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History of Hindu Mathematics: A Source Book v2.c17.s23: 25. Double Equations of the Second Degree

History of Hindu Mathematics: A Source Book v2.c17.s23

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25. Double Equations of the Second Degree

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**First Type.** The double equations of the second degree considered by the Hindus are of two general types. The first of them is $. ax² + by² + c & = u², / a'x² + b'y² + c' & = v².$

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Of these the more thoroughly treated particular cases are as follows :

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*Case i.* x² + y² + 1 = u², / x² - y² + 1 = v²; .

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*Case ii.* x² + y² - 1 = u², / x² - y² - 1 = v².

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It should be noted that though the earliest treatment of these equations is now found in the algebra of Bhâskara II (1150), they have been admitted by him as being due to previous authors (*âdyodâharaṇam*).

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For the solution of (*i*) Bhâskara II assumes¹ $x² = 5z² - 1, y² = 4z²,$ so that both the equations are satisfied. Now, by the method of the Square-nature, the solutions of the equation 5z² - 1 = x² are (1, 2), (17, 38),...Therefore, the solutions of (*i*) are $.x &= 2 / y &= 2, .x &= 38 / y &= 34,$

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¹ *BBi*, p. 108.

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Similarly, for the solution of (*ii*), he assumes $x² = 5z² + 1, y² = 4z²,$ which satisfy the equations. By the method of the Square-nature the values of (z, x) in the equation 5z² + 1 = x² are (4, 9), (72, 161), etc. Hence the solutions of (*ii*) are $.x &= 9 / y &= 8, .x &= 161 / y &= 144,$

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Bhâskara II further says that for the solution of equations of the forms (*i*) and (*ii*) a more general assumption will be $x² = pz² 1, y² = m²z²;$ where p, m are such that $p ± m² = a square,$ the upper sign being taken for Case *i* and the lower sign for Case *ii*. Both the equations are then identically satisfied. Suppose $p + m² = s²,$ $p - m² = t².$

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Whence $s = ½((2m²)/n + n),$ $t = ½((2m²)/n - n),$ where *n* is any rational number. Therefore $p = ¼((4m⁴)/(n²) + n²).$

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Here he observes that m² should be so chosen that p will be an integer.

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Hence $. x² &= ¼((4m⁴)/(n²) + n²)z² 1, / y² &= m²z²; (1)$ the upper sign being taken for Case *i* and the lower sign for Case *ii*.

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Whence $u = ½((2m²)/n + n)z,$ $v = ½((2m²)/n - n)z.$

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Or, we may proceed in a different way, says Bhâskara II :

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Since $(p² + q²) ± 2pq$ is always a square, we may assume $x² = (p² + q²)w² 1,$ $y² = 2pqw².$

v2.c17.s23.16

For a rational value of y, 2pq must be a square. So we take p = 2m², q = n². Hence we have the assumption .x² &= (4m⁴ + n⁴)w² \mp 1, / y² &= 4m²n²w²;} \eqno{(2)} the upper sign being taken for Case *i* and the lower sign for Case *ii*.

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Whence u &= (2m² + n²)w, / v &= (2m² - n²)w.

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It will be noticed that the equations (1) follow from (2) on putting w = z/2n. So we shall take the latter as our fundamental assumption for the solution of the equations (*i*) and (*ii*). Then, from the solutions of the subsidiary equations $(4m⁴ + n⁴)w² 1 = x²,$ by the method of the Square-nature, observes Bhâskara II, an infinite number of integral solutions of the original equations can be derived.¹

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¹ Cf. *BBi*, p. 110.

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Now, one rational solution of (4m⁴ + n⁴)w² + 1 = x² is w &= 2r/((4m⁴ + n⁴) - r²), / x &= ((4m⁴ + n⁴) + r²)/((4m⁴ + n⁴) - r²). Therefore, we have the general solution of .x² + y² - 1 &= v² / x² - y² - 1 &= u²} as . x &= ((4m⁴ + n⁴) + r²)/((4m⁴ + n⁴) - r²), & u &= (2r(2m² + n²))/((4m⁴ + n⁴) - r²), / y &= 4mnr/((4m⁴ + n⁴) - r²), & v &= (2r(2m² - n²))/((4m⁴ + n⁴) - r²), } (A) where m, n, r are rational numbers.

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For r = s/t, we get Genocchi's solution.¹

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¹ *Nouv. Ann. Math.*, X, 1851, pp. 80-85; also Dickson, *Numbers*, II, pp. 479. For a summary of important Hindu results in algebra see the article of A. N. Singh in the *Archeon*, 1936.

