Wei-Hwa Huang, Juha Saukkola, Gareth Rees, and Joseph DeVincentis sent solutions. The solution in the image above is by Jukka-Pekka Ikaheimonen.
This write-up below by Gareth Rees. A similar write-up is at the
website of Torsten
The 28 dominoes can cover 56 squares of the chessboard, leaving 8
"holes". I will focus on the positions of these holes.
No two holes can be adjacent orthogonally or diagonally, because there's
no way to put dominoes in the adjacent squares, so all adjacent squares
must be holes too; hence holes cover the whole board.
I can't put a hole (represented by "O") at a corner, because
in the diagram at the right the domino a is forced, but then ba......
there's no way to cover square b. ........
I can't put a hole on an edge, because dominoes a and b in
the diagram at the right are forced, and then there's no way ..acb...
to cover square c. ........
I can't place two holes with only one square between them
because in the diagram at the right, after placing piece a .bbagf..
(WLOG) pieces b, c, d, e and f are forced, and then there's .cOaOf..
no way to cover sqaure g. .cddee..
I can't place three holes in the configuration at the right
because I must cover the squares marked "a" with dominoes ..aOa...
aligned vertically, and then there's no way to cover the .OabaO..
square marked "b". ........
I believe (but cannot prove other than by exhaustive search) that the
above five conditions are sufficient to ensure that the only possible
arrangements of the 8 holes are
and its reflection. The arrangement of dominoes is then forced to be:
d bqqe g
o muuj l
(The order of the letters - lowercase then uppercase - gives one forcing