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6.AW.3. DIVIDING A SQUARE INTO CONGRUENT PARTS
In the 1960s, a common trick was to give someone a number of quadrisection problems where the parts happen to be congruent to the original figure -- e.g. the quadrisection of the square or the L-tromino. Then ask her to divide a square into five congruent parts. She usually tries to use square pieces in some way and takes a long time to find the obvious answer. c1980, Des MacHale told me that it was a serious question as to whether there was any non-trivial dissection of a square into five or even three congruent pieces. Sometime later, I found a solution -- slice the square into 6 equal strips and say part A consists of the 1st and 4th strips, part B is the 2nd and 5th, part C is the 3rd and 6th. However this is not what was intended by the problem though it leads to other interesting questions. Since then I have heard that the problem has been 'solved' negatively several times on the backs of envelopes at conferences, but no proof seems to have ever appeared. Very little seems to be published on this, so I give what little I know. Much of this applies to rectangles as well as squares. QARCH is an occasional problem sheet issued by the Archimedeans, the Cambridge (UK) student mathematics society.
Gardner, in an article: My ten favorite brainteasers in Games (collected in Games Big Book of Games, 1984, pp. 130-131) says the dissection of the square into five congruent parts is one of his favorite problems. ??locate

David Singmaster. Problem 12. QARCH III (Aug 1980) 3. Asks if the trisection of the square is unique.

David Singmaster. Response to Problem 12 and Problem 21. QARCH V (Jan 1981) 2 & 4. Gives the trisection by using six strips and unconnected parts. In general, we can have an n-section by cutting the square into kn strips and grouping them regularly. For n = 2, k = 4, there is an irregular dissection by using the parts as strips 1, 4, 6, 7 and 2, 3, 5, 8. If p is an odd prime, are there any irregular p-sections?

John Smith, communicated in a letter from Richard Taylor, editor of QARCH, nd, early 1981? Smith found that if you slice a square into 9 strips, then the following parts are congruent, giving an irregular trisection. 1, 2, 6; 3, 7, 8; 4, 5, 9.

David Singmaster. Divisive difficulties. Nature 310 (No. 5977) (9 Aug 1984) 521 & (No. 5979) (23 Aug 1984) 710. Poses a series of problems, leading to the trisections of the cube. No solutions were received.

Angus Lavery asserts that he can trisect the cube, considered as a 3 x 3 x 3 array of indivisible unit cubes. I had sought for this and was unable to find such a trisection and had thought it impossible and I still haven't been able to do it, but Angus swears it can be done, with one piece being the mirror image of the other two.

George E. Martin. Polyominoes -- A Guide to Puzzles and Problems in Tiling. MAA Spectrum Series, MAA, 1991. Pp. 29-30. Fig. 3.9 shows a 5 x 9 rectangle divided into 15 L-trominoes. Shrinking the length 9 to 5 gives a dissection of the square into 15 congruent pieces which are shrunken L-trominoes. Prob. 3.10 (very hard) asks for a rectangle to be dissected into an odd number of congruent pieces which are neither rectangles nor shrunken L-trominoes. He doesn't give an explicit answer, but on p. 76 there are several rectangles filled with an odd number of L, P and Y pentominoes. One might argue that the L and P have the shape of some sort of shrunken L-tromino, but the Y-pentomino is certainly not. Prob. 3.11 (unsolved) asks if a rectangle can be dissected into three congruent pieces which are not rectangles. Prob. 3.12 is a technical generalization of this and hence is also unsolved.

On 19 Jun 1996, I proposed the trisection of the square and Lavery's problem on NOBNET. Michael Reid demonstrated that Lavery's problem has no solution and someone said Lavery had only conjectured it. Reid also cited the following two proofs that the square trisection is impossible.

I. N. Stewart & A. Wormstein. Polyominoes of order 3 do not exist. J. Combinatorial Theory A 61 (1992) 130-136. ??NYS -- Reid says they show that if a rectangle is dissected into three congruent polyominoes, then each is a rectangle.

S. J. Maltby. Trisecting a rectangle. J. Combinatorial Theory A 66 (1994) 40-52. ??NYS -- Reid says he proves the result of Stewart & Wormstein without assuming the pieces are polyominoes.

Martin Gardner. Six challenging dissection tasks. Quantum (May/Jun 1994). Reprinted, with postscript, in Workout, chap. 16. Trisecting the square into congruent parts is his first problem. Cites Stewart & Wormstein. Then asks if one can have three similar parts, with just two, or none, congruent. Then asks the same questions about the equilateral triangle. All except the first question are solved, but the solutions for the fifth and sixth are believed to be unique. In the postscript, Gardner says Rodolfo Kurchan and Andy Liu independently suggested the problems with four parts and cites Maltby.