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MATHEMATISCHE ZEITSCHRIFT, 45 (1939), pp. 607--608.\\
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{\bf An example of a dissection of the square into pairwise unequal squares.}\\
by\\
R. Sprague in Berlin-Charlottenburg.
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Herr St\"ohr\footnote{A. St\"ohr, \"Uber Zerlegungen
von Rechtecken in inkongruente Quadrate.  (Dissertation.)
Schriften des Math.\ Inst.\ und des Inst.\ f.\ angew.\ Math.\ der
Universit\"at Berlin, Band 4 (1939), S.\ 119--140.  Vgl.\ die
Literaturangaben auf S.\ 119 und 120.}
was recently occupied with the question of whether a
rectangle of given aspect ratio, in particular a square, can be
dissected into finitely many pairwise different squares.  Theorem
10 of his dissertation states:

If any rectangle with sides $a$ and $b$, $a<b$, admits two
such dissections, which have no component square in common, and do not
contain the square with side $a$, then the square with side
$a+b$ can be dissected in the desired way (namely, into the
squares with sides $a$ and $b$ and the two given dissected
rectangles.)

In the following we give two dissections of a rectangle with
aspect ratio $13:16$ which satisfy the conditions of this theorem.

% Fig. 1
% Fig. 2

The first is shown in Figure 1; it has been based on 
well-known\footnote{Jaremkewycz, Mahrenholz, Sprague, L\"osungen der
Aufgabe 1242.  Zeitsch.\ f.\ d.\ math.\ u.\ naturwiss.\ 
Unterricht 68 (1937), S.\ 43.}
dissections of the rectangles with aspect ratio $32:33$ and
$47:65$ into pairwise unequal squares, which coincidentally 
have no square in common.
The numbers denote the side lengths of the component squares.

We name the rectangle of Fig.\ 1 $R_1$; it has sides of
$5\cdot 13$ and $5\cdot 16$ and it does not contain the square
with side equal to its smaller side as a component.

In order to find the second decomposition, we first augment $R_1$
with a square along its longer side, thus producing a rectangle
$R_2$ with sides $5\cdot 16$ and $5\cdot 29$.  Furthermore,
a rectangle $R_3$ with sides $16\cdot 16$ and $13\cdot 29$
is used; its dissection is shown in Figure 2.

The sides of the squares are

in $R_2$: 1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 14, 15, 17, 18, 19, 22,
23, 24, 25, 33, 80;

in $R_3$: 7, 10, 28, 54, 61, 68, 75, 113, 115, 123, 133, 141.

Linear enlargement of $R_2$ in the ratio $1:13$ and of $R_3$ in
the ratio $1:5$ yields rectangles $R'_2$ with sides
$5\cdot 13\cdot 16$ and $5\cdot 13\cdot 29$ and $R'_3$ with sides
$5\cdot 16\cdot 16$ and $5\cdot 13\cdot 29$.

Joining $R'_2$ and $R'_3$ at the sides of equal length creates one
rectangle $R_4$ with sides $5\cdot 13\cdot 29$ and $5\cdot 16\cdot 29$.
It does not contain the square with side equal to its smaller
side as a component.  The rectangle $R_4$ contains pairwise
unequal squares. This is because $R_2$ and $R_3$ have this property, and
unlike the squares in
$R'_2$, no square in $R_3$, and hence none in $R'_3$, has side
a multiple of 13.

No side of a square in $R_2$ or $R_3$ is a multiple of 29; therefore,
the same is true for $R_4$.  The enlargement of $R_1$ in the ratio
$1:29$ and $R_4$ thus are two dissections of the rectangle with
sides $a=5\cdot 13\cdot 29$ and $b=5\cdot 16\cdot 29$ which fulfill
the conditions.  

$R_1$ consists of 20, $R_4$ of 33 component squares; two more
component squares also appear, those with sides $a$ and $b$.
Therefore, each square is dissectable into 55 pairwise unequal
squares.

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