History of Hindu Mathematics: A Source Book v2.c17.s20: 22. Single Indeterminate Equations of Higher Degrees

History of Hindu Mathematics: A Source Book v2.c17.s20

Language: English

22. Single Indeterminate Equations of Higher Degrees

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The Hindus do not seem to have paid much attention to the treatment of indeterminate equations of degrees higher than the second. Some interesting examples involving such equations are, however, found in the works of Mahâvîra (850), Bhâskara II (1150) and Nârâyaṇa (1350).

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**Mahâvîra's Rule.** One problem of Mahâvîra is as follows:

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Given the sum (s) of a series in A.P., to find its first term (a), common difference (b) and the number of terms (n).

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In other words, it is required to solve in rational numbers the equation $a + ((n-1)/2)bn = s,$ containing three unknowns a, b and n, and of the third degree. The following rule is given for its solution :

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“Here divide the sum by an optional factor of it ; that divisor is the number of terms. Subtract from the quotient another optional number ; the subtrahend is the first term. The remainder divided by the half of the number of terms as diminished by unity is the common difference.”¹

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¹ *GSS*, vii. 78. There are also other problems where instead of s, the given quantity is s + a, s + b, s + n or s + a + b + n. (*GSS*, ii. 83 ; *cf.* also vi. 80). For such problems also the method of solution is the same as before, *i.e.*, to assume suitable arbitrary values for two of the unknowns so that the indeterminate cubic equation is thereby reduced to a determinate linear equation in one unknown. (*GSS*, ii. 82 ; vi. 317).

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**Bhâskara’s Method.** Bhâskara II proposes the problems :

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“Tell those four numbers which are unequal but have a common denominator, whose sum or the sum of whose cubes is equal to the sum of their squares.”²

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If x, y, z, w be the numbers, then
(1) x + y + z + w = x² + y² + z² + w²,
(2) x³ + y³ + z³ + w³ = x² + y² + z² + w².

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² *BBi*, p. 55.

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Let the numbers be x, 2x, 3x, 4x, says Bhâskara II. That is, suppose y = 2x, z = 3x, w = 4x in the above equations. Then by (1) we get $10x = 30x².$ $∴ x = ⅓.$ Hence x, y, z, w = ⅓, ⅔, 3/3, 4/3, is a solution of (1).

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Again, with the same assumption, the equation (2) reduces to $100x³ = 30x².$ $∴ x = 3/10.$ Hence x, y, z, w = 3/10, 6/10, 9/10, 12/10, is a solution of (2).

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The following problem has been quoted by Bhâskara II from an ancient author :

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"The square of the sum of two numbers added with the cube of their sum is equal to twice the sum of their cubes. Tell, O mathematician, (what are those two numbers)."¹

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¹ *BBi*, p. 101.

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If x, y be the numbers, then by the statement of the question $(x + y)² + (x + y)³ = 2(x³ + y³).$ "Here, so that the operations may not become lengthy," says Bhâskara II, "assume the two numbers to be u + v and u - v." So on putting $x = u + v, y = u - v,$ the equation reduces to $4u³ + 4u² = 12uv²,$ or 4u² + 4u = 12v², or (2u + 1)² = 12v² + 1.

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Solving this equation by the method of the Square-nature we get values of *u*, *v*. Whence the values of (*x*, *y*) are found to be (5, 1), (76, 20), etc.

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**Nârâyaṇa’s Rule.** Nârâyaṇa gives the rule :

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“Divide the sum of the squares, the square of the sum and the product of any two optional numbers by the sum of their cubes and the cube of their sum, and then multiply by the two numbers (severally). (The results) will be the two numbers, the sum of whose cubes and the cube of whose sum will be equal to the sum of their squares, the square of the sum and the product of them.”

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That is to say, the solution of the equations

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are respectively

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| (1.1) x = ((m²+n²)m)/(m³+n³), / y = ((m²+n²)n)/(m³+n³) ; | (4.1) x = ((m²+n²)m)/((m+n)³), / y = ((m²+n²)n)/((m+n)³) ; |
| (2.1) x = ((m+n)²m)/(m³+n³), / y = ((m+n)²n)/(m³+n³) ; | (5.1) x = ((m+n)²m)/((m+n)³), / y = ((m+n)²n)/((m+n)³) ; |
| (3.1) x = (m²n)/(m³+n³), / y = (mn²)/(m³+n³) ; | (6.1) x = (m²n)/((m+n)³), / y = (mn²)/((m+n)³) ; |

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¹ *GK*, i. 58.

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where *m*, *n* are rational numbers.

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It will be noticed that the equation (2) can be reduced, by dividing out by *x* + *y*, to $x² - xy + y² = x + y ;$ and similarly (5) can be reduced to $x + y = 1.$

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With m = 1, n = 2 Nârâyaṇa gives the following sets of particular values : \begin{align*} (1.2) \enspace &x, y = 5/9, 10/9 & (4.2) \enspace &x, y = 5/27, 10/27 \\ (2.2) \enspace &x, y = 1, 2 & (5.2) \enspace &x, y = 1/3 2/3, \\ (3.2) \enspace &x, y = 2/9, 4/9 & (6.2) \enspace &x, y = 2/27, 4/27 \end{align*} He then observes : "In this way one can find by his own intelligence two numbers for the case of *difference*, etc."

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**Form ax2n+2 + bx2n = y².** For the solution of an equation of the form $ax2n+2 + bx2n = y²,$ where *n* is an integer, Bhâskara II gives the following rule :

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"Removing a square factor from the second side, if possible, the two roots should be investigated in this case. Then multiply the greater root by the lesser. Or, if a biquadratic factor has been removed, the greater root should be multiplied by the square of the lesser root. The rest of the operations will then be as before."¹

Linked texts (3)

Gaṇitasārasaṅgraha — Mahāvīra (mathematician), Sanskrit
Bījagaṇita — Bhāskara II, Sanskrit
Gaṇitakaumudī — Nārāyaṇa Paṇḍita (mathematician), Sanskrit