History of Hindu Mathematics: A Source Book v2.c17.s17: 19. General Indeterminate Equations of the Second Degree: Single Equations
History of Hindu Mathematics: A Source Book v2.c17.s17
- Language
- English
Contents
- v2.c17.s17.t1 (i) Solution of ax² + bx + c = y — 18 parts
- v2.c17.s17.t2 (ii) Solution of ax² + bx + c = a'y² + b'y + c' — 22 parts
- v2.c17.s17.t3 (iii) Solution of ax² + by² + c = z — 56 parts
- v2.c17.s17.t4 (iv) Solution of a²x² + by² + c = z — 23 parts
- v2.c17.s17.t5 (v) Solution of ax²+bxy+cy²=z — 30 parts
19. General Indeterminate Equations of the Second Degree: Single Equations
v2.c17.s17.1
The earliest mention of the solution of the general indeterminate equation of the second degree is found in the *Bîjagaṇita* of Bhâskara II (1150). But there are good grounds to believe that he was not its first discoverer, for he is found to have taken from certain ancient authors a few illustrative examples the solutions of which presuppose a knowledge of the solution of such equations.² Neither those illustrations nor a treatment of equations of those types occurs in the algebra of Brahmagupta or in any other extant work anterior to Bhâskara II.
v2.c17.s17.2
² *Vide infra*, pp. 267f.
v2.c17.s17.3
Bhâskara II distinguishes two kinds of indeterminate equations: *Sakṛt samîkaraṇa* (Single Equations) and *Asakṛt samîkaraṇa* (Multiple Equations).³
v2.c17.s17.4
³ *BBi*, pp. 106, 110.
v2.c17.s17.5
**Solution.** For the solution of the general indeterminate equation of the second degree, Bhâskara II (1150) lays down the following rule:
v2.c17.s17.6
"When the square, etc., of the unknown are present (in an equation), after the equi-clearance has been made, (find) the square-root of one side by the method described before for it, and the root of the other side by the method of the Square-nature. Then (apply) the method of (simple) equations to these roots. If (the other side) does not become a case for the Square-nature, then, putting it equal to the square of another unknown, the other side and so the value of the other (*i.e.*, the new) unknown should be obtained in the same way as in the Square-nature ; and similarly the value of the first unknown. The intelligent should devise various artifices so that it may become a matter for (the application of) the Square-nature."¹
v2.c17.s17.7
¹ *BBi*, p. 99.
v2.c17.s17.8
He has further elucidated the rule thus :
v2.c17.s17.9
"When, after the clearance of the two sides has been made, there remain the square, etc., of the unknown, then, by multiplying the two sides with a suitable number and by the help of other necessary operations as described before, the square-root of one side should be extracted. If there be present on the other side the square of the unknown with an absolute term, then the two roots of that side should be found by the method of the Square-nature. There the number associated with the square of the unknown is the *prakṛti* ('multiplier'), and the absolute number is to be considered as the interpolator. What is obtained as the lesser root in this way will be the value of the unknown associated with the multiplier (*prakṛti*); the greater root is (again) the root of that square (formed on the first side). Hence making an equation of this with the square-root of the first side, the value of the unknown on the first side should be determined.
v2.c17.s17.10
"But if there be present on the second side the square of the unknown together with (the first power of) the unknown, or only the (simple) unknown with or without an absolute number, then it is not a case for the Square-nature. How then is the root to be found in that case? So it has been said: 'If (the other side) does not become a case for the Square-nature etc.' Then, putting it equal to the square of another unknown, the square-root of one side should be found in the way indicated before, and the two roots of the other side should then be determined by the method of the Square-nature. There again the lesser root is the value of the unknown associated with the *prakṛti* and the greater root is equal to the square-root of that side of the equation. Forming proper equations with the roots, the values of the unknowns should be determined.
v2.c17.s17.11
"If, however, even after the second side has been so treated, it does not turn out to be a case for the Square-nature, then the intelligent (mathematicians) should devise by their own sagacity all such artifices as will make it a case for the method of the Square-nature and then determine the values of the unknowns."\∗1
v2.c17.s17.12
\∗1 1 *BBi*, p. 100.
v2.c17.s17.13
Having thus indicated in a general way the broad outlines of his method for the solution of the general indeterminate equation of the second degree, Bhâskara II discusses the different types of equations severally, explaining the rules in every case in greater detail with the help of illustrative examples.
