History of Hindu Mathematics: A Source Book v2.c17.s15: 17. Cyclic Method

History of Hindu Mathematics: A Source Book v2.c17.s15

Language: English

17. Cyclic Method

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**Cyclic Method.** It has been just shown that the most fundamental step in Brahmagupta's method for the general solution in positive integers of the equation $Nx² + 1 = y²,$ where N is a non-square integer, is to form an auxiliary equation of the kind $Na² + k = b²,$ where a, b are positive integers and k = ± 1, ± 2 or ± 4. For, from that auxiliary equation, by the Principle of Composition, applied repeatedly whenever necessary, one can derive, as shown above, one positive integral solution of the original Square-nature. And thence, again by means of the same principle, an infinite number of other solutions in integers can be obtained. How to form an auxiliary equation of this type was a problem which could not be solved completely and satisfactorily by Brahmagupta. In fact, he could not do it otherwise than by trial. But Bhâskara II succeeded in evolving a very simple and elegant method by means of which one can derive an auxiliary equation having the required interpolator ± 1, ± 2 or ± 4, simultaneously with its two integral roots, from another auxiliary equation empirically formed with any simple integral value of the interpolator, positive or negative. This method is called by the technical name *Cakravâla* or the "Cyclic Method."¹

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¹ The Sanskrit word *Cakravâla* means "circle," especially "horizon." The method is so called, observes Sûryadâsa, because it proceeds as in a circle, the same set of operations being applied again and again in a continuous round.

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The purpose of the Cyclic Method has been defined by Bhâskara II thus: "By this method, there will appear two integral roots corresponding to an equation with ± 1, ± 2 or ± 4 as interpolator."²

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**Bhâskara's Lemma.** The Cyclic Method of Bhâskara II is based upon the following Lemma: $If Na²+k=b²,$ where a, b, k are integers, k being positive or negative, then $N((am+b)/k)²+(m²-N)/k=((bm+Na)/k)²,$ where m is an arbitrary whole number.

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The *rationale* of this Lemma is simple: We have $Na²+k=b²,$ $and N.1²+(m²-N)=m², identically.$ Then by Brahmagupta's Lemma, we get $N(am+b)²+k(m²-N)=(bm+Na)².$ $∴ N((am+b)/k)²+(m²-N)/k=((bm+Na)/k)².$

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² *BBi*, p. 38.

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**Bhâskara's Rule.** Bhâskara II (1150) says:

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"Considering the lesser root, greater root and interpolator (of a Square-nature) as the dividend, addend and divisor (respectively of a pulveriser), the (indeterminate) multiplier of it should be so taken as will make the residue of the *prakṛti* diminished by the square of that multiplier or the latter minus the *prakṛti* (as the case may be) the least. That residue divided by the (original) interpolator is the interpolator (of a new Square-nature); it should be reversed in sign in case of subtraction from the *prakṛti*. The quotient corresponding to that value of the multiplier is the (new) lesser root; thence the greater root. The same process should be followed repeatedly putting aside (each time) the previous roots and the interpolator. This process is called *Cakravâla* (or the 'Cyclic Method').¹ By this method, there will appear two integral roots corresponding to an equation with ± 1, ± 2 or ± 4 as interpolator. In order to derive integral roots corresponding to an equation with the additive unity from those of the equation with the interpolator ± 2 or ± 4 the Principle of Composition (should be applied)."²

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¹ The original text is *cakravâlamidaṁ jaguḥ*. The commentator Kṛṣṇa explains, "âcâryâ etadgaṇitaṁ cakravâlamiti jaguḥ" or "The learned professors call this method of calculation the *Cakravâla*." So Bhâskara II appears to have taken the Cyclic Method from earlier writers. But it is not found in any work anterior to him so far known.

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Suppose we have an equation of the form $Na² + k = b²,1$ where a, b, k are simple integers, relatively prime, k being positive or negative. Then by Bhâskara's Lemma $N((am+b)/k)² + (m²-N)/k = ((bm+Na)/k)²,2$ where m is an arbitrary integral number. In the above rule, m has been styled the indeterminate multiplier. Now, by means of the pulveriser, its value is determined so that $(am+b)/k is a whole number.$

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² *BBi*, pp. 36ff.

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Again, of the various such values, Bhâskara II chooses that one which will make |m² - N| as small as possible. Let that value of m be n. Now let

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a_1 &= (an + b)/k, \\ b_1 &= (bn + Na)/k, \\ k_1 &= (n² - N)/k. \end{aligned} The numbers a₁, b₁, k₁ are all integral. The equation (2) then becomes Na₁² + k₁ = b₁². 3 Proceeding exactly in the same way, we can obtain from (3) a new equation of the same kind, Na₂² + k₂ = b₂², where again a₂, b₂, k₂ are whole numbers. By repeating the process, we shall ultimately arrive at an equation, states Bhâskara II, in which the interpolator k will reach the value ± 1, ± 2 or ± 4, and in which (a, b) will be integers.

