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Plimpton 322 : A Universal Cuneiform Table for Old Babylonian Mathematicians, Builders, Surveyors and Teachers


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This article deals with the damaged and incomplete Old Babylonian tablet Plimpton 322 which contains 4 columns and 15 rows of a cuneiform mathematical text. It has been shown that the presumed original table with its 7 columns and 39 rows represented: a table of square roots of numbers from 0 to 2 for mathematicians; an earliest rudiments of a trigonometric table for builders and surveyors where angles are not measured as an arc in a unit circle but as a side of a unit right-angled triangle; a list of the 39 exercises on reciprocal pairs, unit and integer-side right triangles (rectangles), factorization and square numbers for teachers. The article provides new arguments in favor of old disputes (squares of diagonals or widths; mistakes in previous analysis of errors in P322 ). Contradictory ideas about P322 are discussed: Is it the table of triangle sides or factorization terms? Was it compiled by a parallel or independent factorization of the sides or of their squares? Are sides of an initial unit triangle enlarged or reduced by such a factorization? Does it contain two or four arithmetical errors? Time and dimensional requirements for calculation and writing of the complete tablet have been also estimated.
Content may be subject to copyright.
DOI: 10.1515/tmmp-2016-0027
Tatra Mt. Math. Publ. 67 (2016), 1–40
Rudolf Hajossy
ABSTRACT. This article deals with the damaged and incomplete Old Babylo-
nian tablet Plimpton 322 which contains 4 columns and 15 rows of a cuneiform
mathematical text. It has been shown that the presumed original table with its 7
columns and 39 rows represented: a table of square roots of numbers from 0 to 2
for mathematicians; an earliest rudiments of a trigonometric table for builders
and surveyors where angles are not measured as an arc in a unit circle but as
a side of a unit right-angled triangle; a list of the 39 exercises on reciprocal pairs,
unit and integer-side right triangles (rectangles), factorization and square num-
bers for teachers.
The article provides new arguments in favor of old disputes (squares of diag-
onals or widths; mistakes in previous analysis of errors in P322). Contradictory
ideas about P322 are discussed: Is it the table of triangle sides or factorization
terms? Was it compiled by a parallel or independent factorization of the sides or
of their squares? Are sides of an initial unit triangle enlarged or reduced by such
a factorization? Does it contain two or four arithmetical errors?
Time and dimensional requirements for calculation and writing of the complete
tablet have been also estimated.
The cuneiform clay tablet Plimpton 322 (P322 ) [1], [8], [10],[11], Fig. 1, is very
likely the most famous mathematical product of Old Babylonian era (1900–1600
BCE), written probably in an ancient city Larsa (southern Iraq) several decades
before the city was conquered (1762 BCE) by Hammurabi of Babylon. Opinions
on the tablet have been gradually changing: After its illegal excavation, it was
sold in 1923 for $ 10 by a dealer E. J. B a n k s to a collector G. A. P l i m p t o n
and in 1936 it was bequeathed to Columbia University, New York.
2016 Mathematical Institute, Slovak Academy of Sciences.
2010 M a t h e m a t i c s Subject Classification: 01A17.
K e y w o r d s: Old Babylonian cuneiform tablet, Plimpton 322, compilation, errors, users, sex-
agesimal numbers, Pythagorean theorem and triplets.
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Figure 1. A. All sides of the cuneiform clay tablet Plimpton 322.
(The photo by Christine Proust from her online article [11] is published
with her permission and by courtesy to Jane Spiegel, the librarian of the
Rare Book and Manuscript Library, University of Columbia, New York.)
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Figure 1.B. The obverse of tablet Plimpton 322.
(The photo taken by Christine Proust is published with her permission.)
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N e u g e b a u e r and S a c h s (1945), [1] revealed mathematical importance
of P322 and supposed that its text deals with “Pythagorean triangles (triplets)”
obtained by generating pairs of regular numbers. B r u i n s (1949), [2] showed
that the triplets could be calculated also from reciprocal pairs and by reduction
of common divisors. D e S o l l a P r i c e (1964), [3] showed how to obtain neces-
sary generating numbers or reciprocal pairs and assumed an original tablet with
38 rows of triangles. His conjecture was supported by the scored reverse side
of the tablet (Fig. 1.A.), suitable to accommodate remaining rows of triangles.
F r i b e r g (1981), [4] reconstructed a complete extended table P322 combin-
ing the proposals of B r u i n s and d e S o l l a P r i c e . H ø y r u p (1990), [5]
gave a geometrical interpretation of the Bruins algebraic relations between re-
ciprocal pairs and right triangles, in accord with other cuneiform texts. J o y c e
(1995), [6] interpreted P322 as a trigonometric table. R o b s o n (2001), [7] recon-
structed the text of damaged headings in P322. She rejected the trigonometric
interpretation and regarded the tablet, together with F r i b e r g , as a teach-
ers’ aid. C a s s e l m a n (2003), [8] briefly and clearly popularized P322,and
showed how to read its cuneiform text in attached photo. B r i t t o n, P r o u s t
and Shnider (2011), [10] offered a detailed review of a current knowledge on
P322. P r o u s t (2015), [11] published new high quality photos of all sides of
P322 enabling to solve long lasting disputes.
The name Plimpton 322 denotes that it is the 322nd item in the catalogue
of the university’s cuneiform tablets where it is described [10] as a very large,
well preserved, burned tablet (measuring some 13×9 cm; its thickness 2–3 cm)
with left-edge broken away, on obverse 4 columns and 16 lines, reverse blank; con-
tent: commercial account. Each of the four columns contains a heading in a mix-
ture of Sumerian and Akkadian and 15 rows of numbers in a sexagesimal posi-
tional system with base 60 (Fig. 2).
Figure 2. The cuneiform notation of sexagesimal digits: Zero is repre-
sented by a blank space. The numbers 1 up to 9 and also 10, 20, 30, 40, 50
have special characters. All other numbers are composed as a sum of these
characters. For example, 11 = 10 + 1 or 59 = 50 + 9.
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Old Babylonians had no signs for zero or floating point – a boundary between the in-
teger and fractional part of a sexagesimal number. The lack makes any interpretation
of a written number to some extent an arbitrary one and leads easily to errors. For
clarity, we write sexagesimal numerals from 0 to 59 as two-place decimal numerals and
denote a floating point as a semicolon “;”.
Multiplication of sexagesimal numbers is laborious as multiplication tables up
to 1 00;×1 00; are beyond the common memory. Division is even more challenging.
This is the reason why the Old Babylonians instead of dividing by the number
(igi)xmultiplied by its inverse (igibi)1/x. They learned by heart The standard
table of reciprocal pairs (Tab. 1).
Tab l e 1 . The standard table of sexagesimal reciprocal pairs xand y=1/x
fulfilling the condition xy = 1. Transcription of the cuneiform tablet
MLC 1670 [12] in its original form: with the blank space instead of zero
and the numbers in their relative form without a floating point. The abso-
lute value of the numbers depends on the value of the number 1. (Whether
it is interpreted as 1;, 1 00;, 0;01 or any other power of 60.)
2/3 of 1 is 40 x1/x x 1/x x 1/x
its 1/2 is 30 igi igibi igi igibi igi igibi
x1/x9640 24 230 45 120
igi igibi 10 625 224 48 115
320 12 527 21320 50 112
415 15 430 254 1640
512 16 345 32 15230 1 1
610 18 320 36 140 14 56 15
8730 20 340 130 121 44 26 40
The table is readable in both directions: e.g., the pair x=3and1/x = 20 corresponds
to the pair x=30and1/x = 2, and vice versa. It enables to extend the table easily by
double-digit numbers, since, e.g., the pair x= 50, 1/x =1 12 corresponds to the
inverted pair x= 1 12, 1/x = 50.
The ancient mathematicians also knew relations between reciprocal pairs x, y
and sides s, d, h of right triangles (rectangles) used by then builders and surveyors
(Fig. 3 and 4). (Old Babylonians preferred rectangles with 2 sides and diagonal,
but for simplicity, we will talk about triangles with 3 sides.)
As it follows from the Fig. 3 and 4, the diagonal d(¸siliptum in Akkadian),
the width s(sag in Sumerian) and the length h(sin Sumerian) of a rectangle
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Figure 3. Right-angled triangles used by Old Babylonian builders or sur-
On the left : A field, a steep wall of a building or a channel with a triangu-
lar cross-section (S, D, H ). A similar unit (measuring) triangle with sides
s, d, h = 1. Each side had a Sumerian-Akkadian naming: sag =width(S, s),
¸siliptum = diagonal (D, d), s=length(H, h).
On the right : Unit triangles with different slopes sof diagonals d: a steep
slope (s<1), a gentle slope (s>1). The diagonal of unit square has the
slope s=1.(The slope=ˇagal in Sumerian.)
In the bottom: A simple turn of a steep triangle changes it to one with a
gentle slope as in a case of embankments or staircases.
or a right triangle obey the following relations
d2s2=xy =(d+s)(ds).(2)
The “diagonal” rule (1) enables to calculate one side of the right-angled triangle.
(1000 years before the birth of Pythagoras, it would seem ridiculous to call the
rule “Pythagorean”.)
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Figure 4. Geometric relations between the sides s, h, d of a right-angled
(blue) triangle obtained by means of the square (s+h)2with the sides
s+h(top picture) or by the difference of squares d2s2(bottom picture).
The specific relations can be briefly, though anachronistically, expressed
by modern algebraic relations:
Upper pictures: (s+h)2=d2+4(1/2sh)=h2+s2+4(1/2sh) and therefore
Lower pictures: d2s2=xy =(d+s)(ds), while x=d+s,y=ds
or d=(x+y)/2,s =(xy)/2.
