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Here my solution Consider the 5 rows of the 5xn chessboard and colour the 1st, the 3rd and 5th row black and the 2nd and the 4th white. Such a tile can cover : A) 3 black squares and 2 white squares. B) 2 black squares and 3 white squares. Suppose that there are x "A" tiles. Counting the covered black squares: $3x + 2(n-x) = 3n \Leftrightarrow n=x$ . This equation tells us that there are just "A" tiles. A tile is composed of a square 2x2 and an unit square. Let's call the unit square "foot" and the 2x2 square "main square". Since there are just "A" tiles, the "foot" must lie on a black square. Hence a row with white squares (for example the 2nd) is only covered with 1 x 2 rectangular parts of the "main square". Conclusion n must be even. For each $5 \times 2k$ rectangle, let $p[k]$ be the number of tilings not containing a smaller $5 \times 2i$ rectangle and let $a_{k}$ be the number of ways to file a fixed $5 \times 2k$ rectangle. Hence $a_{k} = p[k] + p[k-1] a_{1} + ... + p[1] a_{k-1}$, so we just have to show that for every k $p[k] \geq 2$. It's easy to find a periodic construction, for example there is a construction where (after 6 squares) there is the same situation (= 3 cases to consider).Thus we have the desired result. QED P.S. : I'm sorry, the last point is maybe a bit vague, but it would be quite complicated to explain the whole construction in detail.

You can prove that $a_k> 2\cdot 3^{k-1}$ using giancarlo's method, and it's enough to solve the problem. Just for fun, you can analyze the exact values of the $p[k]$: it turns out that $p[k]=4$ whenever $3\mid k$, and $p[k]=2$ otherwise. From this (and using generating functions) you can calculate $a_k$ exactly: it's possible to prove the following identity of formal series: \[\sum_{k=0}^\infty a_kx^k=\frac{x^3-1}{5x^3+2x^2+2x-1}\]It's possible to get a linear recurrence on $a_k$ of degree 3 from the above identity. Namely, it isn't hard to show that the sequence defined by $a_0=1$, $a_1=2$, $a_2=6$, $a_3=20$, and for all $n\geq 4$, $a_n=2a_{n-1}+2a_{n-2}+5a_{n-3}$ satisfies the identity on formal series above.

uhhh I can't see the diagram of the pieces nvm I found it

i got wtv the p[k] thing is and so you can just show that if p[k] always equal to 2 it's actually equal to 2 * 3^k-1 so then it'd have to be larger

ISL Marabot solve $\textbf{Part 1:}$ We can $4$-colour the board like this: Now if $n$ is odd, then on a $5\times n$ board there will be $n-1$ cells with colour $4$. And clearly, each piece must have at least one cell of each colour in its $2\times 2$ part. Which gives the contradiction. $\blacksquare$ $\textbf{Part 2:}$ Let, $T_k$ be the number of ways to tile a $5\times 2k$ board. Now, using these tilings (and their rotations and reflections), we can show that, $T_k \geq 2T_{k-1}+2T_{k-2}+4T_{k-3}$. And the rest is trivial by induction as, $4\times 3^{k-2}+4\times 3^{k-3}+8\times 3^{k-4} = 56\times 3^{k-4} > 54\times 3^{k-4} = 2\times 3^{k-1} \ \ \ \ \blacksquare$

