The Science of the Shulba: A Study in Early Hindu Geometry c19.s1: Rational Right-angled Triangles

The Science of the Śulba: A Study in Early Hindu Geometry c19.s1

Language: English

Rational Right-angled Triangles

c19.s1.1

The formula of Kātyāyana to find the sum of n equal squares of sides a each amounts to $a² (√ n)² + a² ( (n - 1)/2 )² = a² ( (n + 1)/2 )².$

Putting m² for n, in order to make the sides of the right-angled triangle free from the radical, and dividing out by a² we get $m² + ( (m² - 1)/2 )² = ( (m² + 1)/2 )² I$ as the solution of the indeterminate equation of the second degree $x² + y² = z². A$

c19.s1.3

If the sides of the right-angled triangle are to be integral as well as rational, m must be odd. According to Proclus (c. 450 A.D.), this solution was known to Pythagoras (*c.* 540 B.C.).

c19.s1.4

A still more general solution of (A) which includes the solution (I) as a particular case, is furnished by the method of transformation of a rectangle into a square. It is this: $(√mn)² + ((m-n)/2)² = ((m+n)/2)²,$ where m, n are any two arbitrary numbers. Substituting p², q² for m, n respectively, in order to eliminate the irrational quantities, we get $p² q² + ((p²-q²)/2)² = ((p²+q²)/2)². (II)$

c19.s1.5

A further generalisation has been tacitly assumed in some methods employed by Āpastamba for the construction of the *Mahāvedi*. If the sides of a rational right triangle are known, then other rational right triangles can be obtained by multiplying them by any rational integer, or how he puts it, by increasing them by any rational multiples of them. Thus if α, β, γ be a rational solution of x² + y² = z², then other rational solutions of it will be given by la, lβ, lγ where l is any rational number. This is clearly in evidence in the formula of Kātyāyana. It is now known that all rational solutions of (A) can be obtained without duplication in this way.

c19.s1.6

Karavindasvāmī gives the solution,¹ $x, ((m²+2m)/(2m+2))x, ((m²+2m+2)/(2m+2))x.$ It can of course be derived from the solution (I) by multiplying by x and dividing by m² every element of it and then putting m+1 for m.

c19.s1.7

¹ Vide his commentary on *ĀpŚl*, i. 4.

c19.s1.8

*Rational Right Triangles having a Given Leg.*

c19.s1.9

There seems to have been an attempt to find the rational right triangles having a given leg. We find in the *Śulba* at least two such right triangles having a common leg a, *viz.*, $(a, 3a/4, 5a/4) and (a, 5a/12, 13a/12).$

c19.s1.10

The principle underlying these solutions will be easily detected to be that of the reduction of the sides of any rational right triangle in the ratio of the given leg to the corresponding side of it. Thus the sides of a rational right triangle having a given leg a will be (a, aβ/α, aγ/α), where α, β, γ are the sides of any rational right triangle. So that starting with the solution (II), we shall find that all rational right triangles having the leg a will be given by $a, ((p²-q²)/2pq)a, ((p²+q²)/2pq)a.$ But this general solution is not stated anywhere in the *Śulba*. The commentators of Āpastamba, however, give the solution $a, ((m²+2m)/(2m+2))a, ((m²+2m+2)/(2m+2))a.$

c19.s1.11

It should be noted that the above principle for obtaining the rational right triangles having a given leg has been followed expressly in later times in India by Mahāvīra (850) and in Europe by Leonardo Fibonacci of Pisa (1202) and Vieta (*c.* 1580).

Linked texts (1)

Āpastambaśulbasūtra — Āpastamba, Sanskrit