The posted is posted at MO, it is here.
The problem is as this.
Suppose there is a $p\times p$ square grid, and we tile it by a set of $p$ congruent $p$-ominoes (a $p$-omino being a connected piece of $p$ squares). We allow rotation and reflection. Besides splitting the grid into $p$ $1\times p$ bars, is there any other solution?
This is a problem a friend came up when he was trying to solve some problems, like what rectangles can be tiled by $P$, $I$, $L$, and $Y$ pentominoes. Then, the friend found that the smallest odd rectangle that can be tiled by a specified pentomino is quite large for $P, L, Y$, but $I$ can tile the $5\times 5$ square. Additionally, he noticed that the $L$ shaped tromino cannot tile $3\times 3$, and according to https://polyominoes.org/data, he also found out that for $7\times 7$ all the shown heptominoes cannot tile $7\times 7$ except the $I$ shape, and likewise $11$-ominoes and $13$-ominoes. So this concludes the background of this problem.
I am so sorry for that this may be no-clue question which may violate the math SE standard, but we don't know how to prove this and we don't have any idea... For the case $p=3,5$, they are easy and we can make it by brute force. For $p=7$, it is getting harder by really brute-forcing for some shapes, and I believe there is a more generic way to do it... or, is there even a counter-example?
P.S. I remember someone had given an answer a week ago but now the user has deleted the answer... what a pity...