WLOG the grid has 5 rows horizontally. Part 1: Suppose FTSOC that $n$ is odd, and let $n=2m+1$. Then, denote an orz cell as a cell that's in both an even numbered row and an even numbered column (indexing starts from 1). Note that there are $2m$ orz cells. Clearly, any piece must occupy at least one orz cell. However, we need $2m+1$ pieces to tile the entire grid, contradiction. Part 2: Let $f(n)$ be the number of ways to tile a 5 by $2n$. In a tiling, define a wall as a vertical line that does not pass through the interior of any piece. Say that a tiling is creative if there are no walls other than the left and right sides of the grid. Claim: There are at least 2 creative tilings for a 5x2, at least 2 creative tilings for a 5x4, and at least 4 creative tilings for a 5x6. AA AA AB BB BB The second one can be obtained by reflecting across the vertical midline ABBB AABB AACC DDCC DDDC Again, the other one can be obtained by reflecting across the vertical midline. AABBBC AABBCC AFFFCC EEFFDD EEEDDD The other 3 can be obtained by reflecting across the vertical and horizontal midlines (they are all different, consider the orientation of the F-block). It is easy to see that $f(0)=1,f(1)=2,f(2)\geq6$. Consider $f(n)$ for $n\geq3$. If the last wall before the end occurs 2 columns before the end, then there are $f(n-1)$ ways to fill the grid to the left of the wall and 2 ways to fill the right, so this is $2f(n-1)$ ways. If the last wall before the end occurs 4 columns before the end, then there are $f(n-2)$ ways to fill the grid to the left of the wall and at least 2 ways to fill the right (since there are at least 2 creative tilings of a 5x4), so this case has at least $2f(n-2)$ ways. If the last wall before the end occurs 6 columns before the end, then there are $f(n-3)$ ways to fill the grid to the left of the wall and at least 4 ways to fill the right (since there are at least 4 creative tilings of a 5x6), so this case has at least $4f(n-3)$ ways. Hence, $$f(n)\geq 2f(n-1)+2f(n-2)+4f(n-3)$$for $n\geq 3.$ First, we can get that $f(3)\geq20,f(4)\geq60,f(5)\geq 184$, so $f(n)> 2\cdot 3^{n-1}$ for $n=3,4,5$. Now, let's proceed by strong induction. Suppose that $f(n-1)>2\cdot 3^{n-2},f(n-2)>2\cdot 3^{n-3},f(n-3)>2\cdot 3^{n-4}.$ Then, $$f(n)\geq 2f(n-1)+2f(n-2)+4f(n-3)$$$$>2\cdot3^{n-4}(2(9)+2(3)+4)=2\cdot3^{n-4}(28)>2\cdot3^{n-1},$$so by induction we are done.

Suppose we can tile an odd number, $2k+1$, of rows. Repeat the following two rows and end with the first row: \begin{align*} 1~~2~~1~~2~~1\\ 3~~4~~3~~4~~3\\ \end{align*}Then, each piece will cover one of each color and one extra. Note that there are $2k-1$ fours but $2k+1$ pieces. Contradiction. $~$ Now, we claim that the number of ways is more than $2 \cdot 3^{k-1}$ just by using the following "blocks" of $2$, $4$, and $6$ and their reflections: \begin{align*} A~~A~~A~~B~~B\\ A~~A~~B~~B~~B\\ \\ A~~A~~B~~B~~B\\ A~~A~~B~~B~~D\\ A~~C~~C~~D~~D\\ C~~C~~C~~D~~D\\ \\ A~~A~~C~~C~~C\\ A~~A~~C~~C~~D\\ A~~E~~E~~D~~D\\ B~~E~~E~~D~~D\\ B~~B~~E~~F~~F\\ B~~B~~F~~F~~F\\ \end{align*}Where the blocks of two and four have two reflections and the one with six has four. Therefore, if $f(k)=2f(k-1)+2f(k-2)+4f(k-3)$ and $f(0)=1$, $f(1)=2$, $f(2)=6$ and $f(3)=20$. Now, if $f(k-3)\ge 2\cdot 3^{k-4},f(k-2)\ge 2\cdot 3^{k-3},f(k-1)> 2\cdot 3^{k-2}$ then \[f(k)>8\cdot 3^{k-4}+4\cdot 3^{k-3}+4\cdot 3^{k-2}>2\cdot 3^{k-1}\]as desired.

