This is a geometric puzzle though it might conceivably 
also define a special class of Pythagorean triples. 


A *perfect squared square* PSS is a square (as a plane figure) 
partitioned into smaller squares, each of a different size. 
There are other types of squared squares or 
squared rectangles that have been studied (e.g. simple SPSS, vs compound CPSS), see 
[`wikipedia`](http://en.wikipedia.org/wiki/Squaring_the_square), [`wolfram`](http://mathworld.wolfram.com/PerfectSquareDissection.html), and [`squaring.net`](http://www.squaring.net/sq/ss/ss.html). 


Question. Is there a perfect squared square that 
can be split into two perfect squared squares? 
That is, could we use the building blocks (smaller squares) 
that form the given perfect squared square to 
form two smaller perfect squared squares? 


Of course if the given perfect squared square 
has side $c$ and the two smaller ones have 
sides $a$ and $b$ respectively, then the numbers 
$a$, $b$, $c$ would form a Pythagorean triple 
(since the areas of the two smaller squared squares 
sum up to the area of the given bigger squared square). 


Question. Which Pythagorean triples (if any) 
could be represented in the above form? 


For some Pythagorean triples $(a,b,c)$ the numbers $(a^2,b^2,c^2)$ 
seem to sometimes appear as *the sides* of neighboring smaller 
squares forming the partition of a perfect squared square. 
For example, for the Pythagorean triple $(3,4,5)$ the 
squared numbers are $(9,16,25)$ and these appear as 
*the sides* of three neighboring squares from the partition 
of the Lowest-order perfect squared square (i.e. formed by only 21 squares which is smallest possible for SPSS, same links as above). 
Could one say anything more about this (an explanation, or 
a description when it occurs, for which Pythagorean triples $(a,b,c)$)? 


Interestingly, a simple geometric argument shows that 
the analogue in three or more dimensions of squaring the square 
has no solutions, e.g. one cannot partition a cube 
(as a three-dimensional geometric figure) into smaller 
cubes, no two of which are congruent (see first link). 
One is tempted to make a wild guess that this might have 
something to do with Fermat's last theorem (however obvious 
it seems that there could be no actual relation). 


Incomplete history: Roland Sprague published in 1940 the 
first simple squared square 
[link](http://www.squaring.net/history_theory/sprague.html). 
He used squared rectangles found earlier by Zbigniew Moroń, 
plus additional squares. Another important early work was by 
R.L. Brooks, C.A.B. Smith, A.H. Stone and W.T.Tutte 
[link](http://www.squaring.net/sq/ss/spss/spss.html) who 
related the problem to electrical networks (graphs). 


Another post about squared squares is 
[link](https://mathoverflow.net/questions/145356/whats-the-best-way-to-characterise-the-distribution-of-prime-elements-in-simple) 
it has some related numerical data. 



Could a perfect squared square be split into two perfect squared squares?
Nice question. This is not (any longer) an answer, but a strategy. 


First, try to construct 25 mutually disjoint squared squares of the same order. Then arrange them according to a 3,4,5 template.


I initially thought that the 25 mutually disjoint squares should be an easy construction and followed from varying a pair of disjoint squares inside of a larger squared square. But distinctness of the little squares seems tricky to enforce.


Even if this works, I admit it is unsatisfactory. The example squares will be huge and (badly) compound. Simple squares are better.


Incidentally, construction of disjoint squared squares of the same order (number of constituents) and size seems to be a challenging problem;
see <http://www.squaring.net/sq/sr/spsr/spsr_dnt.html>


Edit by Mirko: As indicated in comments below, it is enough to find nine (or even eight) mutually disjoint PSS of the same size. The list of order 25 SPSS at 
<http://www.squaring.net/sq/ss/spss/o25/spsso25.html> 
includes eight squares of size $293,311,317,367,373,421,503,541$, 
which are all prime numbers. (For $421$ and $541$ each, there are versions 
A and B, but we only take version A.) Let $M=293\cdot 311\cdot 317\cdot 367\cdot 373\cdot 421\cdot 503\cdot 541 =$ $453011799002853190123$. 
Inflate each of the above eight squares by the product of the sizes 
of the other seven squares, e.g. inflate the square with size 
$293$ by a factor of $311\cdot 317\cdot 367\cdot 373\cdot 421\cdot 503\cdot 541 =$ $M/293 =$ $1546115354958543311$, and similarly 
for the other sizes. We obtain eight PSS, say $K\_1,...,K\_8$, each of size $M$. 
They are mutually disjoint, e.g. the side of none of the squares in the partitioning of 
$K\_1$ is divisible by $293$, whereas the side of each square in 
the partitioning of each of $K\_2,...,K\_8$ is divisible by $293$. 
So we could put together $K\_1,...,K\_8$, plus a square of side $M$, 
and a square of side $4 M$ to form a CPSS of side (=size) $5 M$. 
Clearly it splits into one PSS of size $4M$ (and of order $1$ ... 
what a cheat), and another CPSS of size $3M$ (and as it happens, 
of order $8\cdot 25 +1= 201$). To make this a bit less of a cheat, we could start with nine primes, that is add e.g. $547$ to the above list, and redefine 
$M=293\cdot 311\cdot 317\cdot 367\cdot 373\cdot 421\cdot 503\cdot 541 \cdot 547=$ $247797454054560694997281$. We get 
$K\_1,...,K\_8,K\_9$ in a similar way as before, but now, in addition, inflate $K\_9$ by a factor of four and use the result in place of the single square of size $4 M$. 


Thank you for all contributions! 


