Mathematical secrets of ancient tablet unlocked after nearly a century of study

[struggle4progress]

40. I suppose that's possible but it's very easy to see, from pebble pictures, that

if (a,b,c) is a Pythagorean triple then so is (n*a,n*b,n*c)

For example, if (3,4,5) is Pythagorean so is (45,60,75)

The "proof" is very simple. Start with (say) a 3 x 3 square of red pebbles and a 4 x 4 square of blue pebbles. The red and blue pebbles can be rearranged as a 5 x 5 square. Now replace every red pebble with a 15 x 15 square of orange pebbles and every blue pebble with a 15 x 15 square of green pebbles. The 3 x 3 square becomes a 45 x 45 square of orange pebbles; and the 4 x 4 square similarly becomes a 60 x 60 square of green pebbles pebbles. Any process of rearranging the red and blue pebbles into a 5 x 5 square implies a process of rearranging 15 x 15 orange squares and 15 x 15 green squares into a 75 x 75 square

The converse -- "if (n*a,n*b,n*c) is a Pythagorean triple then so is (a,b,c)" -- is likewise obvious from pebble pictures

So it seems unreasonable to imagine that competent Babylonian mathematicians were unaware of the relation between (3,4,5) and (45,60,75)

Moreover this relationship is relevant to the first column of Plimpton 322; since here it does not actually matter which of the two common reconstructions of the first column we use, I will illustrate with the simple assumption that the first column gives S^2/L^2 where S and L are the short and long legs of the triple: using more modern concepts, this is the square of the tangent of the smaller acute angle in the triangle. One might want a table of such ratios for astronomical purposes (say); and a natural way to construct such a table would be to compute the ratio for various known right triangles, the easiest being (3,4,5)

To compute 3^2/4^2 in Babylonian sexagesimal notation, one should like the denominator a power of 60: this is accomplished by first multiplying by 15^2/15^2 to obtain (3x15)^2/(4x15)^2 = (45)^2/(60)^2; then dividing 45^2 = 2025 by 60 to obtain a quotient and remainder 2025 = 33 x 60 + 45; and finally noting (33 x 60 + 45)/60^2 = 33/60 + 45/60^2 which in the Babylonian notation is 33 45 -- exactly as reported by the tablet

Similar methods will work whenever L divides a power of 60: that is, whenever L is a product of 2s, 3s, and 5s -- which is the case for EVERY triple in the tablet. For example, the triple (119, 120, 169) could lead to the calculation

119^2/120^2 = (119^2*15)/(120^2*15) = 212415/60^3 = (59 x 60^2 + 15)/60^3 = 59/60 + 15/60^3 which corresponds to the Babylonian notation 59 0 15 reported

Sometimes rather tedious computation with large numbers is required, but the ability involved is mechanical. A more interesting question might be how the Babylonians actually found triples having L a product of 2s, 3s, and 5s: I am not enough of a number theoretician to be sure, but I suspect a hard theory might lie here so that we should think the Babylonians found such triples by trial and error