These days I'm trying to research what's about of relations between prime numbers and the notion of chirality in the $xy-$plane. Wikipedia has the article Chirality.

I don't know if this relation or the problem for which I ask here is in the literarture, 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 to 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 to 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 arangement of $n=A\cdot B$ unit squares. I emphasize with this 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.

Question. I would like to know what work can be done to prove the veracity of previous conjecture. Many thanks.

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've the following obvious statement.

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

Thus we must to deal with squares $S=N\times N$ with $N>7$ an odd integer. Exploting a specialization of a well-known identity $1+3+5=3^2$ I can to tessellate the square $9\times 9$, because this square can be splitted 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 tesellated 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 to get it, and it is required, I can to add in a figure some of the mentioned tessellations.