Is it possible to tile a rectangle by one X-pentomino and any number of I-trominoes?
Consider a $3m+1$ by $3n+2$ rectangle $R$ to be tiled by one X-pentomino and many I-trominoes. Using 3-coloring methods (123 and 321 cyclic), we see that the center of X must be at row $3i-2$, column $3j$. The figure shows an 8 by 10 rectangle and the 3s are the possible locations of the center of X.
Now I want to show the tiling is impossible. The difficulty is, if we remove an I-tromino from the X-pentomino, then it is possible to tile the deficient rectangle say, with the squares right above and below the circled 3 removed, but the I that covers the circled 3 is not centered there.
Is there a proof for this problem? Thanks a lot.
At the risk of killing a housefly with a bazooka, there is a very beautiful solution to this using the tiling group of Conway and Lagarias. I walked through the definition of the tiling group in the beginning of this answer.
Your problem is a particular case of a tiling problem analyzed in the paper Tile Homotopy Groups by Michael Reid.
Proof: The tile homotopy group for I trominos and X pentominos has this presentation: $$ \begin{align} T&=\langle x,y\mid \underbrace{x^3yx^{-3}y^{-1}}_{3\times 1\text{ I tromino}},\; \underbrace{xy^3x^{-1}y^{-3}}_{1\times 3\text{ I tromino}},\; \underbrace{(xy)^2x^{-1}y(x^{-1}y^{-1})^2xy^{-1}}_{\text{ X pentomino}}\; \rangle\\\\ &=\langle x,y\mid [x^3,y],\;\;[x,y^3],\;\;xyxyx^{-1}yx^{-1}y^{-1}x^{-1}y^{-1}xy^{-1} \rangle \end{align} $$ Now, consider the homomorphism $\phi: T\to S_5$, defined as follows (in cycle notation): $$ \phi(x)=(1\;\;2\;\;3),\qquad \phi(y)=(3\;\;4\;\;5) $$ In order for this to be well-defined, we need to show that these two permutations satisfy all of the group relations. Let $X=\phi(x)$ and $Y=\phi(y)$. Since $X^3$ and $Y^3$ are both the identity, they clearly commute with $X$ and $Y$, so the first two relations are satisfied. For the X-pentomino relation, the below pictorial computation proves that $XYXYX^{-1}YX^{-1}Y^{-1}X^{-1}Y^{-1}XY^{-1}$ is the identity permutation. I find it amusing that a picture like this is useful for a tiling proof!
However, letting $R$ be the boundary of the $m\times n$ rectangle, with group word $\partial R=x^my^nx^{-m}y^{-n}$, the image of $\partial R$ under this homomorphism is $$ \phi(\partial R)=X^m Y^n X^{-m} Y^{-n}=[(1\;\;2\;\;3)^m,(3\;\;4\;\;5)^n] $$ Now, assume that $R$ can be tiled. By the fundamental theorem of tiling groups, this means that $\partial R=1$ as an element of $T$, which would imply $\phi(\partial R)$ is the identity permutation, so that $(1\;\;2\;\;3)^m$ and $(3\;\;4\;\;5)^n$ commute in $S_5$. If neither $m$ nor $n$ were multiples of $3$, then $(1\;\;2\;\;3)^m$ and $(3\;\;4\;\;5)^n$ would be two three-cycles which overlap at $3$, which clearly implies they do not commute (look at the image of $3$). We conclude that $3\mid m$ or $3\mid n$.$\tag*{$\square$}$
One might wonder if there are any tilings of a rectangle which use the X tile at all. It turns out that there are. Here is an example taken from the same Reid paper.