(i) Solution of ax² + bx + c = y
v2.c17.s17.t1.1
For the general solution of the quadratic indeterminate equation $ax² + bx + c = y², 1$
v2.c17.s17.t1.2
Bhâskara II gives the following particular rule :
v2.c17.s17.t1.3
“On taking the square-root of one side, if there be on the second side only the square of the unknown together with an absolute number, in such cases, the greater and lesser roots should be determined by the method of the Square-nature. Of these two, the greater root is to be put equal to the square-root of the first side mentioned before, and thence the value of the first unknown should be determined. The lesser will be the value of the unknown associated with the *prakṛti*. In this way, the method of the Square-nature should be applied to this case by the intelligent.”
v2.c17.s17.t1.4
As an illustration of this rule Bhâskara II works out in detail the following example:
v2.c17.s17.t1.5
“What number being doubled and added to six times its square, becomes capable of yielding a squareroot? O ye algebraist, tell it quickly.”
v2.c17.s17.t1.6
*Solution.* “Here let the number be x. Doubled and together with six times its square, it becomes 6x² + 2x. This is a square. On forming an equation with the square of y, the statement is $6x² + 2x + 0y² = 0x² + 0x + y².$ On making equi-clearance in this the two sides are 6x² + 2x and y².
v2.c17.s17.t1.7
“Then multiplying these two sides by 6 and superadding 1, the root of the first side, as described before, is 6x + 1.
v2.c17.s17.t1.8
“Now on the second side of the equation remains 6y² + 1. By the method of the Square-nature, its roots are : the lesser 2 and the greater 5, or the lesser 20 and the greater 49. Equating the greater root with the square-root of the first side, *viz.*, 6x + 1, the value of x is found to be 2/3 or 8. The lesser root, 2 or 20, is the value of y, the unknown associated with the *prakṛti*. In this way, by virtue of (the multiplicity of) the lesser and greater roots, many solutions can be obtained."¹
v2.c17.s17.t1.9
¹ *BBi*, pp. 100-1.
v2.c17.s17.t1.10
² *BBi*, p. 101.
v2.c17.s17.t1.11
¹ *BBi*, p. 101.
v2.c17.s17.t1.12
In other words the method described above is this:
v2.c17.s17.t1.13
Completing the square on the left-hand side of the equation ax² + bx + c = y², we have $(ax + ½b)² = ay² + ¼(b² - 4ac).$ Putting z = ax + ½b, k = ¼(b² - 4ac), we get $ay² + k = z². 1.1$ If y = l, z = m be found empirically to be a solution of this equation, another solution of it will be $y = lq ± mp,$ $z = mq ± alp,$ where ap² + 1 = q². Hence a solution of (1) is $x = - b/2a + 1/a(mq ± alp),$ $y = lq ± mp.$
v2.c17.s17.t1.14
Now suppose x = r, when z = m; that is, let m = ar + b/2. Substituting in the above expressions, we get the required solution of (1) as $. x &= 1/2a(bq - b) + qr ± lp, / y &= lq ± (apr + ½bp); 1.2$ where ap² + 1 = q² and ar² + br + c = l².
v2.c17.s17.t1.15
Thus having known *one* solution of ax² + bx + c = y², an infinite number of other solutions can be easily obtained by the method of Bhâskara II. The method is, indeed, a very simple and elegant one. It has been adopted by later Hindu algebraists. As the relevant portion of the algebra of Nârâyaṇa (1350) is now lost, we cannot reproduce his description of the method. Jñânarâja (1503) says :
v2.c17.s17.t1.16
"(Find) the square-root of the first side according to the method described before and, by the method of the Square-nature, the roots of the other side, where the coefficient of the square of the unknown is considered to be the *prakṛti* and the interpolator is an absolute term. Then the greater root will be equal to the previous square-root and the other (*i.e.*, the lesser root) to the unknown associated with the *prakṛti*."
v2.c17.s17.t1.17
The above solution (1.2), but with the upper sign only, was rediscovered in 1733 by Eluer.¹ His method is indirect and cumbrous. Lagrange's (1767) method begins in the same way as that of Bhâskara II by completing the square on the left-hand side of the equation.²
v2.c17.s17.t1.18
¹ Leonard Euler, *Opera Mathematica*, vol. II, 1915, pp. 6-17 ; Compare also pp. 576-611.