According to the relations (1) and (2), in case of a unit right triangle with
the length h= 1, the sides
x=d+s, y =ds(3)
of a rectangle with the unit area
xy =h2=1
represent a reciprocal pair x, y =1/x determined by the diagonal dand
the width sof the unit triangle.
Inversely, as it follows from (3), the sides
d=(x+y)/2,s=(xy)/2,h=1 (5)
of a unit triangle could be determined by a reciprocal pair x, y.
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The set of relations (4), (5) can be written in a form of modern quadratic
equations x22dx +1=0 or x22sx 1=0 (6)
for an unknown number xand known values of the diagonal dor the width s.
Of course, Old Babylonians solved the quadratic equations (6) for an unknown
reciprocal pair x, y in a different way as we would do:
At first, knowing the diagonal d, they calculated its square d2
. Gradually, using
the “diagonal” rule (1), the square of width s2=d212andthenitssquare
root, i.e., the unknown width s=d21, were calculated. Eventually, using
relations (3) for already known pair of sides d, s, they calculated the searched
reciprocal pair
In another type of equation (6), with the known width s, by a similar procedure and
relations (1) and (3), they obtained the reciprocal pair x=s2+1+s,y=s2+1s.
To obtain these solutions, it was necessary to calculate squares d2or s2and
then values of the square roots s=d212or d=12+s2(probably, using
a suitable table). Since in this case the roots represented sides s, d of a unit tri-
angle then such a table of square roots could be obtained by the relations (5)
from a set of reasonable chosen pairs of reciprocals x, y.
It was natural to use at first the simple, well-known reciprocal pairs x, y from
The standard table (Tab. 1) for calculation of sides s=(xy)/2, d=(x+y)/2,
h= 1 of unit triangles and their squares s2or d2=1+s2
. The results of such
calculations for the values xfrom l; to 3; are shown in Tab. 2. This prelim-
inary table does not represent any known cuneiform tablet. In spite of this,
all its data, except of those yellow colored, can be found in the Old Babylonian
tablets MS 3502 (the 8th row of Tab. 2), MS 3971 (the 2nd to 7th row) and
partly in the P322 (the 8th and 9th row) (F r i b e r g (2007), [9]). The green
numbers or letters, shown in Tab. 2, substitute the missing original numbers or
letters in damaged parts of the corresponding clay tablets. Headings in Tab. 2
(in Sumerian-Akkadian mixture) for the parameters x, y, s, d, use the same words
as tablets MS 3502, MS 3971. The rest of headings is in agreement with P322
and is explained in Fig. 5. The squares of the width s2or the diagonal d2=1+s2
in Tab. 2 could have been calculated in several ways:
By the generating reciprocal pair x, y as
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Tab l e 2 . The universal table of unit triangles s, d, h = 1 and similar triangles with integer sides S, D, H ;squares
of diagonals d2and square roots s, d obtained by pairs of reciprocals (two-digit as maximum) x, y from The standard
table (Tab. 1), wherein xis in the range from 1; to 3;. In accordance with the Old Babylonian administrative
tradition, the ordering number Nfor corresponding data is specified at the right end of the table. The last column
shows numbers of corresponding rows in the extended complete tablet P322 (presented as Tab. 4).
igi igibi sag ˇsil´ıptum ´ıbs´atak´ılti ˇsil´ıptim ´ıbs´a ´ıbs´amu
ina 1uˇssaagal sˇsa 1inas´achu´umasagillu´usag ˇsil´ıptim ım P322
xy=1/xsd H d2=1
2+s2S D N
(xy)/2 (x+y)/2 12+S2(1/H)2sH dH N
1; 1; 0; 1;00 11;00 0 1 1 39
1;04 0;5615 0;03 52 30 1;00 07 30 800 1;00 15 00 56 15 31 801 237
1;12 0;50 0;11 1;01 100 1;02 01 11 101 332
1;15 0;48 0;13 30 1;01 30 40 1;03 02 15 941 431
1;20 0;45 0;17 30 1;02 30 24 1;05 06 15 725 529
1;30 0;40 0;25 1;05 12 1;10 25 513 622
1;40 0;360;32 1;08 30 1;17 04 16 34 718
2; 0;30 0;45 1;15 41;33 45 3 5 8 11
2;24 0;25 0;59 30 1;24 30 200 1;59 00 15 159 249 9 1
2;30 0;24 1;03 1;27 20 2;06 09 21 29 10
3; 0;20 1;20 1;40 32;46 40 4 5 11
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Figure 5. A part of the clay tablet Plimpton 322 with a cuneiform text.
Headings above (in Sumerian-Akkadian transcription) and bellow (in Eng-
lish) the four columns of sexagesimal numbers explain their meaning in the
selected rows with the ordering numbers N= 10 and 11. As s, d, h =1are
denoted sides of unit triangles and S, D, H are integer sides of similar
triangles. Also a detail of the tablet with the broken left edge and the dis-
cussed number 1 in the 12th row is included. (Both the images are cut off
from the photos taken by Christine Proust [11].)
By a direct multiplication s2=s×sor d2=d×dof simple two digit
widths s: (0;11, . . . , 1;20) or diagonals d: (1;01, . . . , 1;40). In simple cases also
the Old Babylonian special table of squares (IM 96183, F r i b e r g (2007),
[9, Appendix 7]) or combined multiplication table ( F r i b e r g (2007),
[9, Appendix 2]) could have been used.
By a parallel factorization of multi-digit sides s, d, h of unit triangle:
A multi-digit number can be decomposed into a product of smaller, ideally single-
digit numbers. Then a multiplication by such a number can be performed as
a succession of simple steps. The decomposition of n-digit sexagesimal fractional
number s=0;s1s2...s
n1sncan utilize that the last digit sn=fngn×60n
can be written as a product of the greatest divisor gnand a factor fn. Multiplying
the number s=0;s1s2...s
by the reciprocal of the greatest divisor hn= 1 00; /gncauses that the last
fraction disappears. It changes to fn×60(n1)
. In this way, the resulting number
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s×hnreduces to n1 fractional digits. Thus repeating the multiplication
by the reciprocals hn1,...,h
1of the last greatest divisors gn1,...,g
the final (least) factor
will be obtained, where the multiplier Hsand its reciprocal 1/Hsare determined
by the relations
Hs=hn×···×h1=(1/gn)×···×(1/g1)×(1 00; )n
and 1/Hs=gn×···×g1×(1 00; )n.(9)
As gn×hn= 1 00;, the reciprocals gn,h
nare single-digit sexagesimals. Then
according to (8) and (9), the inequality Ssis satisfied for the absolute value
of the final factor Sand the original number s.Paradoxically, for their relative
values, the opposite expression is valid. Evidently, in accord with (9)
S=sHs0; s1...s
n×(1 00; )n=s1...s
Friberg calls the factorization method: the trailing part algorithm and the final
factor S:the fact or -reduced core. (F r i b e r g (2007), [9]).
By this factorization method, the n-digit number scan be reduced to the
final factor (8): S=s×Hsand inversely, it can be decomposed to the product
The factorization can simplify calculation of squares of many-digit-num-
bers s:
Inversely, by factorization of a square s2to the reduced factor S2=s2×(Hs×Hs),
supposing that its square root Sis known, the square root s=S×(1/Hs)ofthe
number s2can be calculated ( F r i b e r g (2007), [9]).
It should be mentioned that Old Babylonian scribe had calculated with numbers
in their relative form without the floating point therefore he had to keep the factors
(1 00; )nor (1 00; )nfrom relations (9) (more precisely, an idea of the absolute value
of the calculated numbers) in his mind. It could easily lead to errors.
In technical praxis, Old Babylonians had to deal with both small and large ob-
jects. It is illustrated by their length unit system, inherited after Sumers (F r i b e r g
(2007), [9]):
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Relations 6ˇse =1ˇsusi 12 s=1ninda 6´se =1s
Units 1ˇse 1ˇsusi 1s1ninda 1´se 1s1danna
grain finger cubit reed rope length distance
of barley
1/360 m 1/60 m 1/2 m 6m 60 m 360 m 10,800 m
=2.78mm 16.7 mm
Relations 30 ˇsusi =1s10 ninda =1´se 30 s=1danna
This system of different units assisted them in removing ambiguity of the relative
form of inscription of sexagesimal numbers (without a boundary between their integer
and fractional part). Although calculations were done with relative numbers, Old Baby-
lonians were always interested in the absolute value of results: It mattered whether a
purchased field had a width 1 30 ninda (540 m) or 1; 30 ninda (= 18 s=9m).
The factorization formula (11) can be used independently for a calculation
of squares of width s2=s×s=S2×(1/Hs)×(1/Hs) and also diagonal
d2=d×d=D2×(1/Hd)×(1/Hd), where the multipliers Hs,H
dare products
of reciprocals from the greatest divisors of the last digits determined indepen-
dently for the side sand d, respectively. However, to obtain Tab. 2, only the
common greatest divisors of the last digits of the both sides sand dhave been
used. Therefore, the multipliers Hs=Hd=Hhave the same value and the final
factors S=sH,D=dH,H=hH in Tab. 2 can be interpreted as integer sides
of triangles which are similar to the original unit triangles s, d, h =1.
This parallel factorization of sides sand dhas an advantage against an indepen-
dent factorization of the sides: the result is simpler (only one multiplier Hagainst
two Hs,Hd). It can accelerate calculations (performed by one scribe) and save a space
in a table of final results. (A similar space saving effect has a replacement of the trivial
side h= 1 of a unit triangle by the integer side Hof a similar triangle in Tab. 2.)