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In particular, put m = 2t, n = 1, r = 8t² - 1 in (A). Then, we get the solution . x &= 1/2((8t² - 1)/2t)² + 1, & u &= (64t⁴ - 1)/(8t²), / y &= (8t² - 1)/2t, & v &= 1/2((8t² - 1)/2t)². } (a)

v2.c17.s23.24

Putting m = t, n = 1, r = 2t² + 2t + 1 in (A), we have² . x &= t + 1/2t, & u &= t + 1/2t, / y &= 1, & v &= t - 1/2t. } (b)

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Again, if we put m = t, n = 1, r = 2t² in (A), we get . x &= 8t⁴ + 1, & u &= 4t²(2t² + 1), / y &= 8t³, & v &= 4t²(2t² - 1). } (c)

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² Here, and also in (c), we have overlooked the negative sign of x, y, u and v.

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These three solutions have been stated by Bhâskara II in his treatise on arithmetic. He says,

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“The square of an optional number is multiplied by 8, decreased by unity, halved and then divided by that optional number. The quotient is one number. Half its square plus unity is the other number. Again, unity divided by twice an optional number added with that optional number is the first number and unity is the second number. The sum and difference of the squares of these two numbers minus unity will be (severally) squares.”¹

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¹ *L*, p. 13.

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“The biquadrate and the cube of an optional number is multiplied by 8, and the former product is again increased by unity. The results will be the two numbers (required).”²

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² *L*, p. 14.

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Nârâyaṇa writes :

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“The cube of any optional number is the first number ; half the square of its square plus unity is the second. The sum and difference of the squares of these two numbers minus unity become squares.”³

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³ *GK*, i. 46.

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That is, if m be an optional number, one solution of *(ii)*, according to Nârâyaṇa, is

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x &= (m⁴)/2 + 1, & u &= (m² + 2)(m²)/2, / y &= m³, & v &= (m² - 2)(m²)/2.

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It will be noticed that this solution follows easily from the solution *(c)* of Bhâskara II, on putting t = m/2. This special solution was found later on by E. Clere (1850).⁴

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⁴ *Nouv. Ann. Math.*, IX, 1850, pp. 116-8; also Dickson, *Numbers*, II, p. 479; Singh, l. c.

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Now, let us take into consideration the equation (4m⁴ + n⁴)w² - 1 = x². Its solutions are known to be .w &= 1/(n²) / x &= (2m²)/(n²)}, .w &= 1/(2m²) / x &= (n²)/(2m²)}.

v2.c17.s23.40

From these, by the Principle of Composition, we get respectively two other solutions .w &= (16m⁴ + n⁴)/(n⁶) / x &= (32m⁶ + 6m²n⁴)/(n⁶)}, .w &= (m⁴ + n⁴)/(2m⁶) / x &= (n⁶ + 3n²m⁴)/(2m⁶)}.

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Therefore, the general solutions of .x² + y² + 1 &= u², / x - y² + 1 &= v²;} are .x &= (2m²)/(n²), & u &= (2m² + n²)/(n²), / y &= 2m/n, & v &= (2m² - n²)/(n²);} (a') .x &= 1/(n⁶)(32m⁶ + 6m²n⁴), / y &= 2m/(n⁵)(16m⁴ + n⁴), / u &= 1/(n⁶)(16m⁴ + n⁴)(2m² + n²), / v &= 1/(n⁶)(16m⁴ + n⁴)(2m² - n²);} (a'')

v2.c17.s23.42

$.x &= (n²)/(2m²), & u &= (2m² + n²)/(2m²), / y &= n/m, & v &= (2m² - n²)/(2m²); (b')$

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$.x &= 1/(2m⁶)(n⁶ + 3n²m⁴), / y &= n/(m⁵)(m⁴ + n⁴), / u &= 1/(2m⁶)(m⁴ + n⁴)(2m² + n²), / v &= 1/(2m⁶)(m⁴ + n⁴)(2m² - n²). (b″)$

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Putting n = 1 in (a') and (a″), we have the integral solutions $.x &= 2m², & u &= 2m² + 1, / y &= 2m, & v &= 2m² - 1; (a'.1)[¹]$ and $.x &= 2m⁴(16m² + 3), / y &= 2m(16m⁴ + 1), / u &= (16m⁴ + 1)(2m² + 1), / v &= (16m⁴ + 1)(2m² - 1). (a″.1)$

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Similarly, if we put m = 1 in (b') and (b″), we get $.x &= ½n², & u &= ½(n² + 2), / y &= n, & v &= ½(n² - 2); (b'.¹)$ and $.x &= ½n²(n⁴ + 3), & u &= ½(n⁴ + 1)(n² + 2), / y &= n(n⁴ + 1), & v &= ½(n⁴ + 1)(n² - 2). (b″.1)$

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¹ This solution was given by Drummond (*Amer. Math. Mon.*, IX, 1902, p. 232).