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**Nârâyaṇa’s Rule.** The above rule of Bhâskara II has been reproduced by Nârâyaṇa (1350). He writes :

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“Making the lesser root, greater root and interpolator (of a Square-nature) the dividend, addend and divisor (respectively of a pulveriser), the (indeterminate) multiplier of it should be determined in the way described before. The *prakṛti* being subtracted from the square of that or the square of the multiplier being subtracted from the *prakṛti*, the remainder divided by the (original) interpolator is the interpolator (of a new Square-nature); and it will be reversed in sign in case

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of subtraction of the square of the multiplier. The quotient (corresponding to that value of the multiplier) is the lesser root (of the new Square-nature); and that multiplied by the multiplier and diminished by the product of the previous lesser root and (new) interpolator will be its greater root. By doing so repeatedly will be obtained two integral roots corresponding to the interpolator ± 1, ± 2 or ± 4. In order to derive integral roots for the additive unity from those answering to the interpolator ± 2 or ± 4, the Principle of Composition (should be adopted)."¹

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¹ *NBi*, I, R. 79-82.

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It will be noticed that Nârâyaṇa does not expressly state that the value of the indeterminate multiplier *m* should be so chosen as will make |m² - N| least. It is perhaps particularly noteworthy that he recognised the relation b₁ = a₁n - k₁a. For b_1 &= (bn + Na)/k, / &= (n(a_1k - an) + Na)/k, [\because\ a_1k = an + b] / &= a_1n - ((n² - N)/k)a, / &= a_1n - k_1a, / \therefore a &= (a_1n - b_1)/k_1. Similarly, it will be found that b = (b₁n - Na₁)/k₁. For b₁n = a₁n² - k₁an,

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&= a_1 (N + kk_1) - k_1 an, \\ & \ [\because\ \ kk_1 = n² - N] \\ &= a_1 N + k_1 b, \ [\because\ \ a_1 k = an + b] \\ \therefore b &= (b_1 n - N a_1)/k_1. \end{aligned}

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**Illustrative Examples.** In illustration of the Cyclic Method, Bhâskara II works out in detail the following examples:

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"What is that number whose square multiplied by 67 or 61 and then added by unity becomes capable of yielding a square-root? Tell me, O friend, if you have a thorough knowledge of the method of the Square-nature."¹

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That is to say, we are to solve

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(*i*) 67x² + 1 = y²,
(*ii*) 61x² + 1 = y².

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Leaving out the details of the operations in connection with the process of the pulveriser, Bhâskara's solutions are substantially as follows:

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(*i*) 67x² + 1 = y². We take the auxiliary equation $67.1² - 3 = 8².$ Then, by the Lemma, $67((1.m+8)/(-3)) + (m²-67)/(-3) = ((8m+67.¹)/(-3))². (1)$ By the method of the *Kuṭṭaka* the solution of $(m+8)/(-3) = an integer,$ is m = -3t + 1. Putting t = -2, we get m = 7 which makes |m² - 67| least. On substituting this value, the equation (1) reduces to $67.5² + 6 = 41².$ Again, by the Lemma, we have $67((5n + 41)/6)² + (n² - 67)/6 = ((41n + 67.5)/6)². (2)$ The solution of $(5n + 41)/6 = a whole number,$ is n = 6t + 5. |n² - 67| will be least for the value t = 0, that is, when n = 5. The equation (2) then becomes $67.11² - 7 = 90².$ Now, we form $67((11p + 90)/(-7))² + (p² - 67)/(-7) = ((90p + 67.11)/(-7))². (3)$ The solution of $(11p + 90)/(-7) = an integral number,$ is p = -7t + 2. Taking t = -1, we have p = 9; and this value makes |p² - 67| least. Substituting that in (3) we get $67.27² - 2 = 221².$ By the Principle of Composition of Equals, we get from this equation $67(2.27.221)² + 4 = (221² + 67.27²)²,$ $or 67(11934)² + 4 = (97684)².$ Dividing out by 4, we have $67(5967)² + 1 = (48842)².$

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¹ *BBi*, p. 38. It is remarkable that the equation 61x² + 1 = y² was proposed by Fermat to Frénicle in a letter of February, 1657. Euler solved it in 1732.