To prove the first part, consider a checkerboard coloring. Then each piece either has $1$ more black cell then white cell, or $1$ less black cell then white cell. But we need an equal number of the two, so we need $n$ to be even. Now to prove the second part, we first claim the following: Claim: For any positive integer $k \ge 2$, then there are at least two ways to tile a $5 \times 2k$ grid with the pieces, such that the pieces do not tile any $5 \times i$ subgrid for $2 \le i \le 2k-2$. Proof. Due to symmetry, we only need to show that there exists one way. We will do so by exhibiting a construction. Consider the following: [asy][asy] unitsize(1cm); draw(box( (0,0),(4,5))); draw( (0,2)--(2,2)); draw( (2,2)--(2,1)); draw( (2,1)--(3,1)); draw( (3,1)--(3,0)); draw( (2,2)--(2,4)); draw( (2,3)--(4,3)); draw( (1,4)--(2,4)); draw( (1,4)--(1,5)); [/asy][/asy] We then extend it to a $5 \times 10$ grid as so: [asy][asy] unitsize(1cm); draw(box( (0,0),(10,5))); draw( (0,2)--(2,2)); draw( (2,2)--(2,1)); draw( (2,1)--(3,1)); draw( (3,1)--(3,0)); draw( (1,5)--(1,4)); draw( (1,4)--(2,4)); draw( (2,4)--(2,2)); draw( (2,3)--(4,3)); draw( (4,3)--(4,5)); draw( (4,3)--(5,3)); draw( (5,3)--(5,2)); draw( (5,2)--(4,2)); draw( (4,2)--(4,1)); draw( (4,1)--(2,1)); draw( (5,2)--(6,2)); draw( (6,2)--(6,0)); draw( (6,2)--(8,2)); draw( (8,2)--(8,1)); draw( (8,1)--(9,1)); draw( (9,1)--(9,0)); draw( (8,1)--(8,3)); draw( (8,3)--(10,3)); draw( (8,3)--(8,4)); draw( (8,4)--(7,4)); draw( (7,4)--(7,5)); draw( (5,3)--(6,3)); draw( (6,3)--(6,4)); draw( (6,4)--(7,4)); [/asy][/asy] Clearly this can be extended more and it works. Apply similar logic (but with different constructions) for $5 \times 6$ and $5 \times 8$ to generate constructions for everything. The $5\times 2$ construction is also obvious. $\square$ Now, we consider partitioning a $5 \times 2k$ grid into $m$ blocks, where the $i$th block has dimensions $5 \times 2a_i$ for $a_i \ge 1$; then from stars-and-bars there are $\binom{k-1}{m-1}$ ways to choose the $a_i$, and for each block there are at least $2$ ways to tile them without the pieces within a block also tiling a smaller $5 \times 2j$ block (basically the statement in our claim). But then it follows that there are at least \[ 2^k + \binom{k-1}{k-2} 2^{k-1} + \dots + \binom{k-1}{1}2^2 + 2 = 2 \cdot 3^{k-1} \]ways to tile the block, with the last equality following from Binomial theorem. $\blacksquare$

Proof that $n$ is even: Assume otherwise we can produce a tiling with an odd $n$. Then we have $n$ 2 by 2 squares in a $5$ by $n$ rectangle, or we have $2n$ vertical dominos such that each domino is attached to another horizontally, and moreover the graph of the attachment relation has a perfect matching. Since there are $n$ columns and each can only have $2$ dominoes, this means that each column has $2$ dominoes. Then the $2$ dominoes in the first column must be matched with those in the second, so those in the third must be matched in the fourth, and so on until the $n - 1$ and $n - 2$ columns are paired and the last column cannot be paired. Bound: Let $t_{2k}$ be the answer for a $5$ by $2k$ rectangle. We show the recursion $t_{2k} \ge 2t_{2k - 2} + 2t_{2k - 4} + 4t_{2k - 6}$ holds for all $k > 3$, along with showing the bound holds for $t_2, t_4, t_6$, which is obviously sufficient. We present tilings for the $5$ by $2,4,6$ cases, and the $4$ and $6$ cases are tiled in such a manner such that no smaller tiling is present within them. 11 11 12 22 22 1112 1122 3322 3344 3444 466655 446655 443335 113322 111222 This demonstrates that there at least $2t_{2k -2} + 2t_{2k - 4} + 4t_{2k - 6}$ ways to tile the $5$ by $2k$ rectangle, as we can tile the first two columns using the tiling for $5$ by $2$ (there are two symmetric tilings by flipping horizontally), contributing $2t_{2k - 2}$ ways, alternatively we can tile the first four columns using the tiling for $5$ by $4$ (there are two symmetric tilings by flipping horizontally), contributing $2t_{2k - 4}$ ways (note that all of these ways are distinct from the previously counted ways, as the first two columns are not tiled as they would be in a $5$ by $2$), and lastly we can tile the first six columns using the tiling for the $5$ by $6$ (there are four symmetric tilings by flipping horizontally and vertically), contributing $4t_{2k - 6}$ ways (note that all of these ways are distinct from the previously counted ways, as we cannot tile the first $2$ or $4$ columns separately), so we are done.