For simplicity, only the squares d2are shown in Tab. 2 as the squares s2=
d21dierfromd2only by the initial number 1. Moreover, the square d2is al-
ways unambiguous but not s2as it can be seen in the second row of Tab. 2 where
d2=1;0015005615. Then s2=0;0015005615. Relative values of these parameters
1 00 15 00 56 15 and 15 00 56 15, used by ancient scrib es, illustrate that the initial digit
1ind2does show where the missing floating point should be placed but the miss-
ing beginning zeros in s2make its absolute value uncertain. The statement is justi-
fied by new high-quality pictures of P322 made by C h r i s t i n e P r o u s t (Fig. 1
and 5) published also in her online article [11]. Her picture has definitely solved
a long lasting dispute on the “s2or d2?”. In the 12th row with d2= 1;2921...
in Fig. 1.B and 5, the sequence: the numeral 1 – blank space – the numeral 29 can
be seen. In the case of s2=0;2921..., the numeral 29 ought to be seen just at
the vertical line parting the columns. (All numbers in P322 startattheparting
lines!) The free space and the numbers 33 or 35 can be also clearly seen in the rows
11th or 10th in Fig. 5, respectively. (The old black-and-white picture of P322 in the
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N e u g e b a u e r and S a c h s primary article has unreadable left broken edge [1], [8].
So readers were obliged to believe the lucky persons who had opportunity to see the
original tablet P322.)
As it follows from Tab. 2, the triangles for the simplest integers x=2and3
are similar to the well-known triangle with integer sides 3,4,5 used for delin-
eation of perpendiculars in ancient building or surveying. The two triangles
are only mutually turned. In accord with Fig. 3, the number x=2provides
a steep triangle (s=0;45<1) while the number 3 corresponds to a gentle slope
The exact boundary between the steep and mild slope of diagonal is
determined by the diagonal in the unit square, with its unit width and length
(s=h= 1) or by the square of the diagonal d2=1
2= 2. The Tab. 2
shows that the boundary lies between the row N= 9 with the bold number
x=2; 24 and the row N= 10 with the magenta number x=2; 30.The
latter allows to replace the exact condition (0 <s<1) for the width of steep
triangle by a simpler, though less precise condition (1 <x<2; 30) for the
generating number x. (Thus, when selecting x, the last condition allows to
decide without further computation, whether a calculated triangle will be steep
or mild. It certainly saves useless calculations and time to compile the table.)
The least steep diagonal (d2= 1; 59 00 15) in Tab. 2, close to the diagonal
of the square (d2= 2) is in the ninth triangle (N= 9), calculated for the
number x= 2; 24. The steepest diagonal (d2= 1; ) is in the first (N=1)of
the triangles in Tab. 2, with a zero width (s= 0) calculated for x= 1. Such a
“triangle” was certainly a mystery for Old Babylonians. Everything depended
on their interpretation of zero: Whether it was something awfully small (almost
vertical segment in Fig. 3) or something non-existent – “nu” (not in Sumerian)
(N e u g e b a u e r [15, Glossar p. 30]). On the other hand, Tab. 2 is mainly a
table of square roots. From this point of view, the value zero or one ofarootis
not so puzzling.
Interest in steep triangles is natural, because their widths shave the upper and lower
limit (0 <s<1). The limits are also in the corresponding numbers 1 <x<2; 30.
Triangles with a mild slope are limited only from their lower side by the intervals 1 <s
and 2; 30 <x, so an upper limit of the number xis not known in advance. It will
lead to many useless calculations. It is sufficient to determine only steep triangles since
mild ones can be obtained, in accordance with Fig. 3, by a simple turning of the steep
The sides s, d, h = 1 of calculated unit triangles in Tab. 2 were of interest
to builders. Old Babylonian builders did not know an angle as an arc of unit circle
but they could measure it as a side of unit triangle. They measured the slope
of walls or embankments as a deviation sfrom the vertical or horizontal line
in the unit length h=1s(Fig. 6).
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Figure 6. Expected measurements of a slope of a steep wall or a mild
embankment by a suitable unit (white) triangle with the sides:
width (sag)s, diagonal dand length (s)h= 1 cubit (s).
A suitably orientated right angle is created by the yellow ruler with the scale
1 cubit (s) = 30 fingers (ˇsusi) and set by the (blue) lead line in direction
13 or 42.Aslope(ˇagal ) determined by the width s(12)ismeasuredby
the red line in direction 42.
Aslope(ˇagal in Sumerian) represents a ratio between the width sand
the length hof a right-angled triangle (F r i b e r g: [4, p. 311]). To determine
the inclination of the wall (diagonal d) in Fig. 6, the ancient builder would
have used the formulation:
ina uˇs1k´ssag 16 ˇsusi ˇagal =
in the length 1 cubit the width 16 fingers (is) the slope
It is in agreement with the tablet YBC 4673, R o b s o n [7, p. 183] and
N e u g e b a u e r [15, Glossar p. 32, 12]. Since the width scan serve as an in-
clination rate, the formulation is also included in the headings of the column s
in Tab. 2.
The width sof a unit triangle changes in Tab. 2 continuously with the slope
of diagonal dtherefore it is suitable for measurements in building and surveying.
Just opposite, the similar triangles with integer sides S, D, H change errat-
ically. So they cannot be applied for measurements (with the exception of the
well-known triplet 3, 4, 5). But they can help, using the relation (11), to square
the sides of a similar unit triangle.
Because the length 1 s(cubit) is small (around 1/2 m), the measuring
unit triangles (e.g., 0;45, 1;15, 1;) in sdo not guarantee a sufficient accu-
racy of the inclination measurements in surveying. A higher precision
can be achieved by a similar sixty times (1 00; ×) larger triangles with sides
(45; 1 15; 1 00;) swhere 1 00; s30 m.
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How the tablet Plimpton 322 was compiled?
Extension of The standard table of reciprocals
The tablet P322 comprises 15 steep triangles but only 2 of them (1th and
11th) are also in Tab. 2. It means that the author of P322 wanted to generate a
more detailed table of the triangles (rectangles) than the Tab. 2 which had been
calculated using only the reciprocal pairs from The standard table of reciprocals
(Tab. 1). To obtain further steep triangles, the author of P322 had to extend
The standard table by new reciprocal pairs x, y with numbers xfrom the
interval 1x<2; 30, as it has followed from Tab. 2. Such numbers x=p/q > 1
are produced by division of a greater number pby a smaller q.(Intodaysterms,
numbers xrepresent improper fractions with p>q.) Furthermore, dividing
one numerator pby different denominators q, several fractions x=p/q and
consequently several triangles can be determined.
To minimize a computational time, it was necessary to choose the simplest,
single-digit denominator 1 q<1 00;. (Of course, the numerator pmight be
also two-digital.) If the simple numbers p, q were from The standard table
then a calculation of the improper fraction x=p/q and its reciprocal y=q/p
was within a scribe routine.
The standard table (Tab. 1) shows that the denominator qhas 25 values
(1, 2, . . . , 54) and the numerator pcould have 40 figures (1, 2, . . . , 40 00;).
From these figures, 25×40 = 1000 fractions can be assembled, including
(401)+···+(4025) = (39+15)×25/2=675 improper fractions x=p/q > 1.
If a scribe calculated the fractions xfor a given denominator qand gradually
increased numerators p>quntil he found an invalid fraction x2; 30, then
all his results, except the last one, were correct. However, many of the correct
results (x<2; 30) were calculated in vain, because they were equal to the re-
sults already obtained before, for the other (smaller) values of denominators q.
Namely, a gradual increase of denominator q, 2q, 3q, 4q, 5q,... leads to an in-
dependent increase of numerator p, 2p, 3p, 4p, 5p,... but the value of a fraction
x=p/q =2p/2q=3p/3q=5p/5qdoes not change by this proportional increase
of the numerator and denominator. To avoid the useless calculation of fractions x
with the same value, it is necessary to exclude the numerators and denom-
inators with a common divisors 2, 3 or 5. (Finding a common divisor was
for a scribe a known operation, used also at the parallel factorization of sides
of a unit triangle.)
The gradual elimination of unsuitable pairs p, q with a common divisor and
the subsequent calculation of fractions from suitable pairs until the first invalid
fraction is found x2; 30 save a time and material. (The elimination of a fraction
is faster than its calculation. Moreover, excluded or invalid fractions need not
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be stored for further processing.) The numerator pth of the first invalid fraction
x=pth/q 2; 30 represents the threshold numerator for a given denomina-
tor q. (All fractions p/q pth /q are invalid as they give mild triangles.)
Calculation of improper fractions using criteria of the common divisor and slightly
modified invalid fraction x2; 25 was first done by d e S o l l a P r i c e [3]. F r i -
b e r g [4] used the condition x4. The both authors used, instead of numbers p, q
from The standard table, all regular numbers p2 15; and q<1 00;, including the
numbers 1 36, 1 48, 2 05, 2 08, 2 15 which were not in Tab. 1. (The regular sexagesimal
number xhas a finite reciprocal number 1/x. Such number can be written in the form
a3b5cwhere exponents a, b, c are integers. The numbers in Tab. 1 are regular.)
However, apart from the 2 05, the additional numbers are eliminated as unsuitable for
calculation of improper fractions. So they do not influence the final extension of The
standard table of reciprocals Tab. 3.
A disadvantage of the criterion of invalid fraction x2; 30 is in wasting
time with calculation of the invalid fraction and in the useless elimination of un-
suitable pairs p, q before the threshold numerator pth is reached. The useless
operations can be avoided when the criterion x2; 30 is replaced with a con-
dition for the maximum appropriate fraction xmax:
x=p/q < xmax =pmax/q =2;30.