(ii) Solution of ax² + bx + c = a'y² + b'y + c'
v2.c17.s17.t2.1
Bhâskara II has treated the more general type of quadratic indeterminate equations : $ax² + bx + c = a'y² + b'y + c'. 2$ His rule in this connection runs as follows :
v2.c17.s17.t2.2
² Additions to *Elements of Algebra* by Leonard Euler, translated into English by John Hewlett, 5th edition, London, 1840, pp. 537ff.
v2.c17.s17.t2.3
"If there be the square of the unknown together with the (simple) unknown and an absolute number, putting it equal to the square of another unknown its root (should be investigated). Then on the other side (find)
v2.c17.s17.t2.4
the roots by the method of the Square-nature, as has been stated before. Put the lesser root¹ equal to the root of the first side and the greater root equal to that of the second."²
v2.c17.s17.t2.5
¹ The meaning of the terms ‘lesser root’, ‘greater root’, etc., as used here, will be clear from the illustration and the general solution given below.
v2.c17.s17.t2.6
He further elucidates the rule thus:
v2.c17.s17.t2.7
“In this case, on taking the square-root of the first side, there remain on the other side the square of the unknown and the (simple) unknown with or without an absolute number. In that case forming an equation of the second side with the square of another unknown, the roots (should be found). Of these (roots just determined), making the lesser equal to the root of the first side (of the given equation) and the greater to the root of the second side, the values of the unknowns should be determined.”
v2.c17.s17.t2.8
*Example.* “Say what is the number of terms of a series (in A. P.) whose first term is 3, the common difference is 2; but whose sum multiplied by 3 is equal to the sum of a different number of terms.”³
v2.c17.s17.t2.9
³ *BBi*, p. 104.
v2.c17.s17.t2.10
*Solution.* “Here the statements of the series are: first term =3, common difference =2, number of terms =x; first term =3, common difference =2, number of terms =y. The two sums are (respectively) x²+2x, y²+2y. Making three times the first equal to the second, the statement for clearance is $3x²+6x=y²+2y.$ After the clearance, multiplying the two sides (of the equation) by 3 and superadding 9, the square-root of the first side is 3x+3. On the second side of the equation stands 3y²+6y+9. Forming an equation of this with z², and similarly multiplying the sides by 3 and superadding -18, the root of it is 3y+3. Then the roots of the other side, 3z²-18, by the method of the Square-nature are the lesser =9 and greater =15, or the lesser =13 and greater =57. Equating the lesser root with the square-root of the first side, namely, 3x+3, and the greater root with the square-root of the second side, namely, 3y+3, the values of x, y are found to be (2, 4) or (10, 18). So in every case."
v2.c17.s17.t2.11
² *BBi*, p. 104.
v2.c17.s17.t2.12
In general, on completing the square on the left-hand side, equation (2) becomes $(ax+½b)² = aa'y² + ab'y + ac' + (¼b² - ac).$
v2.c17.s17.t2.13
Put $ax+½b=z,$ @(2.1) and then complete the square on the right-hand side. Thus the given equation is finally reduced to $aa'z² - β = w²,$ @(2.2) where $w = aa'y + ½ab',$ @(2.3) and $β = a²a'c' + (¼b² - ac)aa' - (½ab')².$
v2.c17.s17.t2.14
Now, if z = l, w = m be a solution of the equation (2.2), another solution will be z = lq ± mp, w = mq ± aa'lp; where a'ap² + 1 = q². Substituting in (2.1) and (2.3), we get . x &= -b/2a + 1/a(lq ± mp), / y &= -(b')/(2a') + 1/(aa')(mq ± aa'lp). } @(2.4)
v2.c17.s17.t2.15
Now, let l = ar + ½b and m = aa's + ½ab'. Substituting in the above expressions, we get the required solution of (2) in the form : $. x &= 1/2a(qb ± pab' - b) + qr ± pa's, / y &= 1/(2a')(qb' ± pa'b - b') + qs ± par ; (2.5)$ where aa'p² + 1 = q², and ar² + br + c = a's² + b's + c'.
