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It's rather easy to fill a $7 \times 7$ board with 16 long triominos, leaving the center square void: see the picture below. But if I want to move the void square in another position, where else could I place it?

enter image description here

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The trick to this puzzle is to:

tri-color the board.
enter image description here

Any tromino must cover one square of each color. There are 16 blue squares, 16 yellow squares, but 17 red squares, so a red must be the uncovered one. This is true for the reverse coloring as well, which gives our final result: the only possible squares are the corners, the center of the edges, and the center.

(And here are those tilings: the center was already given, and the rest are obtainable from these by rotation.)

enter image description here

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    $\begingroup$ That's not actually a full solution. You've shown that certain squares are impossible, but you haven't shown that the others are possible. $\endgroup$ – Peter Taylor Jul 10 '17 at 12:57
  • $\begingroup$ why not the square on second row, second column? $\endgroup$ – Marius Jul 10 '17 at 13:00
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    $\begingroup$ @Marius because you can use a mirror-image colouring, which has the same number of red squares but in different places; between them, these two colourings rule out all squares but the ones Deusovi says. $\endgroup$ – Gareth McCaughan Jul 10 '17 at 14:08
  • $\begingroup$ Ah. Makes sense now. $\endgroup$ – Marius Jul 10 '17 at 14:14
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it could be (any) one of the corners:
corner
black one is the empty space.

or

it could be one of the center squares on the first/last row/column.
enter image description here

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  • $\begingroup$ that's right (and if you think about it, there is another position you can have :-) ) $\endgroup$ – mau Jul 10 '17 at 12:27
  • $\begingroup$ @mau I edited the answer and added a new position, also explained the first finding a bit more. $\endgroup$ – Marius Jul 10 '17 at 12:30
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Here is a full solution, explaining why the places that work work, and why the places that don't don't.

I was discussing colourings of the Cartesian grid with @Deusovi, and he posed a question like this to me, and I independently discovered my own colouring. This one is able to be used as an argument without needing to mirror it (it has mirror symmetry):
every third cell, going up and down from the top left is orange, all cells in between orange are green, and the remaining 2x2 blocks are blue
Counting each colour, there are 9 orange cells, 16 blue cells and 24 green cells. Each orange cell must match with exactly two green cells, and you can only match pairs of blue cells to a single green cell. Thus you cannot remove a blue cell, as it would cause an odd parity. If you instead remove a green cell, then you end up with 16 blue cells and 5 green cells after matching a every yellow cell to two green cells, and then you will always be left with 6 blue cells after matching each green cell to two blue cells. Thus the only option left is to remove a yellow cell. It is simple enough to fill in a rectangular region with one dimension as a multiple of three, if you just tile it with all long edges parallel to the dimension that is a multiple of three. Each position has a swastika or windmill pattern that gives four rectangles surrounding the removed cell, each having one dimension that is a multiple of three, which is the dimension that goes between the outer edge and the orange cell (often one or two of the rectangles have zero width): Here are the three symmetrically unique cases:
The three possible cases

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