It is impossible.
Let the $7\times 28$ area be painted with black and white squares in a checkerboard pattern. Every piece will cover $2$ black and $2$ white squares, except the T-piece, which covers $3$ of one color and $1$ of another. Since there are $7$ T-pieces, a tiling that uses every piece cannot cover the same number of black and white ...
TLDR: I'll fill the board and prove that the solution is unique.
First, let's start by:
I'll paint those green:
Let's repeat those steps a few more times, using orange, blue, red and purple, in precisely that order:
We can easily fill the topmost white squares by that reasoning. They can't be filled in any other way:
Now, let's look at:
And by ...
The X-pentomino tiles the plane, so that tiling is a good way to start. There are two ways to cut an 8x8 region out of that tiling. If one of the 4 central squares of the 8x8 region has an X centred on it, you get this
or else you get this
The latter can be easily improved by replacing the ones at the edges to give this
A different way to get the same ...
Challenge 1: Fit the four yellow birds into the tray, no overlapping. Rotation and reflection are allowed.
Challenge 2: Fit the four yellow birds and the blue piece into the tray, no overlapping. Rotation and reflection are allowed.
Challenge 3: Fit the four yellow birds and the red piece into the tray, no overlapping. Rotation and reflection are allowed.
Here is a proof that $12$ is the smallest possible number of regions in any feasible solution.
Consider an arbitrary division of an arbitrary rectangle into $n$ regions, such that every region has exactly five neighboring regions. We translate this picture into a so-called planar graph: each of the $n$ regions then translates into a vertex/point, and there ...
Finally arrived at this solution after playing around on
for way longer than I care to admit!!!! :)
First pair (UI-TF)
Second pair (WX-PY)
Third pair (VZ-LN)
First, a generalizable solution for $1 \times n$, $n$ is even. By halving the rectangles, we can also obtain solutions for odd $n$, and the parts with just rectangles and no W-pentominos can be shortened.
This is a way to tile
This is optimal for this $a \times b$ because
And here is a way to tile
I've found two more, one for 1x4:
and a rather large one ...
They are trying to prove that
How are they doing it?
Will they succeed?
A special case of the result being considered may be found at Maths SE (spoilers, obviously). The L-ish proof here uses the same underlying idea, but with the difference that
I can prove that the answer is exactly
Several people, including Jaap Scherphuis, have shown that the square can be covered with this many pentominoes, so it only remains to show that at least this many pentominoes are needed. (A matching lower bound).
Let us start with the magic board given by A. Rex:
As a first lower bound,
Therefore, we can ...
I think I've found a solution for an 8x8 square. I do not know if it is the minimum solution or how to prove that:
It was definitely fun to try and find this! Took me a while. Excellent puzzle.
Some comments on how I got to the solution (Rather a chronology than a full deduction):
Since a single line doesn't do damage, it is possible to do
To achieve this, there are a couple of requirements:
To get these pieces one after the other
For this to occur so that the final piece also clears the board, we get these constraints on the number of pieces $X$:
Given these, the smallest $X$ that satisfies both requirements is
This is a small ...