v2.c17.s17.t2.16
The form (2.5) shows that having found empirically one solution of ax² + bx + c = a'y² + b'y + c' Bhâskara could find an infinite number of other solutions of it.
v2.c17.s17.t2.17
Jñânarâja (1503) says :
v2.c17.s17.t2.18
"If on the other side be present the square as well as the linear power of the unknown together with an absolute term, put it equal to the square of another unknown and then determine the lesser and greater roots. The lesser root will be equal to the first square-root and the greater to the second square-root."
v2.c17.s17.t2.19
He gives with solution the following illustrative example : $3(x² + 4x) = y² + 4y,$ or (3x + 6)² = 3y² + 12y + 36.
v2.c17.s17.t2.20
Putting 3x + 6 = z, where z is the "first square-root" of Jñânarâja, we get $z² = 3y² + 12y² + 36,$ or 3z² = (3y + 6)² + 72.
v2.c17.s17.t2.21
Now put 3y + 6 = w, where w is the "second square-root." Then $3z² - 72 = w².$
v2.c17.s17.t2.22
Therefore, by the method of the Square-nature, z = 18, w = 30. Whence x = 4, y = 8, is a solution.
(iii) Solution of ax² + by² + c = z
v2.c17.s17.t3.1
Bhâskara II followed several devices for the solution of the equation $ax² + by² + c = z². 3$ In every case his object was to transform the equation into the form of the Square-nature. He says :
v2.c17.s17.t3.2
"In such cases, where squares of two unknowns with (or without) an absolute number are present, supposing either of them optionally as the *prakṛti*, the rest (of the terms) should be considered as the interpolator. Then the roots should be investigated in the way described before. If there be more equations than one (the process will be especially helpful)."¹
v2.c17.s17.t3.3
¹ *BBi*, pp. 105f.
v2.c17.s17.t3.4
He then explains further :
v2.c17.s17.t3.5
"Where on finding the square root of the first side, there remain on the other side squares of two unknowns with or without an absolute number, there consider the square of one of the unknowns as the *prakṛti*; the remainder will then be the interpolator. Then by the rule: 'An optionally chosen number is taken as the lesser root, etc.,'² the unknown in the interpolator multiplied by one, etc., and added with one, etc., or not, according to one's own sagacity, should be assumed for the lesser root; then determine the greater root."³
v2.c17.s17.t3.6
² The reference is to the rule for solving the Square-nature (*vide supra* p. 144) (*BBi*, p. 33).
v2.c17.s17.t3.7
³ *BBi*, p. 106.
v2.c17.s17.t3.8
There are thus indicated two artifices for solving the equation (3). They are :
v2.c17.s17.t3.9
(i) Set x = my; so that equation (3) transforms into $z² = (am² + b) y² + c$ $= α y² + c,$ where α = am² + b. Hence the required solution of ax² + by² + c = z² is $x = my = m(rq ± ps),$ $y = rq ± ps,$ $z = sq ± α pr;$ where s² = α r² + c and q² = α p² + 1.
v2.c17.s17.t3.10
*(ii)* Set x = my ± n; then the equation reduces to $z² = α y² ± 2amny + γ$ where α = am² + b and γ = an² + c.
v2.c17.s17.t3.11
Completing the square on the right-hand side of this, we get $α z² - β = w²,$ where w = α y ± amn and β = γα - a² m² n² = a(bn² + cm²) + bc.
v2.c17.s17.t3.12
If z = s, w = r be a solution of this equation, another solution will be $z = sq ± rp,$ $w = rq ± α sp;$ where q² = α p² + 1. Hence the solution of ax² + by² + c = z² is $x = m/(α) (rq ± α sp amn) ± n,$ $y = 1/(α) (rq ± α sp amn),$ $z = sq ± rp;$ where q² = α p² + 1, r² = α s² - β, α = am² + b and β = a(bn² + cm²) + bc.
v2.c17.s17.t3.13
In working certain problems, Bhâskara II is found to have occasionally followed other artifices also for the solution of the equation (3). For instance :
v2.c17.s17.t3.14
(*iii*)¹ Set w² = by² + c. Then equation (3) becomes $z² - w² = ax².$
v2.c17.s17.t3.15
¹ See *BBi*, p. 108.