Moreover, the last condition can be substituted for simpler criterion for the
maximum numerator
pmax =2;30q>p>q. (12)
The maximum numerator pmax can be calculated in advance, so the elimina-
tions of unsuitable pairs p, q and calculation of searched fractions xstop be-
low the maximum value pmax. There is no need to continue the elimination till
the threshold numerator pth >p
max will be reached. (A modified criterion of the
maximum numerator p<2; 25qwas used by F r i b e r g [9] and also suggested
by Abdulaziz [13].)
Eventually, the numbers x=p/q, calculated from single-digit denominators q
and numerators pfrom the interval (12), are shown in Tab. 3. The pairs p, q
themselves are from The standard table (Tab. 1). (Only the magenta numerators
p=1; and 2 05; in the first and the last row of Tab. 3 have been determined
additionally in an effort to find fractions xwith extreme numerators p.) Sorting
of all calculated fractions xin decreasing order gives them the order number N.
According to Tab. 3, some of denominators q(6,10,30,36,48) produced no frac-
tion x. Also the greatest denominator q= 54 failed as 25 possible numerators (p>54)
from The standard table (Tab. 1) had to be eliminated. In an attempt to obtain the
widest possible spectrum of triangles, the author of P322 tried to find a numerator
beyond Tab. 1 which could not be eliminated. The numerator pcannot have a common
divisor with the denominator q=54=2.33. Only such numerators are the multiples
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Tab l e 3 . The extended table of reciprocal pairs: (igi)x=p/q 1and(igibi )
y=q/p for a calculation of “all” steep right triangles where the pair p, q is from The
standard table (Tab. 1) and has no common divisor 2, 3 or 5, moreover, 1; q54;
and q<p<pmax=2;30q. Sorting of all 39 calculated fractions xin decreasing
order gives them the order number N.
qpmax N235 p x =p/q y=q/p Nth pth N
2; 30qigi igibi mu b´ım
1; 2;30 01; 1; 1; 39
2; 2; 0;30 13; 11
2; 5; 13; 1;30 0;40 25; 22
3; 7;30 14; 1;20 0;45 29
5; 1;40 0;36 18; 18
4; 10; 25; 1;15 0;48 31
9; 2;15 0;26 40 315; 5
5; 12;30 16; 1;12 0;50 32
8; 1;36 0;37 30 20
9; 1;48 0;33 20 15
12; 2;24 0;25 216; 1
6; 15; 4 – 6 25;
8; 20; 49; 1;07 30 0;53 20 34
15; 1;52 30 0;32 325; 13
9; 22;30 310; 1;06 40 0;54 35
16; 1;46 40 0;33 45 16
20; 2;13 20 0;27 225; 6
10; 25; 5 – 2 27;
12; 30; 625; 2;05 0;28 48 33 (2 05;) 9
15; 37;30 716; 1;04 0;56 15 37
32; 2;08 0;28 07 30 30 (2 08;) 8
16; 40; 625; 1;33 45 0;38 24 21
27; 1;41 15 0;35 33 20 245; 17
18; 45; 725; 1;23 20 0;43 12 29 (2 05;) 27
20; 50; 827; 1;21 0;44 26 40 61 21; 28
24; 1 00; 925; 1;02 30 0;57 36 25 (2 05;) 38
25; 1 02;30 527; 1;04 48 0;55 33 20 36
32; 1;16 48 0;46 52 30 30
36; 1;26 24 0;41 40 24
48; 1;55 12 0;31 15 12
54; 2;09 36 0;27 46 40 41 21; 7
27; 1 07;30 632; 1;11 06 40 0;50 37 30 33
40; 1;28 53 20 0;40 30 23
50; 1;51 06 40 0;32 24 14
1 04; 2;22 13 20 0;25 18 45 31 20; 2
30; 1 15; 10 – – 22 (40 00;)
32; 1 20; 845; 1;24 22 30 0;42 40 26
1 15; 2;20 37 30 0;25 36 21 21; 3
36; 1 30; 11 – – 19 (2 05;)
40; 1 40; 10 1 21; 2;01 30 0;29 37 46 40 18 (4 03;) 10
45; 1 52;30 10 1 04; 1;25 20 0;42 11 15 17 (2 08;) 25
48; 2 00; 10 – – 17 (2 05;)
50; 2 05; 91 21; 1;37 12 0;37 02 13 20 16 (4 03;) 19
54; 2 15; 92 05; 2;18 53 20 0;25 55 12 16 (10 25;) 4
Tot al : 152 281
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of the number 5: p=5,5
2= 25, 53=205;or5
4= 10 25;. Due to the additional con-
dition (12) for the denominator q= 54, only the numerators 54 <p<2 15; can o ccur.
In this range, there is the only regular numerator p=5×25 = 205; with its finite
reciprocal 1/p=(1/5)(1/25) = 0;00 28 48. (Of course, the numerator 2 05 could be
also found by a method of trials and errors.)
It is worthwhile to mention that the pair of the reciprocal (relative) numbers 53=205
and 123=28 48 served in the Old Babylonian tablets CBS 1215, CBS 10201 and
UM 29.13.21 from Nippur (F r i b e r g [9]) as a base for calculation of new reciprocal
pairs beyond the standard table. It is questionable whether the author of P322 knew
these tablets or their copies.
Considering the divisors of numerators and denominators, the following thre-
shold numerators pth :5
3=(205;), 35=(403;) or 27=(208;) have been cal-
The yellow denoted parameters in Tab. 3 help to compare calculation de-
mands by the different criteria: the maximum numerator pmax or the thresh-
old value pth.Inthiscase
N235 is the number of eliminated fractions x, whose denominator qand
numerator p(from the interval q<p<p
max =2;30q) have a common
divisor 2,3or5;
Nth is the number of uselessly checked numerators pfrom the interval
pmax ppth.
As it follows from Tab. 3, the condition (12) for the maximum numerator pmax =
2; 30qreduces the number 675 of improper fractions x=p/q > 1to191(=152+39), from
which a checking of common divisors eliminates 152 and the remained 39 fractions x
can later on produce 39 steep triangles. A calculation of mild triangles should be more
demanding because it would need to check not 191 but 484(= 675 191) from the 675
possible fractions x.
Using the threshold numerator pth as a criterion ( d e S o l l a P r i c e [3], F r i -
b e r g [4]) increases, due to Tab. 3, the number of uselessly checked numerators p
by 281 and in vain calculated fractions x(with the threshold numerators) by 14.
Time and spatial demands of The extended tablet of reciprocals
Table 3 enables to asses a time demand on its formation:
Calculation of 25 maximum numerators pmax =2;30q=2q+q/2, election of 191
available numerators from the standard Tab. 1 and the intervals q<p<p
exclusion of 152 pair p, q with the common divisors 2,3 or 5 lasted (to the au-
thor R. H.) 80 minutes.
Calculation of 39 improper sexagesimal fractions x=p/q lasted (without any calcu-
lator) 100 minutes. (For comparison, the same calculation with a simple calculator
lasted only 8minutes – 13 times less.)
Manual ordering of 39 leaflets with the fraction values xin decreasing order and their
numbering lasted 15 minutes.
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Calculation of 39 reciprocals y=q/p lasted 100 minutes.
An ancient scribe could be faster in multiplication but he had also to prepare record-
ing material.
Preparation of 39 round handy tablets (bringing and shaping of a piece of clay) could
take about 39 ×5min200 min.
Thus the calculation and a temporal record of the data represented for one scribe
around 500 min.
Eventually, for a clear arrangement and higher security of obtained results, it was
necessary to rewrite them to a final tablet. It was sufficient to record 39 rows of
the most important parameters x, y, p, q, N . A corresponding time assessment supposes
knowledge of a recording speed (roughly, one edge per second) and a number of
rewritten edges: According to Fig. 2, the smallest sexagesimal digit 1 has two edges
while the greatest numeral 59 has them 28. In average, it is 15 edges per digit.
Due to Tab. 3, the individual columns of parameters have following number of digits:
x(107),y(95),p(45),q(39),N(39). It is altogether 325 digits, 325 ×15 = 4875 edges.
A transcript of the edges on a clay tablet then lasted around
80 min(= 4875 sec ×1min/60 sec)
.If 20 minutes had been spent before preparing the final clay tablet and recording the
headings above individual columns of parameters, then the total time for obtaining
the final version of the extended table of reciprocals (Tab. 3) was approximately
600 min = 10 hours = nearly 2 days of one scribe work.
As it has been already shown, the threshold numerator pth criterion increases
the number of uselessly checked numerators pby 281 and in vain calculated fractions
xby 14. Utilizing the previous time guesses, it will be found that the increased check-
ing needs additional 281 ×80 min/152 150 min and the vain calculation will spend
14×100 min/39 36 min. The less proper criterion protracts the time for obtaining
The extended table of reciprocals (Tab. 3) for approximately 186 min 3hours
(in total, to 13 h 2–3 days). (In this case, the 25 maximum numerators pmax are not
calculated but such saved up 4 minutes are negligible.)
Without any preliminary checking of common divisors, the shorter checking
of N235 = 152 improper fractions will be substituted by a longer calculation of the
fractions. It means that 80 minutes of the checking will be replaced by 152 ×100/39 =
390 minutes of the calculations plus 100 minutes of additional computation of 39 invalid
fractions x2; 30. It represents in total a protraction 410 min 7 h in comparison to
10 hours in the case of the criterion of maximum numerators pmax =2;30q.Thus,
a complete construction of The extended table of reciprocals (Tab. 3) without
preliminary checking will take approximately 17 hours 34daysperascribe.
Of course, the construction could be accelerated by division of its calculations among
a higher number of scribes. (A calculation by rude force [18], of all 675 improper
fractions x>1, lasting 675×100/39 = 1731 min 29 h 6 days per a scribe seems
improbable as it would increase gratuitously a possible 2-day-work to 8 days.)
