3 congruent shapes in a $5 \times 5$ square board Given a $5 \times 5$ board of $25$ cells. Is it possible to color $24$ of them in one of three colors in such a way, that the resulting three shapes of the particular colors are congruent to each other?
 A: Update: Fixed a non-critical bug pointed out by NickG in the comments, which prevented me from finding all possible solutions that didn't cover all but one cell.

This isn't, like, a super satisfying solution or anything...
But I ran a brute-force computer search, and failed to find any near-partition of the $5\times 5$ board into three congruent $8$-cell figures, connected or otherwise.
The search found five different connected $7$-cell figures whose congruent copies can be packed into a $5\times 5$ board (sometimes in more than one way). These are shown below:


My brute-force approach was to make a list of all transformations that can take some cells of the $5\times 5$ board to other cells. So my first list of candidates was all pairs of these transformations.
To each pair of transformations $(\tau_1, \tau_2)$, I associated a list of $25$ triples $(P, \tau_1(P), \tau_2(P))$ where $P$ ranges over all cells on the board. As a preliminary filter, I threw away all triples whose three elements were not pairwise distinct, or did not all fit on the board; then I threw away all pairs $(\tau_1, \tau_2)$ whose remaining triples did not cover at least $24$ cells of the board. (Update: to find all $7$-cell figures, we need to reduce this number to $21$.)
For each of the remaining $939$ pairs, I defined a graph on the triples, where two triples were adjacent if they shared one or more points, and then found a maximum independent set in the graph. Given any such independent set, we can get a packing of three congruent figures by taking the first element of every triple to be one figure, and applying $\tau_1$ and $\tau_2$ to get the other two figures.
None of these graphs had an independent set of size $8$, but some had an independent set of size $7$. (There were lots of solutions, so I filtered for connectedness and found only one remaining, up to symmetry.)
If anyone sees things I might've missed with the search, or can find a different connected $7$-cell figure that can be packed into a $5\times 5$ board (which would also show that my program can miss things), let me know.

Update: Generalizing to the next nontrivial case, my computer thinks that the $7\times 7$ board does not admit a packing of three congruent $16$-cell figures. There are multiple packings of three congruent $15$-cell figures, none of which are connected, though this one comes closest:

A: (1)
(2)
(3)
One of them have two corners. Then in the case of $(1)$, another shape must be two corners, too.
However $(2,1)(4,1)(2,5)(4,5)$ can't devide to three areas.
In the case of $(2)$, we can't draw $(3,3), (5,1)$ or $(5,5)$ into a color, satisfying the condition.
In the case of $(3)$, if we draw the color of $(3,3)$, we can't make congruent shapes.
Therefore three congruent shapes are impossible.