v2.c17.s17.t3.16
Whence $z = ½ (a/m + m)x,$
v2.c17.s17.t3.17
and $w = ½ (a/m - m)x;$ where m is an arbitrary number. Therefore $x = 2mw/(a - m²),$ $z = ((a + m²)/(a - m²))w.$
v2.c17.s17.t3.18
Now, if y = l, w = r be a solution of $w² = by² + c,$ another solution of it will be $y = lq ± pr,$ $w = rq ± blp;$ where ap² + 1 = q². Therefore, the solution of (3) will be $x = 2m/(a - m²) (rq ± blp),$ $y = lq ± pr,$ $z = (a + m²)/(a - m²)(rq ± blp);$ where ap² + 1 = q² and bl² + c = r².
v2.c17.s17.t3.19
(*iv*)² Suppose c = 0; then the equation to be solved will be $ax² + by² = z².$
v2.c17.s17.t3.20
² *BBi*, p. 106.
v2.c17.s17.t3.21
In this case set x = uy, z = vy; so that u, v will be given by $au² + b = v²,$ which can be solved by the method of the Square-nature. Some of these devices were followed also by later Hindu algebraists, for instance, Jñânarâja (1503) and Kamalâkara (1658).¹
v2.c17.s17.t3.22
¹ *SiTVi*, xiii. 260-1.
v2.c17.s17.t3.23
*Example* from Kamalâkara:² $7x² + 8y² = z².$ This is one of a double equation by Bhâskara II.³
v2.c17.s17.t3.24
³ *BBi*, p. 106.
v2.c17.s17.t3.25
To solve ax² + by² + c = z², Kamalâkara observes:
v2.c17.s17.t3.26
² *SiTVi*, xiii. 258.
v2.c17.s17.t3.27
"In this case, suppose the coefficient of the square of the first unknown as the *prakṛti* and the coefficient of the square of the other unknown together with the absolute number as the interpolator to that. The two roots can thus be determined in several ways."⁴ And again:
v2.c17.s17.t3.28
⁴ *SiTVi*, xiii. 264.
v2.c17.s17.t3.29
"(Suppose) the coefficient of the square of one of the unknowns as the *prakṛti* and the rest comprising two terms, the square of an unknown and an absolute number, as the interpolator. Then assume the value of the lesser root to be equal to the other unknown together with an absolute term."⁵
v2.c17.s17.t3.30
⁵ *SiTVi*, xiii. 267 f.
v2.c17.s17.t3.31
He seems to have indicated also a slightly different method:
v2.c17.s17.t3.32
"Or assume the value of the lesser root to be equal to another unknown plus or minus an absolute number and similarly also the value of the greater root. The remaining operations should be performed by the intelligent in the way described by Bhâskara in his algebra."¹
v2.c17.s17.t3.33
¹ *SiTVi*, xiii. 265.
v2.c17.s17.t3.34
That is to say, assume $x = mw ± α, z = nw ± β.$
v2.c17.s17.t3.35
Substituting in the equation ax² + by² + c = z², we get $(am² - n²)w² ± 2w (ama nβ) + by²$ $+ (c + aa² - β²) = 0.$
v2.c17.s17.t3.36
Putting λ = am² - n², μ = ama nβ, ν = c + aa² - β², this equation can be reduced to $- λ by² + (μ² - νλ) = u²,$ where u = λ w ± μ.
v2.c17.s17.t3.37
Kamalâkara gives also some other methods which are applicable only in particular cases.
v2.c17.s17.t3.38
*Case i.* Suppose that b and c are of different signs.² Two sub-cases arise:
v2.c17.s17.t3.39
(1) *Form* ax² + by² - c = z².
v2.c17.s17.t3.40
First find u, v, says Kamalâkara, such that $au² - c = v².$
v2.c17.s17.t3.41
Assuming $x = √(b/ac) vy + u,$ we have $ax² + by² - c = b/c v²y² + 2√(ab/c) uvy + by²$ $+ (au² - c)$ $= b/c (au² - c) y² + 2√(ab/c) uvy + by² + v²$
v2.c17.s17.t3.42
² *SiTVi*, xiii. 285-7.
