History of Hindu Mathematics: A Source Book v2.c17.s16: 18. Solution of Nx² ± c = y

History of Hindu Mathematics: A Source Book v2.c17.s16

Language: English

18. Solution of Nx² ± c = y

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The general solution of the indeterminate quadratic equation $Nx² ± c = y²$ in positive integers was first given by Brahmagupta (628). He says :

"From two roots (of a Square-nature) with any given additive or subtractive, by making (combination) with the roots for the additive unity, other first and second roots (of the equation having) the given additive or subtractive (can be found)."¹

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¹ *BrSpSi*, xviii, 66.

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Eliminating n between $a₁ k = an + b, b₁ k = bn + Na,$ we get $a₁ b - ab₁ = 1.$

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Hence $b₁ = (a₁ b - 1)/a = a whole number.$

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Now $n² - N = ((a₁ k - b)² - Na²)/(a²)$ $= (a₁² k² - 2 b k a₁ + k)/(a²)$ $= (k(a₁² k - 2 b a₁ + 1))/(a²).$

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Therefore k/(a²) (a₁² k - 2ba₁ + 1) is a whole number.

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Since a, k have no common factor, it follows that $(a₁² k - 2 b a₁ + 1)/(a²) = (n² - N)/k = k₁ = an integer.$

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Also $k₁ = (n² - N)/k = (a₁² k - 2ba₁ + 1)/(a²)$ $= (a₁²(b² - Na²) - 2ba₁ + 1)/(a²)$ $= ((a₁ b - 1)/a)² - Na₁².$

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Thus having known a single solution in positive integers of the equation Nx² ± c = y², says Brahmagupta, an infinite number of other integral solutions can be obtained by making use of the integral solutions of Nx² + 1 = y². If (p, q) be a solution of the former equation found empirically and if (α, β) be an integral solution of the latter then, by the Principle of Composition, $x = pβ ± qa, y = qβ ± Npa$ will be a solution of the former. Repeating the operations we can easily deduce as many solutions as we like.

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This method reappears in later Hindu algebras. Bhâskara II says :

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"In (a Square-nature) with the additive or subtractive greater (than unity), one should find two roots by his own intelligence only ; then by their composition with the roots obtained for the additive unity an infinite number of roots (will be found)."¹

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¹ *BBi*, p. 42.

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Nârâyaṇa writes similarly :

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"When the additive or subtractive is greater than unity, two roots should be determined by one's own intelligence. Then, by combining them with the roots for the additive unity, an infinite number of roots can be obtained."²

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² *NBi*, I, R. 86.

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We take the following illustrative examples with solutions from Nârâyaṇa :

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*Example.* "Tell me that square which being multiplied by 13 and then increased or diminished by 17 or 8 becomes capable of yielding a square root."³

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³ *NBi*, I, Ex. 44.

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That is, solve
(1) 13x² ± 17 = y²,
(2) 13x² ± 8 = y².

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*Solution.* "In the first example it is stated that the multiplier = 13 and interpolator = 17.

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"Now the roots for the interpolator 3 are (1, 4). And for the interpolator 51, the roots are (1, 8). For the composition of these with the previous roots (1, 4) the statement will be $m = 13 & l = 1 & g = 8 & i = 51 / & l = 1 & g = 4 & i = 3$

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So, by the Addition Lemma, we get the roots corresponding to the interpolator 153 as (12, 45). The rule says, 'If the interpolator (of a Square-nature) be divided by the square of an optional number etc.' Now take the optional number to be 3, so that the interpolator may be reduced to 17. For 3² = 9 and 153/9 = 17. Therefore, dividing the roots just obtained by the optional number 3, we get the required roots (4, 15).

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"Applying the Subtraction Lemma and proceeding similarly we get the roots for the interpolator 17 as (4/3, 19/3).

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"In the second example the statement is : multiplier = 13, interpolator = -17. Proceeding as before we get (by the Addition Lemma) the roots (147, 530); and (by the Subtraction Lemma), the roots (3, 10)."¹

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¹ Our MS. does not contain the solution of the equations 13x² ± 8 = y².

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**Form Mn²x² ± c = y².** Brahmagupta says :

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"If the multiplier is that divided by a square, the first root is that divided by its root."²

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² *BrSpSi*, xviii. 70.

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That is to say, suppose the equation to be $Mn²x² ± c = y²,1$ so that the multiplier (*i.e.*, coefficient of x²) is divisible by n². Putting nx = u, we get $Mu² ± c = y².2$ Then clearly the first root of (1) is equal to the first root of (2) divided by n. The corresponding second root will be the same for both the equations.

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The same rule is taught by Bhâskara II¹ and Nârâyaṇa. The latter says :

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¹ *BBi*, p. 42.

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"Divide the multiplier (of a Square-nature) by an arbitrary square number so that there is left no remainder. Take the quotient as the multiplier (of another Square-nature). The lesser root (of the reduced equation) divided by the square-root of the divisor will be the lesser root (of the original equation)."²

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**Form** a²x² ± c = y². For the solution of a Square-nature of this particular form, Brahmagupta gives the following rule :

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² *NBi*, I, R. 84.

