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!

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.

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.