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Spatial dimensions of the final version: According to Fig. 1.B and 5, the 21
places for digits, a character and gaps cover the width 127 mm of P322,then6mmis
an average width of one place. 15 rows of a mathematical text and 3 rows of headings
share the height of 88 mm, then, 5 mm is their average height. A necessary dimension
for a cuneiform text of the extended table (Tab. 3) is determined by the dimension
of a digit:width6mm×height 5 mm; the maximum number of digits of individual
parameters in Tab. 3: x(4),y(4),p(2),q(1),N(2); the number of gaps (equal to one
digit) between columns (4 at the 5 columns). In total, it is 17 digits along the width
17×6mm=102 mm. 39 rows of numbers and 3 rows of headings would represent the
height 42 ×5 mm = 210 mm of one sided text or 105 mm of the text on the obverse
and reverse side of the tablet.
It is evident from the order numbers Nin Tab. 3 that The extended table
with the ordered reciprocals had to be constructed en bloc as a complete unit
of all 39 pairs x, y. Individual pairs in the ordered final table cannot be foreseen
in advance (e.g., the first calculated pair: 1; 0,1; 0 is the last 39th one in the
ordered table while the last calculated pair: 2; 18 53 20,0; 25 55 12 in Tab. 3 is
already the 4th one).
Calculations of a complete tablet P322
Using the known reciprocal pairs x, y from the ordered extended table Tab. 3
and applying the same method as in the case of the preliminary table of triangles
Tab. 2, it is possible to make a more detailed table Tab. 4. Actual calculations
can be performed in the following three steps:
1. By means of the reciprocal pairs x, y from Tab. 3, in agreement with the
relations (5), the sides of a unit triangle s=(xy)/2, d=(x+y)/2,
h= 1 are calculated.
2. From the sides s, d, h of the unit triangle by a parallel factorization, the in-
teger sides
H=Hh, S =Hs, D =Hd (13)
of a similar triangle are determined.
3. Eventually, from squares of the integer sides S2and the reciprocals 1/H,
the squares of sides of the unit triangle
are calculated.
During calculation, obtained parameters s, d, h;S, D, H ;S2
have been temporarily registered on at least 8 individual tablets. (Initial pa-
rameters p, q, x, y, N were written before on a special tablet, similar to Tab. 3.)
The constant unit side h= 1 was useless to register. From the squares s2
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it was sufficient to record the more unambiguous d2
. After getting all parame-
ters for a given number x(a row of the complete P322 ), for a transparency and
security reasons, the results were transcribed to a separate, the most likely ver-
tical tablet. Eventually, after a calculation of all 2 ×39 triangles it was supposed
to rewrite the important results s, d, H, d2
,S,D,N on a final (two-sided) tablet
similar to Tab. 4. (The tablet P322 illustrates that already 15 rows, still incom-
plete half of all rows, have been rewritten on the obverse side of the horizontal
The most important part of Tab. 4 is the triple of parameters s, d, H, probably
located on the broken off part of the original complete tablet P322. From the triple,
all remaining parameters could be determined: x, y (by means of (4)); S, D (by means
of (13)); s2
,d2(by means of (14)). Only initial pairs of numbers p, q cannot be deter-
mined. They had to be laboriously compiled in Tab. 3. If there was a desire to made
a universal table with all data interesting for practice but also with those strenuously
obtained then the initializing pair p, q could be placed on the left and right side of the
tablet (similarly, as it is in Tab. 4).
Of course, a selection of parameters for the lost part of P322 is speculative but it
should be done on a reasonable conjecture of the aim of the tablet. The initial words
assign a similar problem in MS 3502 :
igi 2 uˇssasil´ıptum ennam
Number 2. What are the length, width and diagonal?
and the proper text presenting solution of the problem show that its significant pa-
rameters are x, y, s, d, h =1, s2
. It is natural that the missing yellow parameters
occur in the supposed lost part of P322 :x, y, s, d in Friberg [9] and s, d
in Britton, Proust, Shni der [10].Apart ofthe parametersS, D,H, 1/H, S 2
which might be used for the calculation of squares s2
2will still be missing in their
contemplated complete tablet P322. Evidently, the most important of the 5 missing
parameters is the length (multiplier) H. Without it, the sheer integer sides (the factor-
reduced core) S, D are inutile as they are insufficient for a calculation of desired squares
s2=S2(1/H)2or d2=D2(1/H)2
. (The calculation of H=D2S2itself was be-
yond a common scribe knowledge. On the other hand, the calculation of S=sH or
D=dH was within his ability.) For F r i b e r g [9], the key parameter of P322 has
been the square d2from which all remnant parameters could be determined by an
independent factorization of squares d2and s2=d21, including the factors Hd,Hs.
An inclusion of the pair Hd,H
stogether with proposed parameters x, y , s, d would
lead to an extremely large lost part of P322. However, by omitting the pair Hd,H
the factors S, D have become inutile in P322. B r i t t o n et al. [10] have considered
the ordering s, d, H, d2
,S,D unnatural. In contrary, we regard it as a short cut of the
natural set of parameters s, d, h =1,d
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A full agreement of the ordering numbers Nin the calculated tables Tab. 3,
Tab. 4 and the real tablet P322 and a perfect accord of sexagesimal numbers
in the corresponding data in Tab. 4 and P322, demonstrates a correct under-
standing of the technic by means of which the cuneiform tablet had been com-
piled. But it does not mean that the two tables (Tab. 4 and P322 ) have been
obtained by an identical procedure.
For example, the author of P322 could avoid the parallel factorization of
sides s, d during calculation of the integer sides S, D, H.Itwassucientto
realize that the reciprocal pairs
x=p/q, y =q/p
represented fractions. Then a common fractional part of the sides
s=(p/q q/p)/2andd=(p/q +q/p)/2
was the factor
1/H =(1/q)(1/p)(1/2).
To get rid of the common fractional part 1/H in the sides s, d, h, it is enough
to multiply the sides by the reciprocal value of the factor 1/H
H=2pq (15)
(The same effect is also provided by a factorization of the sides.) Usage of the
multiplier (15) can strongly simplify the calculation of integer sides. They
could be calculated independently. There would be no need to follow common
divisors of the last fractional digits. It could accelerate such calculation and
diminish the number of errors that could arise at a factorization.
Short comments on the colors used in the Table 4:
The yellow data are added to the existing extant of the damaged and uncom-
plete tablet P322.
The green data are unreadable in P322 extant.
The magenta data are correct but point out the errors made by ancient scribes
in the real tablet.
The blue values S, D, H of integer sides are obtained by factorization of simple,
two-digit sides s, d, h = 1 of unit triangles. In these cases instead of the blue
factorized values S, D, H shown in Tab. 4, the simple original values s, d, h =1
of the unit triangles are presented in the real tablet P322 and also used for a
direct calculation of the square d2
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Tab l e 4 . Reconstruction of the complete universal table P322 with the absolute sexagesimal data and corrected
errors: s–width,d– diagonal of a unit right triangle (rectangular) with the length h=1;S, D , H are the integer
sides of a similar triangle (rectangle). All the data can be calculated using the initial numbers p, q. The order number
Nis identical with that in preceding tables Tab. 3 and Tab. 2.
Obverse side Right side
s=(xy)/2d=(x+y)/2Hd2=1+s2S=Hs D=Hd N p q
(p/q q/p)/2 (p/q +q/p)/2(2)pq d2=1
0;59 30 1;24 30 200 1;59 00 15 159 249 112 5
0;58 27 17 30 1;23 46 02 30 57 36 1;56 56 58 14 50 06 15 56 07 12025 2104 27
0;57 30 45 1;23 06 45 12000 1;55 07 41 15 33 45 11641 15049 3115 32
0;56 29 04 1;22 24 16 34500 1;53 10 29 32 52 16 33149 50901 4205 54
0;54 10 1;20 50 112 1;48 54 01 40 105 137 59 4
0;53 10 1;20 10 600 1;47 06 41 40 519 801 620 9
0;50 54 40 1;18 41 20 45 00 1;43 11 56 28 26 40 38 11 59 01 754 25
0;49 56 15 1;18 03 45 16 00 1;41 33 45 14 03 45 13 19 20 49 832 15
0;48 06 1;16 54 10 00 1;38 33 36 36 801 12 49 925 12
0;45 56 06 40 1;15 33 53 20 14800 1;35 10 02 28 27 24 26 40 12241 21601 10 121 40
0;45 1;15 41;33 45 3 5 11 2 1
0;41 58 30 1;13 13 30 40 00 1;29 21 54 02 15 27 59 48 49 12 48 25
0;40 15 1;12 15 400 1;27 00 03 45 241 449 13 15 8
0;39 21 20 1;11 45 20 45 00 1;25 48 51 35 06 40 29 31 53 49 14 50 27
0;37 20 1;10 40 45 1;23 13 46 40 28 53 15 9 5
0;36 27 30 1;10 12 30 448 1;22 09 12 36 15 255 537 16 16 9
0;32 50 50 1;08 24 10 14 24 1;17 58 56 24 01 40 753 16 25 17 27 16
Table 4 continues on the next page.
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The continuation of Table 4 from the previous page.
