven number of
0-3's. The rectangle (2(m-1))x2n contains (m-1)n each of 0-3. The
remaining strip 2x2n is divided into a 2x2(n-1) strip with (n-1) each
of 0-3 and a 2x2 square with the numbers 0112 (see bottom right of the
square above), so there will be a surplus of 1's and a deficit of 3's.

This numbering technique is a generalization of the classic proof that
a checkerboard with the two opposite corners cut off can't be tiled by
dominoes.

-- Torben Mogensen <torbenm@diku.dk>