v2.c17.s17.t3.43
$= ab/c u² y² + 2 √(ab/c) uvy + v²$ $= (√(ab/c) uy + v)².$
v2.c17.s17.t3.44
Hence z = √(ab/c) uy + v.
v2.c17.s17.t3.45
The following illustrative example and its solution are given :¹ $5x² + 16y² - 20 = z².$ Its solution is $x = ⅖vy + u,$ $z = 2uy + v;$ where 5u² - 20 = v². An obvious solution of this equation is given by u = 3, v = 5. Hence, we get a solution of the given equation as $x = 2y + 3,$ $z = 6y + 5.$ Therefore (x, y) = (5, 1), (7, 2), (9, 3),
v2.c17.s17.t3.46
¹ *SiTVi*, xiii. 279.
v2.c17.s17.t3.47
(2) *Form* ax² - by² + c = z². In this case first solve $au'² + c = v'².$ Then the required solution is $x = √(b/ac) v'y + u',$ $z = √(ab/c) u'y + v'.$
v2.c17.s17.t3.48
*Example from Kamalâkara* :² $5x² - 20y² + 16 = z².$
v2.c17.s17.t3.49
² *SiTVi*, xiii. 279.
v2.c17.s17.t3.50
Then
v2.c17.s17.t3.51
x &= 1/2v'y + u', \\ z &= 5/2u'y + v'; \end{align*} where 5u'² + 16 = v'². One solution of this equation is u' = 2, v' = 6. The corresponding solution of the given equation is \begin{align*} x &= 3y + 2, \\ z &= 5y + 6. \end{align*}
v2.c17.s17.t3.52
Therefore (x, y) = (5, 1), (8, 2), (11, 3), etc.
v2.c17.s17.t3.53
*Case ii.* Let the two terms of the interpolator be of the same sign and positive.
v2.c17.s17.t3.54
*Example* from Kamalâkara:¹ $5x² + 8y² + 23 = z².$
v2.c17.s17.t3.55
¹ *SiTVi*, xiii. 296.
v2.c17.s17.t3.56
Assume arbitrarily a value of x or y and then find the other by the method of the Square-nature.²
(iv) Solution of a²x² + by² + c = z
v2.c17.s17.t4.1
Let the coefficient of x² (or y²) be a square number. The equation is of the form $a²x² + by² + c = z².$
v2.c17.s17.t4.2
For this case Bhâskara II observes:
v2.c17.s17.t4.3
"If the *prakṛti* is a square, then obtain the roots by the rule: 'The interpolator divided by an optional number is set down at two places, etc.' "³
v2.c17.s17.t4.4
³ *BBi*, p. 106.
v2.c17.s17.t4.5
Thus, according to Bhâskara II, the solution of the above equation is $x = 1/2a((by² + c)/m) - m,$
v2.c17.s17.t4.6
² *SiTVi*, xiii. 298.
v2.c17.s17.t4.7
$z = ½((by²+c)/m+m);$ where m is an arbitrary number.
v2.c17.s17.t4.8
Kamalâkara divides equations of this form into two classes according as c is or is not a square.¹
v2.c17.s17.t4.9
¹ *Vide* his gloss on *SiTVi*, xiii. 275.
v2.c17.s17.t4.10
(1) Let c be a square (= d², say). That is to say, we have to solve $a²x² + by² + d² = z².$
v2.c17.s17.t4.11
The solution of this particular case, says Kamalâkara, is given by $x = b/2ady².$
v2.c17.s17.t4.12
For, with this value, we have $z² &= (b²y⁴)/(4d²) + by² + d² / &= 1/(4d²) (by² + 2d²)².$
v2.c17.s17.t4.13
Hence $z = (by²)/2d + d.$
v2.c17.s17.t4.14
(2) When c is not a square, Kamalâkara first finds α, β such that $α² + c = β².$
v2.c17.s17.t4.15
He next obtains n such that the value of $(αβ n - b/2)/(a⁴. α/a)$ is also n; and then says that $x = ny² + (α)/a;$ whence will follow the value of z.
