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.

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As discussed after my incomplete solution: The answer is yes if we have 25 mutually disjoint perfect squared squares of some order. This page discusses disjoint families: squaring.net/sq/sr/spsr/spsr_dnt.html. All indications are that a family of 25 should exist. But, unfortunately, I see no elementary way to get the family (although at first I thought it should be easy via compound squares). –  Peter Dukes Mar 25 '14 at 3:53
The tables at squaring.net might already have 25 suitable mutually disjoint squares, of various orders but same side length, though I do not know how to filter this information out. –  Mirko Mar 25 '14 at 4:08
What we could really use is, for each square, a list of multiples of that square which are mutually disjoint. With enough such examples, we can recursively split up squares into smaller and smaller squares and guarantee they are all disjoint. –  Peter Dukes Mar 26 '14 at 2:25
Mirko: a formal comment. It's a bit hard for me to read a term like perfect squared square. Could you replace it with psquare--it'd really help me, and perhaps others would not mind? –  Włodzimierz Holsztyński Mar 26 '14 at 4:00
@ Wlodzimierz Here abbreviations SPSS and CPSS are used for Simple Perfect Squared Squares and Compound Perfect Squared Squares, seems a standard terminology. So my question is about PSS (without specifying SPSS or CPSS, but assuming, as the following discussion shows, that CPSS are fine). –  Mirko Mar 26 '14 at 13:02

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!

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Well, a simple square doesn't have to be a simple example ;) Your proof shows there are infinitely many Pythagorean triples $(a,b,c)$ that are sides of perfect squared squares $A,B,C$, respectively, where $C$ may be compound, and is built from the squares forming $A$ and $B$ ... actually are there infinitely many that are essentially different (no common factor)? A computer search may work, but it may also be slow, if large numbers are involved: It might be interesting if one could find any shortcuts to help the computer. –  Mirko Mar 24 '14 at 21:49
I hesitated, initially, when I read step (2), but since 1429 is a prime, it seemed in this case it presented no problems. So I did not follow why you needed a large square: You need to get at least 25 squares, so it is enough if $S$ is formed by at least 5 squares (since $2^5=32 > 25$), but this is a vacuous requirement since the Lowest-order perfect squared square has 21 squares, well above 5. Perhaps it is not as easy and I am missing something, will review it again. I am starting to question what was the relevance of 1429 being a prime ... –  Mirko Mar 25 '14 at 0:42
I no longer follow (2), neither what I meant in my previous comment. What I had "understood", was that given any one of the 25 squares, the "inflated" squares inside it are mutually disjoint (which is easy using 1429 is a prime), and not that these 25 squares are mutually disjoint ... I do not see how to get the latter, even if $S$ is large. Of course, if the 25 squares are not mutually disjoint then this solution would not be interesting: We could as well have split 5x5 into 3x3+4x4, if we allowed repetitions of the same building block. I do not even follow your idea in (2). –  Mirko Mar 25 '14 at 2:32
You don't need 25, just 9 of the same size and one that is four times the size. Again a cheat, but with fewer requirements. Also, one of the nine need only be a single square. –  The Masked Avenger Mar 26 '14 at 3:41
Say $S,T$ are PSS (perfect squared squares) of size $s,t$. We want to make them disjoint, keeping $s/t$ constant, as follows. First zoom in, i.e. multiply each $S$ and $T$ by the same factor $c$ (e.g. the size of some PSS $C$). Then replace each (or some) constituent square of $T$ with a PSS (e.g. using $C$, though there may be other ways). Do this finitely many times, if necessary (zoom into $S,T$, then refine $T$), till the largest square in the modified $T$ is smaller than the smallest square in the modified $S$. Is it possible to do this, so in the end the modified $T$ remains perfect? –  Mirko Mar 26 '14 at 13:48

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.

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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.

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Marvelous! A related problem on squaring of rectangles with an asnwer from the same person at mathstackexchange. –  Mirko Apr 6 '14 at 6:04