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Hence x = 5967, y = 48842 is a solution of (i). $(ii) 6Ix² + I = y².$ Here we start with the auxiliary equation $6I.I² + 3 = 8².$ By the Lemma, we have $6I((m+8)/3)² + (m²-6I)/3 = ((8m+6I)/3)². (I)$ Now the solution of $(m+8)/3 = an integer,$ is m = 3t + I. Putting t = 2, we get the value m = 7 which makes | m²-6I | least. On substituting this value in (1), it becomes $6I.5² - 4 = 39².$ Dividing out by 4, we get $6I(5/2)² - I = (39/2)². (2)$ By the Principle of Composition of Equals, we have $6I(2.5/2.39/2)² + I = (39/2)² + 6I(5/2)²²,$ $or 6I(195/2)² + I = (1523/2)². (3)$ Combining (2) and (3), $6I(3805)² - I = (29718)².$ Composing this with itself, we get $6I(226I53980)² + I = (I7663I9049)².$ Hence x = 226I53980, y = I7663I9049 is a solution of (ii).

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The following two examples have been cited by Nârâyaṇa : $(iii) I03x² + I = y²,$ $(iv) 97x² + I = y².$

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Their solutions are given substantially as follows : For (*iii*) we have the auxiliary equation $103.1² - 3 = 10².$ By the Lemma, we get $103((m+10)/(-3))² + (m²-103)/(-3) = ((10m+103)/(-3))².$ The general solution of $(m+10)/(-3) = an integer,$ is m = -3t + 2. Putting t = -3, we get m = 11. Then $103.7² - 6 = 71².$ Again, by the Lemma, $103((7n+71)/(-6))² + (n²-103)/(-6) = ((71n+103.7)/(-6))².$ The solution of $(7n+71)/(-6) = a whole number,$ is n = -6t + 1. Taking t = -1, we get $103.20² + 9 = 203².$ Next, we have $103((20p+203)/9)² + (p²-103)/9 = ((203p+103.20)/9)².$ Now, (20p+203)/9 = an integral number for p = 9t + 2. When t = 1, p = 11. On taking this value we find $103.47² + 2 = 477².$

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Applying the Principle of Composition of Equals, we get $103(2 . 47 . 477)² + 4 = (477² + 103 . 47²)²,$ $or 103(44838)² + 4 = (455056)².$ Hence 103(22419)² + 1 = (227528)², which gives x = 22419, y = 227528 as a solution of (*iii*).

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For the solution of (*iv*) the auxiliary equation is $97 . 1² + 3 = 10².$ Therefore $97((m+10)/3)² + (m² - 97)/3 = ((10m + 97)/3)².$ The solution of $(m+10)/3 = an integer,$ is m = 3t + 2. Taking t = 3, we have m = 11. Then $97 . 7² + 8 = 69.²$ Next, we have $97((7n+69)/8)² + (n² - 97)/8 = ((69n+97 . 7)/8)².$ The solution of $(7n+69)/8 = an integer,$ is n = 8t + 5. Taking t = 1, that is, n = 13, we get $97 . 20² + 9 = 197².$ Whence $97((20p + 197)/9)² + (p² - 97)/9 = ((197p + 97 . 20)/9)².$ The solution of $(20p + 197)/9 = a whole number,$ is p = 9t + 5. Putting t = 1, we get p = 14. With this value of p we have $97·53² + 11 = 522².$ Whence $97 ((53q + 522)/11)² + (q² - 97)/11 = ((522q + 97·53)/11)².$ The solution of $(53q + 522)/11 = an integer,$ is q = 11t + 8. The appropriate value of q is given by t = 0. So, taking q = 8, we have $97·86² - 3 = 847².$ Next, we find $97 ((86r + 847)/(-3)) + (r² - 97)/(-3) = ((847r + 97·86)/(-3))².$ The solution of $(86r + 847)/(-3) = a whole number,$ is r = 3t + 1. Putting t = -3, we get r = 10. Taking this value, we have $97·569² - 1 = 5604².$ By the Principle of Composition of Equals, we find $97 (6377352)² + 1 = (62809633)².$ Hence x = 6377352, y = 62809633 is a solution of (*iv*).

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**Proofs.** It has been stated by Bhâskara II that:

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(1) when a₁ is an integer, k₁ and b₁ are each a whole number;

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(2) his Cyclic Method will in every case lead to the desired result.

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He has not adduced proofs. We presume that he knew a proof at least of the first proposition. For he must have recognised the simple relation $b₁ = a₁n - k₁a,$ which has been expressly stated by Nârâyaṇa (1350). This shows at once that b₁ will be a whole number, if k₁ is so. This is also evident from the equation, Na₁² + k₁ = b₁², itself. Hence, it now remains to prove that k₁ is an integral number.

Linked texts (2)

Bījagaṇita — Bhāskara II, Sanskrit
Bījagaṇitāvataṃsa — Nārāyaṇa Paṇḍita (mathematician), Sanskrit