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Aaron Meyerowitz
Aaron Meyerowitz
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A LITTLE MORE WORK

We know how to get all squares which are of even side at least $4$ using just the $L$-teromino (no $J$-tetrominos)

Here are a $5 \times 5, 7 \times 7$ and $9 \times 9$. The solid rectangles are all ones which we know we can fill. For what it is worth, the $5 \times 5$ uses only $L$'s. I did not try to do that for the rest, it might not be hard. In fact the $9 \times 9$ has that property,

enter image description here

With four $3 \times 8$ rectangles around a $5 \times 5$ square we can make an $11 \times 11$ square.

Finally, if $k \geq 5$ is odd then we can make a a square of side $k+8$ by using a $k\times k$ in the upper left, and $8 \times 8$ on the lower right, a $k \times 8$ in the upper right and an $8 \times k$ in the lower left.

A LITTLE MORE WORK

We know how to get all squares which are of even side at least $4$ using just the $L$-teromino (no $J$-tetrominos)

Here are a $5 \times 5, 7 \times 7$ and $9 \times 9$. The solid rectangles are all ones which we know we can fill. For what it is worth, the $5 \times 5$ uses only $L$'s. I did not try to do that for the rest, it might not be hard. In fact the $9 \times 9$ has that property,

enter image description here

With four $3 \times 8$ rectangles around a $5 \times 5$ square we can make an $11 \times 11$ square.

Finally, if $k \geq 5$ is odd then we can make a a square of side $k+8$ by using a $k\times k$ in the upper left, and $8 \times 8$ on the lower right, a $k \times 8$ in the upper right and an $8 \times k$ in the lower left.

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Aaron Meyerowitz
Aaron Meyerowitz
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Here is a different answer. You mention a particular interest in the $L$-$n$omino: a column of $n-1$ squares with one more square to the right on the bottom row. It has $4$-rotations and another $4$ obtained by rotation and reflection. The reflection might be called the $J$-$n$omino if one wanted to distinguish rotations from reflections (more often, people do not.) It is true that $n-1=\varphi(n)$ for $n$-prime, but I don't see that you have motivated your focus on primality.

I will conjecture that if $\gcd(m,n)=1$ then using just the $L$-$m$omino and $L$-$n$omino you can tile any rectangle with both sides above some $N=N(m,n).$ In fact it might be that you could make do with just one orientation of each one in all but a few cases (that is to say, $L$s but not $J$s). But then you would need a larger $N.$

In spite of your disinterest in tiling with rectangles, that is the key. Once you have a few rectangles you can use those as blocks to make others.

A coloring argument shows that any rectangle tiled by the $L$-tetromino uses an even number so has area a multiple of $8$. One easily finds how to get a $2\times 4$ and once you know to look for a $3\times 8$ you can find that too.

Aside: These two are called the primes of this tetromino because any other rectangle you can get can be made from them.

Claim: any rectangle with area a multiple of $8$ and shortest side at least $2$ can be tiled.

Sketch: if both sides are even, the $2 \times4$ alone is sufficient. If one side is odd, the other is a multiple of $8$. Stacking $k-1$ $2 \times 8$ and a single $3 \times 8$, you can get $(2k+1) \times 8$

You can find a great deal of information about rectangle tilings at this page

For the $L$-pentomino one finds figures showing that the $2 \times 5$ and $7 \times 15$ can be tiled

Claim: Any rectangle with area a multiple of $5$ can be tiled as long as the one side is at least $15$ and the other at least $7.$

Using just the $2 \times 4$ and $2 \times 5$ you can get any $2 \times k$ except for $k=1,3,6,7,11$ and from that any even area rectangle with shortest side at least $12.$ One could probably improve that.

And using just those two $L$-ominos and the four prime rectangles mentioned, one can get a $15 \times k$ and an $8 \times k$ for any $k \geq 7$ and hence also a $j \times k$ as long as $j \geq 98.$ Probably much better results are possible with a little work.

A link on that pentomino page leads to the claim (perhaps with illustrations proving it) that using only the $L$ (and not the $J$) one can get $2\times 5, 13\times 55, 15\times 39, 17 \times 35$ and $19 \times 25.$ Those are the one handed primes and with them you can get any rectangle with area a multiple of $5$ as long as the shortest side is at least $19$.