These days I'm trying to research relations between prime numbers and the notion of chirality in the $xy$-plane. Wikipedia has the article [*Chirality*](https://en.wikipedia.org/wiki/Chirality).


I don't know if this relation or the problem for which I ask here is in the literature; please add a comment in such case.


**Definition.** In this post, I consider a definition ad hoc, defining as **admissible polygons** $P$ the connected regions in the $xy$-lattice that are arrangements of unit squares with $\text{area}(P)=p>3$ a prime number, and also by definition I consider that these polygons must have a chiral shape, that is these pieces are chirals. A tessellation of a connected region $R$ by means of admissible polygons is called an **admissible tessellation of** $R$ (in our problem $R=S$ will be a square $S$ in the $xy$-lattice).


(The rationale for previous definitions is that any polygonal representation of the integers $1,2$ or $3$ aren't chirals, and on the other hand, again by definition one can consider that a given composite integer can be represented as a rectangle or square $A\times B$, that isn't chiral polygon, of sides $1<A\leq B$ in the $xy$-lattice as an arrangement of $n=A\cdot B$ unit squares. I emphasize with these brackets that these pieces will not be used in the tessellations in this post).


**Conjecture.** *There exists a constant* $N\_0$ *such that every square* $S=N\times N$ *of side* $N$ *with* $N>N\_0$ *can be represented by admissible polygons.*



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> **Question.** I would like to know what work can be done to prove the veracity of the previous conjecture. **Many thanks.**
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With admissible polygons it is easy get admissible tessellations for the square $4\times 4$, $5\times 5$, $6\times 6$ and $7\times 7$. We have the following obvious statement.


**Proposition.** *Given an admissible tessellation for the square* $S=N\times N$ *of* $N^2$ *unit squares, it is immediate to get an admissible tessellation for the square* $\hat{S}$ *of side* $2N$, *since this second square* $\hat{S}$ *admits the obvious decomposition into four squares* $S$.


Thus we must deal with squares $S=N\times N$ with $N>7$ an odd integer. Exploiting a specialization of a well-known identity $1+3+5=3^2$ I can tessellate the square $9\times 9$, because this square can be split as two copies of the square $(1+3)\times(1+3)$, plus the tessellated region corresponding to the square $5\times 5$ plus a region that can be tessellated with two admissible polygons with areas $7$ and $17$, being these chirals (I just emphasize it to provide details).


I think that the conjecture isn't obvious.


If you have problems getting it, and it is required, I can add in a figure some of the mentioned tessellations.



Primes and chirality: a definition and question in the context of tessellations for squares
I claim that there is an $N$ so that any rectangle with both sides at least $N$ can be decomposed into squares of sides $4,5,6$ and $7.$ If we show that, then the same applies squares of sides at least $N.$ I will show this for $N=1178.$ With a little more work that number could be decreased to $90.$ Although that is probably not optimal.


Let $a,b>0$ be relatively prime integers.


It is a result of [Frobenious](https://en.wikipedia.org/wiki/Coin_problem) that any integer $m\geq f(a,b)=ab-a-b+1$ can be written in the form $m=as+bt$ with $s,t$ non-negative.


Using just $a\times a$ squares we can make a rectangle $a\times ab$ and using just $b \times b$ squares we can make a rectangle $b \times ab.$ Using those blocks we can make a rectangle $m \times ab$ for any $m\geq f(a,b)=ab-a-b+1$


Hence


* Using $4 \times 4$ and $5 \times 5$ squares we can make a rectangle $20 \times m$ for $m \geq f(4,5)=12.$


-Using $7 \times 7$ and $9 \times 9$ squares we can make a rectangle $63 \times m$ for $m \geq f(7,9)=48$


* Using all four sizes of squares we can make any rectangle $m \times n$ provided that $m \geq 48$ and $n \geq f(20,63)=1178.$


Sticking to squares and rectangles which can be decomposed into vertical rows each of which itself can be decomposed into horizontal columns of widths $4,5,6,7$ or $9$ (with the further restriction that each row uses only two widths) , is probably far from optimal. However, by the method above we can get


* Any row $2x \times 2y$ with $\min(2x,2y) \geq 12$ using $(a,b)=(4,6).$
* Any row $3x \times 3y$ with $\min(3x,3y) \geq 18$ using $(a,b)=(6,9).$
* Any row $ab \times m$ with $m \geq f(a,b)=ab-a-b+1$ for any of the eight relatively prime pairs drawn from $\{4,5,6,7,9\}$


Ignoring the first two types, we can get any rectangle $p \times q$ with with $p \in \{20,28,30,35,36,42,45,63\}$ and $q \geq 48.$


Combining those we can get any rectangle $r \times q$ with


$r \in \tiny{\{ 20, 28, 30, 35, 36, 40, 42, 45, 48, 50, 55, 56, 58, 60, 62, 63, 64, 65, 66, 68, 70, 71, 72, 73, 75, 76, 77, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88\}}$ or $r \geq 90$ and $q \geq 48.$