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"If the multiplier be a square, the interpolator divided by an optional number and then increased and decreased by it, is halved. The former (of these results) is the second root ; and the other divided by the square-root of the multiplier is the first root."³

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³ *BrSpSi*, xviii. 69.

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Thus, it is stated that $x = 1/2a ((± c)/m - m),$ $y = ½ ((± c)/m + m),$ where m is an arbitrary number, is a solution of the equation $a²x² ± c = y².$

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The same solution has been given by Bhâskara II and Nârâyaṇa.¹ Bhâskara's rule runs as follows :

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¹ *NBi*, I, R. 85.

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"The interpolator divided by an optional number is set down at two places; the quotient is diminished (at one place) and increased (at the other) by that optional number and then halved. The former is again divided by the square-root of the multiplier. (The quotients) are respectively the lesser and greater roots."²

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The *rationale* of the above solution has been given by the commentators Sûryadâsa and Kṛṣṇa substantially as follows : ± c & = y² - a²x² / & = (y - ax)(y + ax). Assume y - ax = m, m being an arbitrary rational number. Then y + ax = (± c)/m.

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Whence by the rule of concurrence, we get $x = 1/2a ((± c)/m - m),$ $y = ½ ((± c)/m + m).$

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**Form c - Nx² = y².** Though the equation of the form c - Nx² = y² has not been considered by any Hindu algebraist as deserving of special treatment, it occurs incidentally in examples. For instance, Bhâskara II has proposed the following problem :

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² *BBi*, p. 42.

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“What is that square which being multiplied by -5 becomes, together with 21, a square? Tell me, if you know, the method (of solving the Square-nature) when the multiplier is negative.”[¹]

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Thus it is required to solve $-5x² + 21 = y². 1$ Nârâyaṇa has a similar example, *viz.*,² $-11x² + 60 = y². 2$

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Two obvious solutions of (1) are (1, 4) and (2, 1). Composing them with the roots of $-5x² + 1 = y²,$ says Bhâskara II, an infinite number of roots of (1) can be derived.

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**Form Nx² - k² = y².** Bhâskara II observes:

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“When unity is the subtractive the solution of the problem is impossible unless the multiplier is the sum of two squares.”[³]

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Nârâyaṇa writes:

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“In the case of unity as the subtractive, the multiplier must be the sum of two squares. Otherwise, the solution is impossible.”[⁴]

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Thus it has been said that a rational solution of $Nx² - 1 = y²,$ and consequently of $Nx² - k² = y²$ is not possible unless N is the sum of two squares.

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¹ *BBi*, p. 43.

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² *NBi*, I, Ex. 43.

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³ “Rûpaśuddhau khiloddiṣṭaṁ vargayogo guṇo na cet”—*BBi*, p. 40.

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⁴ *NBi*, I, R. 83.

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For, if x = p/q, y = r/s be a possible solution of the equation, we have $N(p/q)² - k² = (r/s)²,$ or $N = (qr/ps)² + (qk/p)².$

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Bhâskara II then goes on :

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"In case (the solution is) not impossible when unity is the subtractive, divide unity by the roots of the two squares and set down (the quotients) at two places. They are two lesser roots. Then find the corresponding greater roots at the two places. Or, when unity is the subtractive, the roots should be found as before."

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Thus, according to Bhâskara II, two rational solutions of $Nx² - 1 = y²,$ where N = m² + n², will be $.x &= 1/m / y &= n/m, .x &= 1/n / y &= m/n.$

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So two rational solutions of $(m² + n²) x² - k² = y²,$ will be $.x &= k/m / y &= kn/m, .x &= k/n / y &= km/n.$

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The following illustrative example of Bhâskara II¹ is also reproduced by Nârâyaṇa :² $13x² - 1 = y².$

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¹ *BBi*, p. 41.

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² *NBi*, I, Ex. 38.

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The former solves it substantially in the following ways :

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(1) Since 13 = 2² + 3² two rational solutions are (1/2, 3/2) and (1/3, 2/3).

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(2) An obvious solution of $13x² - 4 = y²$ is x = 1, y = 3. Then dividing out by 4, as shown before, we get a solution of the equation 13x² - 1 = y² as (1/2, 3/2).

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(3) Again, since an obvious solution of $13x² - 9 = y²$ is x = 1, y = 2, we get, on dividing out by 9, a solution of our equation as (1/3, 2/3).

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(4) From these fractional roots, we may derive *integral roots* by the Cyclic Method. Since $13(½)² - 1 = (3/2)²,$ we have, by Bhâskara's Lemma, m being an indeterminate multiplier, $13((m/2 + 3/2)/(-1))² + (m² - 13)/(-1) = ((3m/2 + 13/2)/(-1))²,$ or $13((m + 3)/(-2))² + (m² - 13)/(-1) = ((3m + 13)/(-2))².$ The suitable value of m which will make (m+3)/2 an integer and |m² - 13| minimum is 3. So that we have $13.3² + 4 = 11².$ From this again we get the relation $13((3n + 11)/4)² + (n² - 13)/4 = ((11n + 13.3)/4)².$

Linked texts (2)

Brāhmasphuṭasiddhānta — Brahmagupta, Sanskrit
Bījagaṇita — Bhāskara II, Sanskrit