For which $n$ does a y-formed $n$-polyomino tile a $n \times n \times n$-cube?

Question

I got from my children as a gift a puzzle consisting of 25 y-shaped 5-polyominoes that form a $5 \times 5 \times 5$-cube (see picture).

I'm wondering for which $n$ does a y-formed $n$-polyomino tile a $n \times n \times n$-cube?

By a y-formed $n$-polyomino I understand a generalisation of the y-formed $5$-polyomino (made of the $5$ squares $(0, 0), (1, 0), (2, 0), (3, 0), (1, 1)$), i.e., the center of the $n$ squares can be chosen as $(0,0), (1,0), (1,1), (2,0), \dotsc, (n-2,0)$. The polyomino has a thickness of 1.

For $n=5$ I have the (wooden) proof that it's possible before me, for $n=4$ it's trivial (as one can tile one $4 \times 4 \times 1$-slice of the $4 \times 4 \times 4$-cube already with four y-formed $4$-polyominoes), what about $n=6, 7 , \dotsc$?

Any ideas for patterns that work for whole sequences of $n$, any proofs that for some $n$ it's not possible?

Answer 1

There are no solutions for $n\ge6$. For $n=6$ one checks that the exact cover formulation of the problem has no solution, as ho boon suan did in the comment.

For $n\ge7$, the following argument works: Let $(i,j,k)$ for $1\le i,j,k\le n$ be the $n^3$ positions of the cube to be covered. Consider the subset $S$ of positions $(i,j,k)$ where $i,j,k\in\{1, 4, n\}$. Each polyomino intersects $S$ in either $0$ or $2$ positions. As $\lvert S\rvert=27$ is odd, there is no cover.

There are also choices of $S$ which are symmetric under the symmetries of the cube:

If $n$ is odd, then for $S=\{(i,j,k)\;|\;i,j,k\in\{1,(n+1)/2,n\}\}$ we get $\lvert S\rvert=27$, while $\lvert S\cap P\rvert=0$ or $2$ for each polyomino.

If $n$ is even, then for $S=\{(i,j,k)\;|\;i,j,k\in\{1,n/2,n/2+1,n\}\}$ we get $\lvert S\rvert=64$, while $\lvert S\cap P\rvert=0$ or $3$ for each polyomino.