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Is it possible to completely tile a square with different rectangles of integer sides but all with the same area?

The original problem, not requiring integer sides for rectangles, was proposed by Joe Fendel in Gary Antonick´s New York Times Monday puzzle column (June 13, 2013). A tiling with seven rectangles was subsequently provided.

Here is an image from Nick Baxter's solution (see Fendel's article link above), but with lengths normalized such that the outer square has side $1$. (Each "v" represents $\sqrt{19}$). It is easy to check that each rectangle has area $\dfrac17$. enter image description here

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The Nick Baxter solution is actually Blanche's Dissection, published in 1971.

I've outlined a general solution method at my Commuunity post Blanche Dissections. As a proof of concept, here are 16 dissections of a square into equal-area non-congruent rectangles:

blanche dissections

So far, none of these gives a rational solution. But many polyhedral graphs will give a Blanche-style solution. Since we have an infinite number of polyhedral graphs to choose from, there is very likely an integral solution. I just need to devote a bit more programming and computer power. If a solution doesn't pop out, then at least we'll have a few million more non-integral solutions.

Relax the problem to allow non-congruent integer-sided rectangles of different areas, but minimize the difference between the largest and smallest rectangle. This is the Mondrian Art Puzzle. Here are best solutions for 10x10 to 17x17.

Mondrian Art Solutions

We could also attack this from the numerical side. Some good candidate squares are the following:

Side 2520 into 35 rects of area 181440 (35 available).
Side 4080 into 34 rects of area 489600 (34 available).
Side 3960 into 33 rects of area 475200 (40 available).
Side 3780 into 30 rects of area 476280 (32 available).
Side 2520 into 30 rects of area 211680 (35 available).
Side 3120 into 26 rects of area 374400 (32 available).
Side 2340 into 26 rects of area 210600 (27 available).
Side 1560 into 26 rects of area 93600 (26 available).
Side 2760 into 23 rects of area 331200 (26 available).
Side 3300 into 22 rects of area 495000 (22 available).
Side 3080 into 22 rects of area 431200 (23 available).
Side 2772 into 22 rects of area 349272 (25 available).
Side 2640 into 22 rects of area 316800 (31 available).
Side 2310 into 22 rects of area 242550 (22 available).
Side 1980 into 22 rects of area 178200 (26 available).
Side 1848 into 22 rects of area 155232 (24 available).
Side 1320 into 22 rects of area 79200 (25 available).
Side 3150 into 21 rects of area 472500 (21 available).
Side 3024 into 21 rects of area 435456 (21 available).
Side 2940 into 21 rects of area 411600 (23 available).
Side 2772 into 21 rects of area 365904 (23 available).
Side 2730 into 21 rects of area 354900 (21 available).
Side 2520 into 21 rects of area 302400 (37 available).
Side 2310 into 21 rects of area 254100 (21 available).
Side 1890 into 21 rects of area 170100 (22 available).
Side 1680 into 21 rects of area 134400 (23 available).
Side 1260 into 21 rects of area 75600 (28 available).
Side 2280 into 19 rects of area 273600 (25 available).
Side 1140 into 19 rects of area 68400 (19 available).
Side 2856 into 17 rects of area 479808 (23 available).

The 3960x3960 square seems especially promising.

For 9 rectangles, square sizes {360, 504, 540, 630, 720, 756, 840, 990, 1008, 1080, 1170, 1188, 1260, 1386, 1404, 1440, 1512, 1530, 1584, 1620, 1638, 1680, 1800, 1872, 1890, 1980, 2016, 2100} are promising.

For 10 rectangles, square sizes {360, 420, 600, 660, 720, 840, 900, 1050, 1080, 1200, 1260, 1320, 1400, 1440, 1560, 1680, 1800, 1890, 1980, 2040, 2100, 2160} are promising.

For 11 rectangles, square sizes {330, 660, 792, 924, 990, 1320, 1386, 1540, 1584, 1650, 1716, 1848, 1980, 2244, 2310} are promising.

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    $\begingroup$ For the 16 Blanche dissections you exhibit, what are the algebraic degrees of the rectangles' sides? If they're bounded, or at least small, it would be easier to believe that a rational dissection will eventually turn up. $\endgroup$ – Noam D. Elkies Aug 11 '16 at 15:19
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    $\begingroup$ The polynomial degrees of root objects defining the rectangles in the given squares are as follows: 6 8 7 8 9 11 8 10 10 11 12 12 12 15 16 14 $\endgroup$ – Ed Pegg Jr Aug 11 '16 at 15:34

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