City Planning: Inca City Planning

Abstract

Let us consider the equations (i) 3x = 6 and (ii) 5y = 2 in the two variables x, y. We recognize at once the (integer) solution x = 2 for equation (i) and further note that there can be no y in the set N: = {1, 2, 3,…} of natural numbers which satisfies equation (ii), since the constant term 2 on the right-hand side of equation (ii) is not an integral multiple of (i.e., is not divisible by) the coefficient 5 of the variable y therein. Leaving these two simple examples of linear equations in a single variable, let us take up the following linear equation involving two variables x and y: px À qy ¼ m ð1Þ where the coefficients p, q and the constant term m belong to the ring Z of the usual integers.…-3,-2,-1, 0, 1, 2, 3, …. Clearly, for (1) to have a solution x = a, y = b with a, b in Z, it is necessary that any common divisor t of p and q (i.e. any t in Z dividing both p and q) must be a divisor of m as well. Thus the greatest common divisor (g.c.d. in brief) d of p and q denoted as (p, q) in symbols must also divide m. Now, a well-known algorithm of Euclid's for finding (in finitely many steps) the g.c.d. of any two given integers p, q enables us to write their g.c.d. d in the form d = pa + qb with suitable a, b in Z, and we are led to the solution for (1) in integers x = am/d, y =-bm/d. We should not fail to mention that an alternative approach to solving (1) is to expand the rational number q/p (for p ≠ 0) as a simple continued fraction. For example, if p = 3, q = 11 in (1), then q=p ¼ 11=3 has the continued fraction expansion ð3+1Þ=ð1+1=2Þ whose "penultimate conver-gent" ð3+1Þ=ð1 ¼ 4=1Þ solves this special case of (1) with x = 4, y = 1. Solving linear equations such as (1) in 2 variables or in more variables often opens the door to cracking interesting puzzles or even settling serious problems! A clear description of the method of solving (1) is available in Āryabhat. īya, a Sanskrit text of the (fifth or) sixth century AD. (According to Ian Pearce (2002), the mathematical part of Āryabhat. īya covers arithmetic, algebra, plane trigonometry and spherical trigonometry and also contains continued fractions.) As observed by André Weil, this is also "the first ever explicit description" of the general solution in integers for (1) from anywhere, except China. In subsequent Sanskrit treatises, this method came to be known as the kuttaka (pulveriser) method, but it might not be fair to attribute the same with authority to Greek mathematicians granting the familiarity of Indian astronomers with developments in Greek mathematics until then. It is interesting to learn from Weil how, in utter disregard or (possible) ignorance of the remarkable application of the kuttaka in India and of the connection with the seventh book of Euclid's Elementa, Bachet ventured to insert (in the second edition of his book Problèmes plaisants et délectables) a strong claim to the kuttaka method as his own. Before attempting to say anything about the cakravāla method, let us see how the kuttaka method is applied. Equation (1) can be rewritten as an (equivalent) "congruence" relation modulo the integer q, namely as px ≡ m(mod q); we might recall here the customary notation, for a, b, c in Z, that a ≡ b(mod c) exactly when c divides a-b. Whenever d, the g.c.d. of p and q divides m, we can find, by Euclid's algorithm, an integer x such that px-m (equals qy and hence) is divisible by q, i.e., px ≡ m (mod q). The kuttaka precisely enables one to find the form of the general solution of this congruence (whenever (p, q) divides m). Take, by way of an example, the congruence 3x ≡ 1 (mod 7). Since the g.c.d. (3, 7) equals 1, the congruence is indeed solvable. Now 3 × (−2) = −6 ≡ 1(mod 7) and so, on multiplying both sides of the congruence by-2, we get −6x ≡ −2 (mod 7) which is the same as x ≡ 5 (mod 7). The kuttaka thus tells us that any integer of the form 7n + 5 for arbitrary n in Z satisfies the congruence 3x ≡ 1(mod 7). From linear equations in x and y, let us move on to (much more non-trivial) equations of degree 2 in two variables, e.g. x 2 − Ny 2 = 1 for N in N. If N = M 2 with M ∈ Z, the solution in integers x, y of this equation can be immediately found by factorising the left hand side and rewriting the equation as (x + My)(x − My) = 1 and hence as (x + My) = (x-My) = ± 1. However, if N = 2 or any square-free integer (i.e. not divisible by the square of any prime number), then one could ask for all