v2.c17.s17.t4.16
Since & & (a\beta n - b/2)/(a\alpha) &= n, \\ we get & & n &= b/(2a(\beta - \alpha)); \\ so that & & x &= (by²)/(2a(\beta - \alpha)) + (\alpha)/a. \end{aligned} Therefore z² = a²x² + by² + c &= (b²y⁴)/(4(\beta - \alpha)²) + \alpha² + (\alpha by²)/((\beta - \alpha)) + by² + c / &= (b²y⁴)/(4(\beta - \alpha)²) + (\beta by²)/((\beta - \alpha)) + \beta² / &= { (by²)/(2(\beta - \alpha)) + \beta }². Hence z = (by²)/(2(β - α)) + β.
v2.c17.s17.t4.17
*Example from Kamalâkara* : 4x² + 48y² + 20 = z². Since 4² + 20 = 6², and the solution of (12n - 24)/8 = n is n = 6, we get the required solution of the given equation as x &= 6y² + 2, / z &= 12y² + 6.
v2.c17.s17.t4.18
It may be noted that the solution stated by Kamalâkara follows easily from that of Bhâskara II, on putting therein m = β - α, where α² + c = β².
v2.c17.s17.t4.19
In particular, if we put c=0 and a=1, Bhâskara's solution reduces to
v2.c17.s17.t4.20
x &= 1/2(b/my²-m), \\ z &= 1/2(b/my²+m); \end{align*} where m is arbitrary.
v2.c17.s17.t4.21
If b=αβ, taking m=2β p², we easily arrive at \begin{align*} x &= \alpha q²-\beta p², \\ y &= 2pq, \\ z &= \alpha q²+\beta p², \end{align*} where p, q are arbitrary integers, as the solution in positive integers of z²=x²+by². This solution was given by A. Desboves (1879).¹
v2.c17.s17.t4.22
¹ Dickson, *Numbers*, II, p. 405.
v2.c17.s17.t4.23
Taking m=2ν², we can derive Matsunago's (*c.* 1735) solution of z²=x²+by², *viz.*, \begin{align*} x &= b\mu²-\nu², \\ y &= 2\mu\nu, \\ z &= b\mu²+\nu²; \end{align*} where μ, ν are arbitrary integers.
(v) Solution of ax²+bxy+cy²=z
v2.c17.s17.t5.1
For the solution of the equation $ax²+bxy+cy²=z² 5$ Bhâskara II lays down the following rule :
v2.c17.s17.t5.2
"When there are squares of two unknowns together with their product, having extracted the square-root of one part, it should be put equal to half the difference of the remaining part divided by an optional number
v2.c17.s17.t5.3
and the optional number.”¹
v2.c17.s17.t5.4
¹ Y. Mikami, *The Development of Mathematics in China and Japan*, Leipzig, 1913, p. 231.
v2.c17.s17.t5.5
It has been again elucidated thus :
v2.c17.s17.t5.6
“Where there exists also the product of the unknowns (in addition to their squares), by the rule, ‘when there are squares etc.,’ the square-root of as much portion of it as affords a root, should be extracted. The remaining portion, divided by an optional number and then diminished by that optional number and halved, should be put equal to that square-root.”²
v2.c17.s17.t5.7
The above rule, in fact, contemplates a particular case of equation (5) in which a or c is a square number.
v2.c17.s17.t5.8
(*i*) Suppose a = p². The equation to be solved is then $p²x² + bxy + cy² = z². 5.1$
v2.c17.s17.t5.9
Therefore (px + by/2p)² + y²(c - (b²)/(4p²)) = z².
v2.c17.s17.t5.10
Putting px + by/2p = w, we get $z² - w² = y²(c - (b²)/(4p²)).$
v2.c17.s17.t5.11
Whence $z - w = λ,$ $z + w = (y²)/(λ)(c - (b²)/(4p²)),$ where λ is an arbitrary rational number. So $w = ½(y²)/(λ)(c - (b²)/(4p²)) - λ,$ as stated in the rule. Therefore, $x = 1/2p(y²)/(λ)(c - (b²)/(4p²)) - λ - by/(2p²),$
v2.c17.s17.t5.12
² *BBi*, p. 106.