Reverse side Left side
s=(xy)/2d=(x+y)/2Hd2=1+s2S=Hs D=Hd N p q
(p/q q/p)/2 (p/q +q/p)/2(2)pq d2=1
2+(S/H )2p2q2p2+q2
0;32 1;08 15 1;17 04 817 18 5 3
0;30 04 53 20 1;07 07 06 40 21500 1;15 04 53 43 54 04 26 40 10741 23101 19 121 50
0;29 15 1;06 45 120 1;14 15 33 45 39 129 20 8 5
0;27 40 30 1;06 04 30 13 20 1;12 45 54 20 15 609 14 41 21 25 16
0;25 1;05 12 1;10 25 513 22 3 2
0;24 11 40 1;04 41 40 36 00 1;09 45 22 16 06 40 14 31 38 49 23 40 27
0;22 22 1;04 02 30 00 1;08 20 16 04 11 11 32 01 24 36 25
0;21 34 22 30 1;03 45 37 30 13600 1;07 45 23 26 38 26 15 34 31 14201 25 104 45
0;20 51 15 1;03 31 15 48 00 1;07 14 53 46 33 45 16 41 50 49 26 45 32
0;20 04 1;03 16 15 00 1;06 42 40 16 501 15 49 27 25 18
0;18 16 40 1;02 43 20 18 00 1;05 34 04 37 46 40 529 18 49 28 27 20
0;17 30 1;02 30 24 1;05 06 15 725 29 4 3
0;14 57 45 1;01 50 15 26 40 1;03 43 52 35 03 45 639 27 29 30 32 25
0;13 30 1;01 30 40 1;03 02 15 941 31 5 4
0;11 1;01 100 1;02 01 11 101 32 6 5
0;10 14 35 1;00 52 05 28 48 1;01 44 55 12 40 25 455 29 13 33 32 27
0;07 05 1;00 25 224 1;00 50 10 25 17 225 34 9 8
0;06 20 1;00 20 300 1;00 40 06 40 19 301 35 10 9
0;04 37 20 1;00 10 40 11 15 1;00 21 21 53 46 40 52 11 17 36 27 25
0;03 52 30 1;00 07 30 800 1;00 15 00 56 15 31 801 37 16 15
0;02 27 1;00 03 20 00 1;00 06 00 09 49 20 01 38 25 24
0;00 1;00; 11;00 0 1 39?? 1 1
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It is even less probable that the author of P322 was known with the relations
between the triangle sides and the initial numbers p, q which are today easy
obtainable from the relations (5) and (13) by simple algebraic operations
s=(xy)/2=(p/q q/p)/2=(p2q2)/(2pq),
d=(x+y)/2=(p/q +q/p)/2=(p2+q2)/(2pq),
H=2pq, S =p2q2
When the initial numbers p, q (underlined italics in Tab. 4) are odd, then the integer
length is calculated as H=pq. The relations (16) are currently used for calculation
of the “Pythagorean triangles (triples). They were also used by N e u g e b a u e r and
S a c h s [1] for proving a hypothesis that the tablet P322 has dealt with 15 Pythagorean
Errors in the previous error analysis
Comparing the numbers from Tab. 4 and from P322, it follows that Old Baby-
lonians have made 6 errors in their tablet. (The original numbers of P322 can
be found in Fig. 1. B or in the black-and-white photo of P322 in [8].) A position
of the errors is denoted in Tab. 4 by the magenta color. Two of the errors are re-
sults of an insufficient attention during a transcription of data from an auxiliary
to the final tablet:
–Inthe9th row in the parameter S, there is, instead of the correct
numeral 8, a similarly looking but incorrect digit 9 (Fig. 2).
–Inthe2nd row in the parameter d2
, there is, instead of the correct pair
of digits 50 06, an incorrect numeral 56. The error should be caused by
atoo small work tablet where the blank space representing zeros in the
numbers 50 06 was too small. So, two numbers 50 06 have been misread
as one number 56.
The remnant 4 errors are of an arithmetic character:
In spite of this, the mistakes are extremely useful since they prove that the data in
P322 have been obtained in the order and way described above by the three steps.
The 1st error:Inthe8th row in the parameter d2
, there is, instead of the
correct pair of digits 45 14, an incorrect numeral 59 (= 45 + 14).
This parameter should be gained by the relation (14) as d2=1
where values of the sides S, H are provided by factorization of the sides s, d.
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The factorization has given for the 8th row in Tab. 4 values S= 13 19; and
H=1600=4×4×100 and also 1/H =0;15×0; 15 ×0; 01 = 0; 00 03 45. Then
S2×(1/H)2= (13 19)2×(0; 15 ×0; 15 ×0; 01)2
= 2 46 15 00 56 15 ×(15 ×0; 016) = 0; 41 33 45 14 03 45.
Eventually, the correct result for the square d2=1
2+0; 4133 45 14 03 45, shown
in Tab. 4, is obtained.
However, if the blank space (zero 00 ) in the last multiplication
24615 00 56 15 ×(15 ×0; 016)
is overlooked, the telescoped number will be multiplied and it will lead to the
faulty result
s2=24615 56 15 ×(15 ×0; 016) = 0; 00 41 33 59 03 45.
The whole calculation of P322 is made with the reciprocal pairs x=p/q and
y=q/p, therefore the product xy = 1; 00 is exactly equal to the unit. According
to (2), the unit had to be respected also in the calculation of the square d2=
1; 00 + s2
. So an incorrect result
d2=1;00+0; 00 41 33 59 03 45 = 1; 00 41 33 59 03 45
should be obtained.
As the ancient scribe made the calculations with relative sexagesimals,he
obtained the result d2=1+413359 03 45 = 1 41 33 59 03 45, where he had done
a second error – omitted the second (yellow) zero. Thus, due to the double error,
he finally obtained the incorrect result written in P322.
Evidently, a calculation with relative numbers and blank spaces instead of zero
had to be a frequent source of errors. However, the error connected with an incor-
rectly determined fractional part of the calculated number is done by nowadays
authors, too:
F r i b e r g [4, p. 297] calculated squares s2and d2directly as S2(1/H )2and
D2(1/H)2. After an omission of zero and calculation with relative numbers, he ob-
tained the wanted results s2=413359 03 45 and d2= 1 41 33 59 03 45 which, as he
stated, obey the condition d2=s2+ 1, so the incorrect results cannot be revealed. But
respecting absolute values and the unit 1; 00, he could reveal the mistake. He would
have to calculate the values s2= 0;004133 59 03 45 and d2= 0;014133 59 03 45 which
do not obey the condition d2=s2+ 1; 00.
Similarly, wanting to get the incorrect result d2= 1 41 33 59 03 45 by the N e u g e -
b a u e r and S a c h s procedure (16), the authors ( B r i t t o n et al., [10, p. 537] show
at first the correct result d2=(p2+q2)2/(4p2q2)=
(7 13 2 0 01) ×(0;0003305615)×(0;0004)=1;413345140345
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for p=32 andq= 15. However, if 7 13 2 0 01 is mistaken as 7 13 21 then, according to
the authors, the wanted incorrect result is obtained
d2= (7 13 21) ×(0;0003305615)×(0;0004) = 1;4133 59 03 45.
But they are wrong. They should have got the value d2= 0;014133 59 03 45 which
should warn them about a mistake, since a diagonal of a unit triangle must be greater
than 1; 00.
The discussed errors of the authors do not reduce the quality of their articles. They
only document that who does numerical calculations he makes the numerical errors
today, just like 4,000 years ago.
The bad value of the square d2in the 8th row of P322, due to omission of zeros,
can be obtained even by two other ways:
1. Bruins [17] started from the value s= 0; 49 56 15 (Tab. 4) of which square he
counted directly as
s2= (0; 50 0; 00 03 45)2
=0;41400; 00 06 15 + 0; 00 00 00 14 03 45
=0;413345 14 03 45.
If a scribe omitted one of the zeros in the third term and telescoped it to the
value 0; 00 00 14 03 45, then he gained the wrong result s2=0;413359 03 45.
This error would be transferred to square d2=1;+s2as well.
2. A n a g n o s t a k i s and G o l d s t e i n [14] applied the relations (7) between
squares s2
,d2and reciprocal pairs x, y. The triangles in the 8th row of Tab. 4
have the ordering number N=8, for which Tab. 3 gives the values x=2;08and
y= 0;280730. Then, according to (7)
d2= (2; 082+0;280730
= (4; 33 04 + 0; 13 11 00 56 15)/4 + 0; 30
=1;413345 14 03 45.
If a scribe overlooked the zero in the second term and telescoped it to the value
0; 13 11 56 15, then he got the wrong result d2=1;413359 03 45.
Theincorrectvalueofsquared2while the sides S, D arecorrectinthe8throw
of P322, suggests that the squares s2
,d2were calculated either later than S, D, H
or independently: from the numbers p, q, reciprocal pairs x, y or as a direct square
of the side s. The reverse steps (from d2to S, D), promoted by F r i b e r g [9] and
R o b s o n [7], should lead to wrong sides S, D, but those are, however, in the
8th row right.
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The 2nd error: In the 13th row in the parameter S,there is, instead
of the correct number 241, the incorrect value 712 01 =241
This error relates to the fact that for calculating a square s2=S2/(1/H)2
of a unit triangle, it is necessary to calculate and store a square S2of a similar
integer triangle. During transcription of results from auxiliary tablets
to the final tablet,thevalueS2was mistakenly written instead of S.
This error is explained similarly by F r i b e r g [4] but R o b s o n [7] and B r i t t o n
et al. [10] suggest another explanation: They suppose that during a factorization instead
of the sides s, d, the pair s2
,d was mistakenly factorized. It should lead to the incorrect
result S2
,D. Their conjecture is wrong. As it follows from (13), the parallel factorization
is a linear process: the sides s, d increase linearly to S, D. After such factorization, the
pair s2
,denlarges to some value S2=Hs2and D=Hd.ThevalueS2=Hs2will differ
from the wanted S2=H2s2
,showninP322. (It is proven by direct calculations, too.)
Also none of the other methods that explained successfully the previous error in the
2nd row (numbers p, q; reciprocal pairs x, y;directsquareofs) can explain this error
in the 13th row.
The 3rd error: In the 2nd row in the parameter D, there is, instead of the
correct number 12025, the incorrect value
312010012025×12×12 = 3 13 00 00.
