Could a perfect squared square be split into two perfect squared squares? 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, wolfram, and squaring.net. 
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. 
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 who 
related the problem to electrical networks (graphs). 
Another post about squared squares is 
link 
it has some related numerical data. 
 A: Mirko's solution is of order 226, A solution of order 202 can be constructed with a different choice of SPSSs (from http://www.squaring.net/ ).
The five order 22 SPSSs with sides 110(A), 139, 154, 172 and 192, together with the three order 23 SPSSs with sides 188, 208 and 257, are mutually disjoint (with a total of 179 subsquares). To see this, express the side of each subsquare as a fraction of the side of the SPSS. List the denominators associated with each SPSS. The only one associated with more than one SPSS is 16, for 22:192A and 23:208A, but those SPSSs are mutually disjoint. Let M be the LCM of the sizes of the eight SPSSs. We can put together these eight SPSSs, each of side 7M, a square of side 7M, and SPSS 22:147A of side 28M to form a CPSS of side 35M. It can be split into the order 22 SPSS of side 28M and an order 180 CPSS of side 21M. 
A: My 2nd example is a PSS of order 90 with side 14137200.
Let k = lcm(110,112,135,136) = 2827440. Multiply the elements of SPSSs 22:110A and 21:112A by 25704 and 75735 respectively, and of SPSRs 22:272x136 and 23:270x135 (the one with corner square 70) by 20790 and 41888 respectively. This gives SPSSs with sides k and 3k, and SPSRs 2k x k and 4k x 2k. Together with squares with sides k and 2k, the pieces tile an order 90 CPSS of side 5k or, alternatively, the order 21 SPSS with side 3k and an order 69 CPSS with side 4k.
The 2x1 SPSRs can be found in A. J. W. Duijvestijn's "Simple Perfect Squared Squares and 2x1 Squared Rectangles of Orders 21 to 24", J. Combinatorial Theory, Series B 59, 26-34 (1993) at http://doc.utwente.nl/17948/1/Duijvestijn93simple.pdf
My 3rd example is a PSS of order 77 with side 1054680.
Let k = 210936. Create (1) a CPSR 3k x k by juxtaposing SPSRs 14:533x376A and 15:595x376A and multiplying their elements by 561; (2) an SPSR 2k x k from 22:272x136 by multiplying its elements by 1551; (3) a perfect squared hexagon (a square 4k x 4k with a k x k square missing from a corner) by removing corner element 66 from SPSS 25:264C and multiplying its other elements by 3196. These three pieces, together with two squares (sides k and 2k), form an order 77 CPSS with side 5k or, alternatively, the order 25 SPSS with side 4k and an order 52 CPSS with side 3k.
The lowest order of a 3x1 PSR may be 26, which may enable an example of order 74 to be constructed. 