v2.c17.s17.t5.13
$z = ½(y²)/(λ)(c - (b²)/(4p²)) + λ.$
v2.c17.s17.t5.14
Now, if we suppose y = m/n, where m, n are arbitrary integers, we get the solution of (5.1) as x &= 1/(8\lambda p²n²) {m²(4cp² - b²) - 4\lambda² p²n² - 4\lambda bmn}, / y &= m/n, / z &= 1/(8\lambda p²n²) {m²(4cp² - b²) + 4\lambda² p²n²}.
v2.c17.s17.t5.15
Since the given equation is homogeneous, any multiple of these values of x, y, z will also be its solution. Therefore, multiplying by 8λ p² n², we get the following solution of the equation p²x² + bxy + cy² = z² in integers : .x &= m² (4cp² - b²) - 4\lambda²p²n² - 4\lambda bmn, / y &= 8\lambda mnp², / z &= m² (4cp² - b²) + 4\lambda²p²n²,} (5.2) where m, n, are arbitrary integers.
v2.c17.s17.t5.16
In particular, putting a = b = c = 1, and λ = p = 1 in (5.2), we get x &= 3m² - 4n(n + m), / y &= 8mn, / z &= 3m² + 4n², as the solution of the equation x² + xy + y² = z².
v2.c17.s17.t5.17
Dividing out by 8n, the above solution can be put into the form $x = ½((3m²)/4n - n - m),$
v2.c17.s17.t5.18
$y = m,$ $z = 1/8n (3m² + 4n²) ;$ as has been stated by Nârâyaṇa :
v2.c17.s17.t5.19
"An arbitrary number is the first. Its square less by its (square's) one-fourth, is divided by an optional number and then diminished by the latter and also by the first. Half the remainder is the second number. The sum of their squares together with their product is a square."¹
v2.c17.s17.t5.20
¹ *GK*, i. 55.
v2.c17.s17.t5.21
It is noteworthy that in practice Nârâyaṇa approves of only integral solutions of his equation. For instance, he says :
v2.c17.s17.t5.22
" 'Any arbitrary number is the first.' Suppose it to be 12. Then with the optional number unity, are obtained the numbers (12, 95/2). For *integral values*, they are doubled (24, 95). With the optional number 2, are obtained (12, 20). It being possible, these are reduced by the common factor 4 to (3, 5). In this way, owing to the varieties of the optional number, an infinite number of solutions can be obtained."²
v2.c17.s17.t5.23
(*ii*) If neither *a* nor *c* be a square, the solution can be obtained thus :
v2.c17.s17.t5.24
Multiplying both sides of the equation (5) by *a* and then completing a square on the left-hand side, the equation transforms into $(ax + ½by)² + (ac - ¼b²)y² = az².$ Putting ax + ½by = w and β = ¼(b² - 4ac), we get w² = az² + β y². (5.3)
v2.c17.s17.t5.25
² See the example in illustration of the same.
v2.c17.s17.t5.26
The method of the solution of an equation of this form, according to Bhâskara II, has been described before.
v2.c17.s17.t5.27
Assume w = vy, z = uy; so that the values of u, v will be given by $v² = α u² + β. 5.4$
v2.c17.s17.t5.28
If u = m, v = n be a solution of (5.4), another solution will be $u = mq ± pn,$ $v = nq ± α mp;$ where α p² + 1 = q². Therefore, a solution of (5) is $x = y/2a 2(nq ± α mp) - b,$ $z = y(mq ± pn);$ where α p² + 1 = q² and α m² + β = n².
v2.c17.s17.t5.29
Put n = ar + ½b and y = s/t; then we have $x = s/2at q(2ar + b) ± 2α mp - b,$ $y = s/t,$ $z = s/2t 2mq ± p (2ar + b).$
v2.c17.s17.t5.30
Multiplying by 2at, we get the following solution of ax² + bxy + cy² = z² in integers : $x = s q (2ar + b) ± 2α mp - b,$ $y = 2as,$ $z = as2mq ± p(2ar + b);$ where α p² + 1 = q² and m² = ar² + br + c.
Linked texts (3)
- Bījagaṇita — Bhāskara II, Sanskrit
- Siddhāntaśiromaṇi — Bhāskara II, Sanskrit
- Gaṇitakaumudī — Nārāyaṇa Paṇḍita (mathematician), Sanskrit