- INDIA 177
On that basis Jupiter was given (inaccurately) a sidereal period of 10 years. Again
inaccurately, using a year of 3651 days, we find that Jupiter undergoes three cycles
in 10,960 days. Thus, in one day, Jupiter moves 3/10960 of a revolution. Since
there are 360 · 60 = 21600 minutes in a revolution, we find that each day Jupiter
moves 64800/10960 = 810/137 minutes. The problem, then, is to solve 810÷/137 =
21,600y+1350 or dividing out the common factor of 270, 3x = 137 · (80y + 5); that
is, 3a; = 10,960y + 685. Here χ will be the number of days elapsed in the cycle and
y the number of revolutions that Jupiter will have made. The Euclidean algorithm
yields 10960 = 3 · 3653 + 1, so that we have(^3) ^ (^3) C -2502305
-685 —>.
o ~
685
That is, χ = -2502305 + 10960* and y = -685 + 3i. The smallest positive solution
occurs when t = 229: χ = 7535, y = 2, which is Brahmagupta's answer.
Brahmagupta illustrated the formulas for right triangles by creating Pythago-
rean triples. In Chapter 12 of the Brahmasphutasiddhanta (Colebrooke, 1817, p.
306) he gives the rule that "the sum of the squares of two unlike quantities are
the sides of an isosceles triangle; twice the product of the same two quantities is
the perpendicular; and twice the difference of their square is the base." This rule
amounts to the formula (a^2 + 6^2 )^2 = (2ab)^2 + (a^2 - b^2 )^2 , but it is stated as if the
right triangle has been doubled by gluing another copy to the side of length a^2 — b^2 ,
thereby producing an isosceles triangle. The relation stated is a purely geometric
relation, showing (in our terms) that the sides and altitude of an isosceles triangle
of any shape can be generated by choosing the two lengths a and b suitably.
Brahmagupta also considered generalizations of the problem of Pythagorean
triples to a more general equation called^11 Pell's equation and written y^2 —Dx^2 = 1.
He gives a recipe for generating a new equation of this form and its solutions from
a given solution. The recipe proceeds by starting with two rows of three entries,
which we shall illustrate for the case D = 8, which has the solution χ = 1, y = 3.
We write
1 3 1
1 3 1'
The first column contains x, called the lesser solution, the second contains y, called
the greater solution, and the third column contains the additive term 1. From these
two rows a new row is created whose first entry is the sum of the cross-multiplied
first two columns, that is 1-3+3-1 = 6. The second entry is the product of the second
entries plus 8 times the product of the first entries, that is 3-3 + 8-1-1 = 17, and
the third row is the product of the third entries. Hence we get a new row 6 17 1,
and indeed 8 • 62 + 1 = 289 = 17^2. In our terms, this says that if 8a;^2 + 1 = y^2 and
8u^2 + 1 = v^2 , then 8(xv + yu)^2 + 1 = (8a;ii + yv)^2. More generally, Brahmagupta's
rule says that if ax^2 + d = y^2 and au^2 + c = í^2 , then
(1) a(xv + yu)^2 + cd = (axu + yv)^2.
(^11) Erroneously so-called, according to Dickson (1920, p. 341), who asserts that Fermat had studied
the equation earlier than John Pell (1611-1685). However, the website at St. Andrew's University
gives evidence that Euler's attribution of this equation to Pell was accurate. In any case, everybody
agrees that the solutions of the equation were worked out by Lagrange, not Pell.