For what n and t can a square be partitioned into n similar rectangles in t congruence classes? It is known that a square can be partitioned into three similar rectangles, all mutually non-congruent. I don't think it's possible with four. With what numbers of rectangles can this be achieved? And what if we relax the congruence condition?
So far, for $n$ similar rectangles in $t$ congruence classes, I have got
$$\begin{array}{cc|cccccc|}
   & & \text{no. of rectangles, n}\\
   & & 1& 2& 3& 4& 5& 6\\
\hline 
  \text{no. of congruence classes, t}& 1& yes& yes& yes& yes& yes& yes\\  
  & 2& -& no& yes& yes& yes& yes& \\ 
  & 3& -& -& yes& yes& ?& ?& \\ 
  & 4& -& -& -& (no?)& ?& ?& \\ 
  & 5& -& -& -& -& ?& ?& \\ 
  & 6& -& -& -& -& -& ?& \\ 
\hline 
\end{array}$$
 A: Here is the proof that square can be partitioned into $n$ non-congruent similar rectangles for any $n\ge 3$.
Consider this modification of the @florian-lehner's construction.

*

*fix $A > 1$, we will choose it later

*start with a $1\times 1$ square and $A \times A$ square lined up so that their upper borders are on the same height

*alternate in attaching the square of appropriate size above and on the right of the drawing until you have $n-1$ squares in total

It should now look like this (here $n=5$):

From here on there is a slight difference between odd and even case.
For odd $n$ you get the rectangle of the size $(F_{n-2} A + F_{n-3}) \times (F_{n-1} A + F_{n-2})$ with a missing rectangular corner in the lower-left, of the size $(A-1) \times 1$. Here $F_n$ denote Fibonacci numbers.
Rescale horizontally by factor $\frac{F_{n-2} A + F_{n-3}}{F_{n-1} A + F_{n-2}}$ so that the big figure is a square, and now we need to find such $A$ that the missing corner rectangle is similar to all the present ones (but has different orientation).
This gives us
$$ \frac{F_{n-2} A + F_{n-3}}{F_{n-1} A + F_{n-2}} : (A-1) = 1 : \frac{F_{n-2} A + F_{n-3}}{F_{n-1} A + F_{n-2}}$$
or the following cubic equation:
$$f(A):=(A-1)(F_{n-1} A + F_{n-2})^2 - (F_{n-2} A + F_{n-3})^2 = 0$$
It is easy to see that $f(1)<0$ and $f(A)\rightarrow +\infty$ as $A$ grows. Thus by Intermediate value theorem the equation has a solution greater than $1$.
For even $n$ big rectangle is of the size $(F_{n-1} A + F_{n-2}) \times (F_{n-2} A + F_{n-3})$, but the argument is essentially the same.
Also here is Sage code that generates such partitions for any $n\ge 3$:
n=3

move = lambda P,v: map(lambda x: (x[0]+v[0],x[1]+v[1]), P) #transport by vector
times = lambda P,c: map(lambda x: (c*x[0],c*x[1]), P) #homothety

A=var('A')

if not n%2:
    A=solve((A-1)*(fibonacci(n-2)*A + fibonacci(n-3))^2 - (fibonacci(n-1)*A + fibonacci(n-2))^2 == 0, A, solution_dict=True)[2][A] #real solution of the equation in the even case
    W=(fibonacci(n-1)*A + fibonacci(n-2))/(fibonacci(n-2)*A + fibonacci(n-3)) #scaling factor in the even case
else:
    A=solve((A-1)*(fibonacci(n-1)*A + fibonacci(n-2))^2 - (fibonacci(n-2)*A + fibonacci(n-3))^2 == 0, A, solution_dict=True)[2][A] #real solution of the equation in the odd case
    W=(fibonacci(n-2)*A + fibonacci(n-3))/(fibonacci(n-1)*A + fibonacci(n-2)) #scaling factor in the odd case
H=A-1 #height of the "missing" corner

#now to just plot all the rectangles

horizontal=[(0,0),(W,0),(W,1),(0,1)]
vertical=[(0,0),(W,0),(W,H),(0,H)]

sumplot=polygon(vertical,fill=False)
v=A-1
c=1
for i in range(n//2):
    sumplot+=polygon(move(times(horizontal,c),(0,v)),fill=False)
    v=v+c
    c=v+c
v=1
c=A
for i in range((n-1)//2):
    sumplot+=polygon(move(times(horizontal,c),(W*v,0)),fill=False)
    v=v+c
    c=v+c

sumplot.show()

A: Here is a partial answer for when we relax the congruence relation: If the number  $n$ of rectangles and the number $t$ of congruence classes satisfy $n \geq \frac{t (t+1)}2$, then the following construction gives a tiling (see the sketches below for $t=3$ and $n \in \{6,8\}$):

*

*start with any rectangle (we'll rescale the drawing later, so the exact dimensions don't matter for now)

*for $2 \leq j \leq t-1$, take $j$ copies of the rectangle scaled by $\frac 1j$ drawn above each other (so the total height is the same as the total height of the original rectangle) and to the right of what was drawn so far.

*same as the second step, but with $j = n - \frac{t (t-1)}{2}$.

This construction doesn't give a tiling of a square, but we can always scale the coordinate system of our drawing so that it does. Since the orientation of all rectangles is the same, similarity is preserved under this scaling.



EDIT:
A similar construction shows that as long as $n > t$, a tiling exists. Start with $n-t+1$ congruent rectangles drawn above each other. Then alternate in attaching a scaled (but not rotated) copy with the same height as the current drawing on the right, and attaching a scaled copy with the same width as the current drawing above (see below for $t=5$, $n=6$). Finally scale the axes so that the whole drawing fills a square.

