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Mathematics in Ancient and Medieval India c5.s13: Factors of Positive Integer

Mathematics in Ancient and Medieval India c5.s13

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Factors of Positive Integer

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The ordinary method of factorisation by successive division by 2, 3, 5 etc, was well-known in ancient India. The rule for the first time, occurs in the *Siddhāntaśekhara* of Śrīpati (1039 A. D.) and also appear in later works.² In addition to this method he gives also rule for factorising a non-square number, as follows :

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² Shukla, K. S. “Hindu Methods for finding factors or divisors of a number”, *Gaṇita*, 17, No. 2, pp. 109–117. 1966.

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*Rule I.* “When the dividend is even, it is divided by 2 again and again, and divided by 5 when 5 occurs in the unit place; this is continued until the dividend is reduced to an odd number; then the (prime number) 3 etc are to be tried as divisors”.³

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³ *dvābhyāṁ dvābhyāṁ bhājyarāśiṁ*,

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*same tadādisthāne pañcake pañcakena* ।
*evaṁ kuryādyāvadojaṁ tu tāvat*,
*tryādyaihārairbhājyarāśiṁ bhajettu* ॥
(*Siddhāntaśekhara*, xiv, 36–37 (Babuaji Miśra’s Edition)

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*Rule II.* “When the dividend is a perfect square, its square root itself is a divisor, if it is not a perfect square its nearest square root is multiplied by 2, increased by 1 and diminished by the residue of the square root, then it is a perfect square. The square root of this perfect square and also the square root of the dividend as increased by this perfect square, added and subtracted, (will give the two factors of the dividend)”.¹

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¹ *vargaścettanmūlamevāsya hāram,* *no ced āsannaṃ padaṃ dvighṇamasmin* । *rūpaṃ yuktvā śeṣahīne kṛtiḥ syāt-* *tanmūlaṃ tadyuktamūlaṃ yutone* ॥ (*Siddhāntaśekhara*, xiv, 36–37).

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Besides Rule I, Nārāyaṇa (1356 A. D.) gives a fuller treatment of the Rule II in his *Gaṇitakaumudī* rediscovered later by French mathematician Fermat² in 1643 A. D. Nārāyaṇa's rule runs as follows. “The nearest square root of the non-square number, is multiplied by 2, increased by 1, and diminished by the residue of the square root. If it is a perfect square, it is added to the dividend to make it a perfect square. If the above number is not a perfect square, it should be increased by the successive terms of the arithmetic series whose first term is twice the nearest square root (of the non-square number) increased by 3 and common difference 2, until it becomes a perfect square. The unequal (successive) divisors thus obtained are multiplied by the preceding ones and by the product of the preceding ones taking one, two, three……all at a time and are placed ahead of those divisors. In case of equal divisors, each is multiplied by the (product of the preceding ones. Each of these is to be multiplied by the others (to get the divisors of the given number".¹

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² Dickson, L. E. *History of the Theory of Numbers*, 1, p. 357.

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¹ *apadapradasya rāśeḥ padamāsunnaṃ dvisaṃ guṇaṃ saikaṃ ।। mūlāvaśeṣahīnaṃ vargaścet kṣepakaśca kṛtisiddhayai । vargo na bhavet pūrvāsannapadaṃ dviguṇitaṃ trisaṃyuktamādyäd dyuttara vṛddhyā tāvad yāvad bhaved vargaḥ । asamānānāṃ purvahatāḥ pare puraḥsthāstathā cānye tulyānāṃ pūrvaghnaḥ paraḥ pṛthak te’nyaharanighnāḥ* । ( *Gaṇitakaumudī*, II, xi, 6½ - 8½, pp. 245-47, Padmākara Dvivedi’s edition ).

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The rule 2 in symbols may be expressed as follows :

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Let N = a² + r
if 2a + 1 - r = b²
than N + b² = a² + r + (2a + 1 - r) = (a + 1)²
so that N = (a + 1)² - b²
= (a + b + 1) (a-b + 1)

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If however, 2a + 1-r = c, where c is not a perfect square, then some terms of the series (2a + 3), (2a + 5), (2a + 7)......are added to c, so as to make the resulting sum a perfect square.

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Let c + [(2a + 3) + (2a + 5) + ...r terms ] = k²
or c + (2a + 3) + (2a + 5) + ... +(2a + 2r + 1) = k²
Then N + k² = (a + r + 1)²
or N = (a + r + 1)²-k²
= (a + r + k + 1) (a + r-k + 1)

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Last part of the rule says if N = p × p × p × q × s × t then the divisors of N are :

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q, s, t, qs, qt, st, qst; p, p², p³; qp, qp², qp³
sp, sp², sp³, tp, tp², tp³, qsp, qsp², qsp³, qtp, qtp²,
qtp³, stp, stp², stp³, qstp, qstp², qstp.³

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The following two example used by Nārāyaṇa will illustrate the above the rule.

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*Example 1.* “O expert in mathematics, tell me if you know the subject of finding divisors, the numbers by which the number 1161 is exactly divisible”

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Now 1161 = 34² + 5 and 2 × 34 + 1 5 = 64 = 8² ∴ 1161 + 8² = 35² or 1161 = 35²-8² = 43 × 27 = 3 × 3 × 3 × 43 ∴ Possible divisors of 1161 are : 3, 9, 27, 43, 129, 387, 1161

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*Example 2* : “If you are fully proficient in finding divisors, quickly tell me the numbers by which 1001 is exactly divisible ”

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Now 1001 = 31² + 40 and 2 × 31 + 1–40 = 23, not a perfect square. ∴ 23 + (2.31 + 3) + (2.31 + 5) + …… = k², say. i.e. 23 + 65 + 67 + …… + 89 = k² = 32². hence by the rule, 1001 + k² = 1001 + 32² = 45² or, 1001 = 45²–32² = 77 × 13 = 7 × 11 × 13 The possible divisors are : 7, 11, 13, 77, 91, 143, 1001

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