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The solution (b'.¹) is stated by Nârâyaṇa thus:

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"Any optional number is the first and half its square is the second. The sum and difference of the squares of these two numbers with unity become capable of yielding a square-root."¹

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¹ *GK*, i. 45.

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*Case iii.* Form $.ax² + by² = u², / a'x² + b'y² + c' = v².$

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For the solution of double equations of this form Bhâskara II adopts the following method:

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The solution of the first equation is x = my, u = ny; where $am² + b = n².$

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Substituting in the second equation, we get $(a'm² + b')y² + c' = v²,$ which can be solved by the method of the Square-nature.

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*Example* from Bhâskara II:² $.7x² + 8y² = u² / 7x² - 8y² + 1 = v².$

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He solves it substantially as follows:

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In the first equation suppose x = 2y; then u = 6y. Putting x = 2y, the second equation becomes $20y² + 1 = v².$ By the method of the Square-nature the values of y satisfying this equation are 2, 36, etc. Hence the solutions of the given double equation are $x = 4, 72, / y = 2, 36,$

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² *BBi*, p. 119.

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*Case iv.* Form .a(x² ± y²) + c = u², / a'(x² ± y²) + c' = v².}

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Putting x² ± y² = z Bhâskara II reduces the above equations to .az + c = u², / a'z + c' = v²;} the method for the solution of which has been given before.

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*Example* with solution from Bhâskara II:¹ .2(x² - y²) + 3 = u² / 3(x² - y²) + 3 = v²}. Set x² - y² = z; then 2z + 3 = u², / 3z + 3 = v². Eliminating z we get 3u² = 2v² + 3, or (3u)² = 6v² + 9. Whence v = 6, 60, 3u = 15, 147, \dots Therefore u = 5, 49, Hence x² - y² = z = 11, 1199,

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¹ *BBi*, p. 119.

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Therefore, the required solutions are .x = 1/2(11/m + m) / y = 1/2(11/m - m)}, .x = 1/2(1199/m + m) / y = 1/2(1199/m - m)}, \dots where m is an arbitrary rational number.

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For m = 1, the values of (x, y) will be (6, 5), (600, 599), ... For m = 11, we get the solution (60, 49), ...

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*Case v.* For the solution of the double equation of the general form $.ax² + by² + c &= u² / a'x² + b'y² + c' &= v²$ Bhâskara II's hint¹ is: Find the values of x, u in the first equation in terms of y, and then substitute that value of x in the second equation so that it will be reduced to a Square-nature. He has, however, not given any illustrative example of this kind.

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¹ *Vide supra*, pp. 190f.

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**Second Type.** Another type of double equation of the second degree which has been treated is $.a²x² + bxy + cy² &= u², / a'x² + b'xy + c'y² + d' &= v².$

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The solution of the first equation has been given before to be $x &= 1/2a(y²)/(λ)(c - (b²)/(4a²)) - λ - by/(2a²), / u &= ½(y²)/(λ)(c - (b²)/(4a²)) + λ,$ where λ is an arbitrary rational number. Putting λ = y, we have $x &= y/2a(c - (b²)/(4a²) - 1) - by/(2a²) = α y, / u &= y/2(c - (b²)/(4a²) + 1);$ where $α = 1/2a(c - (b²)/(4a²) - 1) - b/(2a²).$

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Substituting in the second equation, we get $(a'α² + b'α + c')y² + d' = v²,$ which can be solved by the method of the Square-nature. This method is equally applicable if the unknown part in the second equation is of a different kind but still of the second degree.

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Bhâskara II gives the following illustrative example together with its solution:¹ .x² + xy + y² &= u² / (x + y)u + 1 &= v²}. Multiplying the first equation by 36, we get (6x + 3y)² + 27y² = 36u². Whence 6x + 3y &= 1/2l((27y²)/(\lambda) - \lambdar), / 6u &= 1/2l((27y²)/(\lambda) + \lambdar), where λ is an arbitrary rational number. Taking λ = y, we have 6x + 3y &= 13y, / x &= 5/3y, / u &= 7/3y. Substituting in the second equation, we get 56y² + 9 = 9v². By the method of the Square-nature the values of y are 6, 180, ...

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¹ *BBi*, pp. 107f.

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Hence the required values of (x, y) are (10, 6), (300, 180), ...

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