B. Smith, A. H. Stone and W. T. Tutte (writing under the collective pseudonym "Blanche Descartes") at Cambridge University between 1936 and 1938.
[1] Martin Gardner published an extensive article written by W. T. Tutte about the early history of squaring the square in his Mathematical Games column of November 1958.
Theophilus Harding Willcocks, an amateur mathematician and fairy chess composer, found another.
Willcocks in 1946 and has 24 squares; however, it was not until 1982 that Duijvestijn, Pasquale Joseph Federico and P. Leeuw mathematically proved it to be the lowest-order example.
In other words, the greatest common divisor of all the smaller side lengths should be 1.
[7] Computer searches have found exact solutions for small values of
This problem was later publicized by Martin Gardner in his Scientific American column and appeared in several books, but it defied solution for over 30 years.
The recursive scaling process increases the sizes of the squares exponentially – skipping most integers – a feature which they note was true of all perfect integral tilings of the plane known at that time.
In 2008 James Henle and Frederick Henle proved Golomb's heterogeneous tiling conjecture: there exists a tiling of the plane by squares, one of each integer size.
Their proof is constructive and proceeds by "puffing up" an L-shaped region formed by two side-by-side and horizontally flush squares of different sizes to a perfect tiling of a larger rectangular region, then adjoining the square of the smallest size not yet used to get another, larger L-shaped region.
The squares added during the puffing up procedure have sizes that have not yet appeared in the construction and the procedure is set up so that the resulting rectangular regions are expanding in all four directions, which leads to a tiling of the whole plane.
Now suppose that there is a perfect dissection of a rectangular cuboid in cubes.
The base is divided into a perfect squared rectangle R by the cubes which rest on it.
The smallest square s1 in R is surrounded by larger, and therefore higher, cubes.
By the claim above, this is surrounded on all 4 sides by squares which are larger than s2 and therefore higher.