History of Hindu Mathematics: A Source Book v2.c17.s24: 26. Double Equations of Higher Degrees

History of Hindu Mathematics: A Source Book v2.c17.s24

Language: English

26. Double Equations of Higher Degrees

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There are a few problems which involve double equations of degrees higher than the second. The following examples are taken from Bhâskara II :

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*Example 1.* "The sum of the cubes (of two numbers) is a square and the sum of their squares is a cube. If you know them, then I shall admit that you are a great algebraist."¹

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We have to solve the equations $. x²+y² &= u³, / x³+y³ &= v².$

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The solution of this problem by Bhâskara II is as follows :

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"Here suppose the two numbers to be z², 2z². The sum of their cubes is 9z⁶. This is by itself a square and its square-root is 3z³.

"Now the sum of the squares of those two numbers is 5z⁴. This must be a cube. Assuming it to be equal to the cube of an optional multiple of 5z and removing the factor z³ from both sides (we get z = 25p³, where p is an optional number); so, as stated before, the numbers are (putting p = 1) 625, 1250. The assumption should be always such as will make it possible to remove (the cube of) the unknown."²

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In general, assume x = mz², y = nz²; substituting in the second equation, we have $x³ + y³ = (m³ + n³)z⁶ = v².$ If m³ + n³ = a square = p², say, then v = pz³.

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² *BBi*, pp. 56f.

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Now, from the first equation, we have $(m² + n²)z⁴ = u³.$ Assume u = rz; then $z = (r³)/(m² + n²).$ Hence we get $x = (mr⁶)/((m² + n²)²), y = (nr⁶)/((m² + n²)²);$ where r is any integer and m, n are such that $m³ + n³ = a square.$ One obvious solution¹ of this equation is m = 1, n = 2. Hence we get the solution $x = (r⁶)/25, y = (2r⁶)/25.$

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¹ Now m³ + n³ can be made a square by putting $m = (p³ + q³)p, n = (p³ + q³)q,$ so that $m^3 + n^3 = (p^3 + q^3)^4. Hence the solution of our equation will be x = (pr⁶)/((p² + q²)² (p³ + q³)³), y = (qr⁶)/((p² + q²)² (p³ + q³)³).$ Putting $r = (p^2 + q^2)(p^3 + q^3)s, we have the solution in positive integers as x = p(p² + q²)⁴ (p³ + q³)³ s⁶, y = q(p² + q²)⁴ (p³ + q³)³ s⁶;$ where $p$, $q$, $s$ are any integral numbers.

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This particular solution has been given by Nârâyaṇa, who says :

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"The square of the cube of an optional number is the one and twice it is the other. These divided by 25 will be the two numbers, the sum of whose squares will

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be a cube and the sum of whose cubes will be a square.”¹ He then adds by way of illustration :

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¹ *GK*, i. 50.

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“With the optional number 1, we get the two numbers (1/25, 2/25); with 2, (64/25, 128/25); with 5, (625, 1250); with 1/2, (1/1600, 1/800); with 1/3, (1/18225, 2/18225). Thus by virtue of (the multiplicity of) the optional number many solutions can be found.”

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*Example 2.* “O most learned algebraist, tell me those various pairs of whole numbers whose difference is a square and the sum of whose squares is a cube.”²

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That is to say, solve in positive integers .y - x &= u², / y² + x² &= v³.}

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Bhâskara II’s process of solving this problem is as follows :

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“Let the two numbers be x, y. Putting their difference, y - x, equal to u², we get the value of x as y - u². Having thus found the value of x, the two numbers become y - u², y.

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“The sum of their squares = 2y² - 2yu² + u⁴. This is a cube. Making it equal to u⁶ and transposing we get u⁶ - u⁴ = 2y² - 2yu². Multiplying both sides by 2 and superadding u⁴, we get the square-root of the second side = 2y - u², and the first side = 2u⁶ - u⁴. Dividing out by u⁴ (and putting w for 2y/u² - 1), we get 2u² - 1 = w². By the method of the Square-nature the roots of this equation are u &= 5, 29, &\dots / w &= 7, 41, &\dots "Then by the rule, 'Or, if a biquadratic factor has been removed, the greater root should be multiplied by the square of the lesser root,'¹ we get

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² *BBi*, p. 103.

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¹ The reference is to the rule on p. 249.

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$2y - 25 = 175,$ or $2y - 841 = 34481.$ Therefore $y = 100, 17661,$

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"Finding the respective values of the numbers, they are (75, 100), (16820, 17661), etc."

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*Example 3.* "Bring out quickly those two numbers of which the sum of the cube (of one) and the square (of the other) becomes a square and whose sum also is a square."²

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That is to say, solve $x³ + y² = u², & (1) / x + y = v². & (2)$

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This problem has been solved by Bhâskara II in two ways, which are substantially as follows:

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*First method.* From (1) we get

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$u = ½((x³)/(λ) + λ), y = ½((x³)/(λ) - λ),$

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where λ is an arbitrary number. Putting λ = x, we get $u = ½(x² + x), y = ½(x² - x).$ Substituting this value of y in (2), we get $x² + x = 2v²,$ or $(2x + 1)² = 8v² + 1.$

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² *BBi*, p. 107.

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By the method of the Square-nature we have $.v &= 6 / 2x + 1 &= 17, .v &= 35 / 2x + 1 &= 99,$ Whence the values of (x, y) are (8, 28), (49, 1176),

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*Second Method.* Assume x = 2w², y = 7w². Then $x + y = 9w² = (3w)².$ So the equation (2) is satisfied. Now, substituting those values in (1) we get $8w⁶ + 49w⁴ = u²,$ or $w⁴(8w² + 49) = u².$ If $8w² + 49 = z²,$ then $u = zw².$ Now the values of w making 8w² + 49 a square are 2, 3, 7 Hence the required numbers (x, y) are (8, 28), (18, 63), (98, 343),

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*Example 4.* "What is that number which multiplied by three and added with unity becomes a cube; the cube-root squared and multiplied by three becomes, together with unity, a square."¹

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¹ *BBi*, p. 119. This problem is admittedly taken by Bhâskara II from an earlier writer.

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That is to say, solve $3x + 1 = u³, & (1) / 3u² + 1 = v². & (2)$ It has been solved by Bhâskara II thus:

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From (2), by the method of the Square-nature, we get the values of (u, v) as (1, 2), (4, 7), (15, 26), Whence the values of x are 21, 3374/3, Alternatively¹ we assume u = 3y + 1; then from the equation (1) we get x = 3y(3y² + 3y + 1). Also from (2) we have 27y² + 18y + 4 &= v² / &= (my - 2)², say. Hence y = (18 + 4m)/(m² - 27). Therefore, the required value of x is x = 9((18 + 4m)/(m² - 27))³ + 9((18 + 4m)/(m² - 27))² + 3((18 + 4m)/(m² - 27)), where m is a rational number greater than 5. The first of the previous solutions is given by m = 9.

Linked texts (2)

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