This error can be explained as a consequence of two mistakes: a different
factorization of the initial sides s=0; 58 27 17 30 and d=1; 23 46 02 30 from
Tab. 4 and a subsequent omission of zero during a calculation of the value
D= 3 13 00 00.
After two factorization steps, the initial (yellow) sides s, d, h of a unit triangle
changed to the values
0;58 27 17 30 ×2×12 = 23; 22 55 ,
1;23 46 02 30 ×2×12 = 33; 30 25 ,
1×2×12 = 24.
The last (green) numbers have a common divisor 5, and then in the third factorization
step it was necessary to multiply by a reciprocal value of the divisor – by the number 12.
From some reason (darkness, a higher priority work, exchange of scribes), the third
round of factorization was interrupted. Before the interruption, the calculator just
managed to multiply the diagonal
1;23 46 02 30 ×2×122= 6 42; 05.
Restoring the calculations (perhaps next day, with new tablets), the scribe con-
tinued to multiply by 12. From the previous records, he found that the sides s, h had
already been multiplied by 12 (forgetting that it had been done in the second round).
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Thus, by mistake, instead of the sides s, h, he again multiplied the diagonal d.Thus,
the third factorization round ended with the final triplet
0;58 27 17 30 ×2×12 = 23; 22 55 ,
1;23 46 02 30 ×2×123=120 25 ,
1×2×12 = 24 .
After another two rounds of a correct factorization of this incorrect triplet, he finally
obtained for integer sides of a triangle the result
S=0;58 27 17 30 ×2×123=5607,
D=1;23 46 02 30 ×2×125=3130000,
with the correct values of Sand Hshown in Tab. 4 and the incorrect value D=
3130000 slightly different from D= 3 12 01 00 which is written in P322.Thelast
erroneous value could occur again due to omission of the blank space (zero) during the
fourth round of factorization:
D=1;23 46 02 30 ×2×124×12 = 16 05 00 ×12 = (16 00 00 + 5 00) ×12
= 3 12 00 00 + 1 00 00 = 3 13 00 00 a “correct” result in Tab. 4
= 3 12 00 00 + 1 00 = 3 12 01 00 an “ incorrect” result in P322.
A probability of this error is increased by the fact that it could be produced also
by interruptions in the second or fourth round of factorization.
This erroneous diagonal Dwas explained by B r u i n s [17] and later
by F r i b e r g [4], R o b s o n [7], B r i t t o n et al. [10] as an excessive factorization
made by a calculator who did not realize that he had already removed all fractional
digits. It is implausible as he knew in advance how many fractional digits should be
removed (only four in this case). Moreover, at the excessive factorization, the number 5
is no more a common divisor. Therefore its reciprocal 12 is no more suitable for such
factorization. On the other hand, the incorrect relative value D= 3 12 01 is explained
by them as a transcript mistake because the sexagesimal numerals 1201 and 13 can be
with a little incaution easily confused.
The wrong diagonal Dsuggests that the square d2of a unit triangle is deter-
mined using the equation (14) only from the square S2of integer triangle and not
from D2because in the latter case the wrong value of Dwould lead to a signifi-
cant error in the calculated square d2in the 2nd row of P322. The wrong diago-
nal Dalso shows that no accuracy checking of the calculated squares d2=s2+1;
envisaged by F r i b e r g [4], has been performed. (It is understandable. It would
have significantly prolonged calculations.)
The wrong value of diagonal D= 3 13 00 00 is 122-times larger than the cor-
rect value 1 20 25. It suggests that the author of P322 did not know the rela-
tion (15) for the multiple H=2pq between the sides of unit and integer triangles.
The relation could help him to get rid of errors connected with a factorization.
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Actually, due to Tab. 4, the numbers p, q in the 2nd row have values 1 04 and 27,
respectively. Then H=2pq =2×1 04; ×27 = 2 ×82×33=2×12 ×122= 57 36.
Evidently, using the multiple (15), the wrong result Dshould preferably consist
of the elementary multipliers 1 04,27,9,8,3 but not of the numbers 12 or 122
typical for factorization in this case.
The 4th error: In the 15th row in the parameter S,there is, instead of the
correct number 28, the incorrect value 56(= 28 ×2).
There is a general meaning (B r u i n s [17], F r i b e r g [4], R o b s o n [7],
B r i t t o n et al. [10]) that the error is a consequence of a different factoriza-
tion of the individual sides s=0;3720andd= 1; 10 40 from Tab. 4. We suppose
that it should be provoked by a time distress when the data of the last (15th) row
were calculated in the last moment, just during transcription of results to the
tablet P322. Such distress should require dividing the factorization between two
scribes. The scribes factorized the sides sand din the following two steps:
The first scribe The second scribe The comment
0; 37 20 ×3 = 1; 52 1; 10 40 ×3 = 3; 32 20 is a divisor and 3its reciprocal.
52 ×30 = 56 D=3;32 ×15 = 53 2or 4are divisors,
Hs=3×30 = 90 Hd=3×15 = 45 30 and 15 t he ir reciprocals .
Ds=dHs=146 Sd=sHd=28
Evidently, the first scribe overlooked the divisor 4 for the last fractional digit 52 and
obtained twice larger values S=56 and Hs=90 instead of 28 and 45 shown in Tab. 4.
Nevertheless, he could obtain the correct value of the square
d2=1;+S2(1/Hs)2=1;+(56×0; 20×0; 02)2=1;23134640
shown both in Tab. 4 and P322. The same correct value might be calculated also by
the second scribe: d2=D2(1/Hd)2=(53×0; 20 ×0; 04)2= 1; 23 13 46 40. Of course, the
values S=56 and D=53 in P322 do not correspond to the same integer triangle.
Paradox of the tablet P322
Just this opinion, that P322 is not the table of integer triangles (Py-
thagorean triplets) S, D, H but only factors S, D provided by independent
factorizations of squares s2
,d2of sides of a unit triangle (rectangle), is hold
by F r i b e r g [9].
Paradoxically, F r i b e r g in his earlier article ( [4, pp. 290–292]) still talked about
Pythagorean triangles (triplets). He compiled the extended tablet P322 in agreement
with the relations (5), (13), (14) and by application of the parallel factorization of
sides s, d. He still considered errors in line 2 and 15 as a result of an incorrect parallel
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The key position of the squares s2
,d2in later Friberg’s interpretation means
that P322 should be completed just in an opposite way as in the case when it
represented a table of triangles:
2−→ S2=s2×H2
−→ S=S2
d−→ s=S/Hs,d=D/Hd.
The way of obtaining the squares s2
,d2is not discussed by him but there is a possi-
bility to calculate them, due to (7), from reciprocal numbers x, y as (x2+y2)/4±1/2.
Individual members S, D, Hs,1/Hs,H
d,1/Hdof the factorization are interesting
in calculation of square roots of multidigital numbers (s2
,d2) which occur
when quadratic problems are solved. Multidigital squares are decomposed into less-
digital multiples with easier computed roots. For example, in the line 10 in Tab. 4, the
square d2is a 9-digit number; side dhas 5 digits; factorization terms D, H
Friberg’s reduced factorization cores have 3 digits. (This Old Babylo-
nian problem of multiplication of huge numbers by their decomposition to smaller
multipliers has remained until now. Only today’s computers with limited registers use
fast, more sophisticated algorithms of decomposition, for example, K a r a t s u b a mul-
tiplication [19].)
In agreement with the procedure, F r i b e r g [9] proposes the following content
of P322 extended by its lost (yellow) part:
x, y, s, d, d2
Factorized multiples Hs,Hdare missing in the supposed table, though they would
be more interesting for a table owner than the reciprocal pair x, y, which could be
easily calculated as a sum or difference of the sides d,susing the relation (3). Without
the multiples Hs,Hd, the meaning of the reduced cores S, D is doubtful because only
a simultaneous knowledge of the multiples and values S, D enables to calculate the
sides s, d or their squares s2
F r i b e r g [9, p. 436] promotes factorization by squares. Thus the mistake in line 13,
the number S2= 7 12 01 in column 3 is a result of such factorization. Then, after
F r i b e r g : “the author of Plimpton 322 may have forgotten to compute the square
side of 7 1201. The correct entry in line 13 of column 2 would be 2 41(= 71201).
There is only a question whether such type of factorization was performed indeed.
The discussion of the 1st error (d2in 8th row) has shown that it was not.
According to F r i b e r g [9], in the case of the independent factorization, the discussed
data from the 2nd and 15th row are already no more errors. There is only a difference
in the number of factorization steps: The square s2in the 2nd row is factorized four
times with the overall multiple H2
, while the square of diagonal d2
is factorized 6 times with the overall multiple H2
According to F r i b e r g [9, p. 436], this should be similarly in the case of line 15”, too.
An independent factorization of squares s2and d2should give the entries S=56and
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D= 53 in the line 15. We will show, however, that such factorization cannot give
such numbers:
Evidently, in this case due to Tab. 4, the square d2=1;23134640. Then, square
s2=d21=0;231346 40 . The equal trail number 6 40 = 400= 202has the recipro-
cal 32
. Then, from the initial squares after the first round of factorization the following
numbers are obtained
1;23134640×32= 12; 29 04and0;23134640×32=3;2904.
Again the common new trail number 4=2
2has the reciprocal 302
. Then, after the
second round of factorization, the numbers
12; 29 04×302= 3 07 16 =D2
3; 29 04×302=5216=S2
are obtained. Their square roots give the factorization terms
d=Hs=3×30 = 130,
in accord with the above discussed results of the first scribe for the line 15. How-
ever, the “independent” factorizations of the squares s2
,d2fail to explain why
only the value D= 1 46 can be computed but not the correct value 53, envisaged by
F r i b e r g [9] and found by the second scribe.
