Cutting and reassembling squares Is there a general way to cut a square into polygonal pieces so that the pieces can be assembled into n equally sized squares for each n? For example, 2 and 4 and n=k^2 is obvious (2 by the diagonals and the n=k^2 by k rows and k columns)
 A: By the Bolyai-Gerwien Theorem. if $A$ and $B$ are polygonal regions of equal area, then $A$ can be decomposed into a finite number of polygonal pieces which can be reassembled to form $B$.  So of $A$ is our square, divide it into $n$ rectangles of equal area in the obvious way. Then these rectangles can be divided into polygonal pieces which can be reassembled to make our $n$ squares.
Remark: Decomposing a rectangle into polygonal pieces that can be reassembled into a square is relatively simple. If the length is much greater than the width, cut into congruent rectangles and stack the pieces until you have a rectangle which is no worse than $1\times 4$. Then by drawing a single line one can decompose that rectangle into pieces that can be reassembled to make a square. Please see this, Figure 9.1.
A: Here is a systematic method.  I am not good at drawing diagrams online so please draw them for yourself - I hope my instructions are clear.
Given a square $ABCD$ of side length $\sqrt n$, we want to cut it up and arrange the pieces into $n$ unit squares.  Let $k=\lfloor\sqrt n\rfloor$ be the largest integer not greater than $\sqrt n$.  Locate $F$ on $CD$ such that
$$DF=\sqrt{\frac{n^2}{k^2}-n}\ .$$
Draw $AF$, and find $E$ on $AF$ such that $BE\perp AF$.
Now cut off the triangle $ADF$ and attach the side $AD$ to $BC$, thus forming a parallelogram with altitude $BE$.  You can check that the length of $BE$ is $k$, that is, it is an integer, so this parallelogram can be cut into parallelograms of altitude $1$; these can then be placed side by side to make one long parallelogram of altitude $1$.
Then cut a bit off one side and transfer it to the opposite side to form an $n\times1$ rectangle.  It is then easy to cut this into $n$ unit squares.
Sorry if this is confusing but if you draw the diagrams I think you will find it's not too hard to follow.
Comments.


*

*It is possible that $DF=0$, that is, $F=D$.  This is not a problem.

*You can in fact take other integer values for $k$.  I'm not sure if all of them work.

A: Let's say we can "$n$-cut" a square if we can cut a square into polygonal pieces that can be assembled into $n$ equally sized squares. Note that if you can $a$-cut a square and also $b$-cut a square, then you can $ab$-cut a square (just $b$-cut each of the $a$ smaller squares, or vice versa). Therefore it suffices to $p$-cut a square for every prime $p$.
I'm guessing this is possible. $3$-cutting a square is a classical problem (see especially page $4$). I also found a picture $5$-cutting a square. I imagine that these diagonal-lattice cutting strategies can be generalized quite substantially, perhaps to all primes $p$.