Of course, it is possible to obtain the value D= 53, but it will need squares of the divi-
sor 4 and his reciprocal 15, used by the second scribe in the second round of factorization
of the current numbers 1;23134640×32= 12; 2904 and0;23134640×32=3;2904 .
It will need a new trail number 9 04 = 544 = 16 ×34 which contains the searched
divisor 16 = 42with its reciprocal 152
. Then, after the second round of factorization,
the numbers 12; 2904 ×152=4649=D2and 3; 2904 ×152=1304=S2are ob-
tained. Their square roots give the results D=53,S=28,Hd=Hs=3×15 = 45,
of the second scribe for the line 15. The reproduction of the second scribe results is
very artificial and depends on the previous knowledge, in contrast to a very natural
procedure of the first scribe. So the independent factorizations of squares s2
cannot explain the error in the line 15 because it will provide only results of the first
The observed equality of the multiples Hd=Hsis no surprise. It could be ex-
pected even at the independent factorization of squares s2
2because the factorized
fractional parts of these squares are exactly equal. Inequality of multiples Hd,Hs
occurs in P322 only in two cases (line 2 and 15). Therefore, it is plausible that the in-
equality of Hd,Hsis a result of an accidental miscalculation and not of an intentionally
independent factorization.
There is also an additional argument against the independent factorization of the
squares s2
2:Fromthewrongd2in the lines 2 and 8 of P322, also incorrect values of
parameters S, D should be expected. However, three of them are correct and also the
incorrect value of the fourth parameter (Din the line 2) has already been explained by a
forced interruption of calculations and not as an intentional independent factorization.
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The width S= 56 and diagonal D= 53 could appear in the line 15 only when
the sides s, d themselves are independently factorized but not the squares s2
the factorization of the sides s, d cannot be considered intentionally independent. If it
actually was, the author of P322 ought to continue in factorization of the side S=105
in the 5th row or of the sides S= 45, D= 1 15 in the line 11 of the P322.
Besides the qualitative difference between the triangle and factorization core interpre-
tation of P322,thereisalsoa paradoxical quantitative difference: From one point of
view, the factorization enlarges fractional sides s, d, h = 1 of a unit triangle H-times
to the integer sides S=sH,D=dH,Hof a similar triangle. After F r i b e r g [4],
however, the same original sides s=S×1/H,d=D×1/H are reduced 1/H-
-times to the factorization cores S, D. For example, in the line 15 in Tab. 4, the sides
S= 28, D=53are3×15 = 45-times larger than the initial sides s=0;3720and
d= 1; 10 40. On the other hand, according to F r i b e r g [4], [9], the factorization cores
S= 28, D=53are20×4=80-times smaller than the original side s=3720and
d= 1 10 40 expressed in relative numbers. Which of the contradictory statements is
A brief mathematical answer to the question is given by the inequality (10). However,
a deeper insight in the paradox is provided by the relation (2) between sides of a unit
triangle d2s2=xy. In the first case, a calculation with absolute values gives the result
20; 37 202=1;231346400; 23 13 46 40 = 1;00.(17)
In the second case, a calculation with the relative values gives
237 202= 1 23 13 46 40 23 13 46 40 = 1(00 00 00 00).(18)
Since the tablet P322 has been obtained from reciprocal pairs x=p/q and y=q/p,
then their product xy can only have one possible absolute value xy = 1;00.Thus,
the first interpretation with integer triangles is correct. It does not mean that the
second version with the relative values is principally faulty. It only solves another task:
Instead of the initial side d= 1; 10 40, it starts implicitly with the 10000-times larger
integer side 1 10 40. Subsequently, this too large side is reduced 80-times to the final
integer value D= 53. Taking into account the implicit increase and the subsequent
explicit reduction, the correct final 1 00 00/80 = 45-times increase, forecasted by
the absolute values, is obtained.
In case of the independent factorization of the initial sides s= 0; 37 20, d= 1; 10 40, the
final values S= 56, D=53inP322 are not the sides of a triangle and therefore they
do not obey to relation (2). In this case, the version with the increase of initial sides by
factorization is supported by unambiguous values of reciprocal pairs: Actually, due
to Tab. 3 in the row N= 15, there are generating numbers p=9,q=5andthe
reciprocal pair x=p/q = 1; 48, y=q/p= 0; 33 20. From the pair and the relation (5), the
width s=(xy)/2 = 0; 37 20 <1; of a steep unit triangle, is determined. The width s
can be written in an ambiguous relative form as 37 20, but the form cannot change its
unambiguous absolute value which must be respected at any factorization.
It must be strictly distinguished between the formation of the tablet P322 and its later
application. The case: F r i b e r g (1981), [4] versus F r i b e r g (2007), [9] covers the
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whole span between the formation and the application of P322. Analysis of the errors
supports the simpler, more natural F r i b e r g [4] version, suitable for construction
of P322 as a table of ordered roots and squares s, d, d2
,(s2)usingTheextended table
of reciprocals. As it will be shown later, this simpler construction of P322 would
last around a month. The latter version [9] would be much more time consuming
as it supposed calculation of roots S, D of some apparently random squares S2
which was beyond the common scribe skill. It would be a great luxury to perform
calculations lasting for several month (without permanent registration of the crucial
parameters S2
,D2) just for a list of 39 numerical exercises. For this purpose, The
standard table of reciprocals would be absolutely sufficient.
The majority of arguments shows that the author of P322 used parallel
factorization of sides s, d with the same multiples Hs=Hd=Hwhich
together with the factorization terms S, D represent sides of an integer triangle
It is undeniable that original sides s, d, h = 1 are the sides of a unit right-angled
triangle. (It is indicated by their names sag, ˇsil´ıptum, uˇsin P322 or MS 3052.) Then
proportional multiples of these sides S, D, H represent sides of a similar triangle (re-
gardless of considering them triangles or not). Although P322 gives only 2 sides S, D,
the author of the tablet had known also the third side H. (By factorization, all three
sides were determined simultaneously. In addition, the side Hshould be displayed
in the lost part of the tablet.)
Spatial and temporal requirements
for a complete tablet P322
Dimensions of a clay tablet containing the parameters of triangles s, d, H,
,S,D,N from Tab. 4 can be determined in a similar manner as the di-
mension of the tablet with parameters of Tab. 3 – from the average width 6 mm
of a sexagesimal digit, the height 5 mm of a row and the height 15 mm of head-
According to Tab. 4, the individual parameters are expressed by the following
number of digits: s(4),d(5),H(2),d
2(9),S(3),D(3),N(3). Adding yet 8 blank
one-digit-spaces between columns and at the edges of tablet, the width of the
complete tablet should be less than 37 ×6mm = 222 mm. (The width of the
missing and remnant part should take up 15 ×6 mm = 90 mm and 22 ×6mm=
132 mm, respectively.)
A heading equivalent to 3 rows and 39 rows of triangles written on the obverse
and reverse side would need a tablet less than 42 ×5/2 mm = 105 mm high.
The left and right side of the tablet might be hypothetically utilized for
a record of generating integers p, q with Akkadian headings igibu (= numerator)
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and igu (= denominator) [15], respectively. (Numbering of rows Non the ob-
verse, reverse and flank sides would be shared.) The two-place numerator, one-
place denominator and 2 blank spaces suppose less than 5×6mm=30 mm of
thickness, similar to that of P322.
A complete tablet, with a horizontal writing of text, could have a shape
of a clay brick less than 222 mm ×105 mm ×30 mm. Initial dimensions should
be larger, as the wet clay volume shrinks 7–14 % by drying [16].
Similar d imensions (223 ×90 ×32 in mm) has also a completely filled vertical tablet
MS 3052, as it follows from its photo ( F r i b e r g [9]). Its reverse side is oval, just
like P322, but not quite symmetrical. The oval shape is no problem for a vertical
tablet with a dominating verbal text. For long horizontal rows with a numerical text,
however, a planar shape of all sides is preferable. (Also on the unfinished P322,the
mathematical text is on its strait not oval side.) A finished table with a full text should
therefore have a shape of a thin rectangular parallelepiped.
Togetatleastamistyinsightintothetime demand of calculations,
Tab. 4 was obtained from relations between individual parameters and the ini-
tial integers p, q, using decimal addition, subtraction, multiplication and division
by a simple Microsoft calculator. The same way was also used at the final trans-
formation of decimals to sexagesimals. The results of the calculations were later
checked by Excel computations which repeatedly used the worksheet functions
INT(number)&“;”&INT(60*MOD(number;1)) for the transformation.
Computing of 15 rows of Tab. 4 took 3 hours. The decimal operations need
only a “click”. All results are initially registered in the calculator memory and
display, and finally simply transcribed to Word tables. On the contrary, clumsy
sexagesimal operations, inscriptions and transcriptions into sets of clay tablets
had to last many times longer.
An experience with 39 improper fractions x=p(1/q) of Tab. 3 has shown that their
calculation by a simple calculator lasted 8 min while by a simple manual multiplication
of two sexagesimal numbers, without any technical aid, lasted 100 min (about 13 times
longer). Then a manual computing of 15 rows of Tab. 4 could last as minimum
13 ×3 hours = 39 hours. Preparation of, maybe, 6 handy tablets for parameters
(s, d;S;D;H;S2;s2) of the same row needed at least additional 6 ×5 min = 30 min
plus 10 min for one final complete tablet for the given row N. Then, completion of
the whole Tab. 4 with its 39 rows could last as minimum
127 h (= 39 ×39 h/15 + 39 rows ×(40/60) h/row),
i.e., more than 3hoursperrow.
So a compilation of Plimpton 322, often being interrupted, lasted not hours
but weeks. A rude guess at velocity of the compilation could be 1–2 rows of the
complete table of triangles per day and per one scribe.
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