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Although there were great advancements in many areas of algebra, and the solution of many different forms of equations, from simple forms such as *ax* + *c* = *by*, to forms as complex as *ax*^{2} + *bxy* + *cy*^{2} = *z*^{2} , I have chosen to look in slightly more detail at solution(s) to the so called Pell's equation, *Nx*^{2} + 1 = *y*^{2}. The equation was so called due to a mistake on the part of Euler, who attributed the solution of the equation to John Pell, a 17^{th} century English scholar, who actually only referred to the equation in a text he wrote on algebra.

The equation was nearly solved by Brahmagupta (628 AD) and the solution was improved by Bhaskara II (1150 AD), leading some historians, including C Srinivasiengar, to suggest:

The complete theory underlying the solution was expounded by Lagrange in 1767, and rests on the theory of continued fractions. It must be briefly noted how remarkable the achievements of Indian scholars were, given the time period in which equations of the Pell's type were studied. The Indian method involves an element of trail-process but contains no mention of continued fractions. Further to solving equations of the Pell's type to obtain solutions for the unknowns, Brahmagupta extended his method of solution to find square roots. This contribution is of huge interest as it is essentially the same method rediscovered and used by Newton and Raphson around 1690, which is known as the Newton-Raphson iterative method. Contained within this brief discussion is a small computer code for the...It is therefore fitting that this equation be called the Brahmagupta-Bhaskara equation.[CS, P 110]

The Pell's type of equation was known in India as *Varga Prakriti*, or "equation of the multiplied square", where prakriti means coefficient and refers to the coefficient *N* (where *N* is a positive integer). As previously mentioned, Bhaskara developed a Chakravala or cyclic method of solution.

Regrettably, due to constraints of space, I will have to forego discussion of the general solutions derived by Brahmagupta and Bhaskara, and will include only an example to illustrate Bhaskara's improved 'cyclic' method. The following example is of great historical interest. It is found in the Bijaganita of Bhaskara and is also of the form of a problem Fermat set as a problem to fellow mathematician Frenicle in 1657. The smallest solution for *x* and *y* runs into 4 and 5 digits respectively. The *chakravala* method is remarkable, as it requires just a few 'easy' steps, while Lagrange's solution required complex use continued fractions.

Example 8.6.1: Solution of 67*x*^{2} + 1 = *y*^{2}.

67By the Principle of Composition of Equals, we get from the above equation:x^{2}+ 1 =y^{2}.

Firstly the auxiliary equation 67 12 - 3 = 82 is taken.

Then using Bhaskara's lemma, whereNa^{2}+k=b^{2}, wherea,b,kare the integers (1, -3 and 8) in the auxiliary equation above, (kbeing positive or negative) then:Thus:N((am+b)/k)^{2}+ ((m^{2}-N)/k) = ((bm+Na)/k)^{2}67((1Then, by the method ofm+ 8)/-3)^{2}+ ((m2 - 67)/-3) = ((8m+ 67 1)/-3)^{2}(1)Kuttakathe solution of (m+ 8)/-3 = an integer, ism= -3t+ 1.

Puttingt= -2, we getm= 7, which makes [m^{2}- 67] least.

On substituting this value, the equation(1)reduces to:

67 52 + 6 = 412

Again, by the lemma:67((5Then the solution of (5n+ 41)/6)^{2}+ ((n^{2}-67)/6) = ((41n+ 67 5)/6)^{2}(2)n+ 41)/6 = a whole number, isn= 6t+ 5. [n^{2}- 67] will be least for the valuet= 0, that is, whenn= 5. The equation(1)then becomes:67 112 - 7 = 902Now we form:67((11The solution of (11p+ 90)/-7)^{2}+ ((p^{2}- 67)/-7) = ((90p+ 67 11/-7)^{2}(3)p+ 90)/-7 = an integral number, isp= -7t+ 2. Takingt= -1, we have

p= 9; and this value makes [p^{2}- 67] least. Substituting that into(3)we get:67 272 - 2 = 2212

67(2.27.221)Dividing out by 4 we have:^{2}+ 4 = (2212 + 67 272)^{2}

Or 67(11934)^{2}+ 4 = (97684)^{2}

67(5967)Hence^{2}+ 1 = (48842)^{2}

Example 8.6.2: Finding the square root of

IfN=5, theny^{2}= 1 + 5x^{2}. It can be observed that 5 = (y^{2}- 1)/x^{2}and that (y^{2}- 1)/x^{2}»y^{2}/x^{2}.

This is the key to finding the square root of 5, as 5 =y/x. With ease we can identifyy= 9 andx= 4 as solutions to this equation, and we see 5 » 9/4 = 2.25, (5 = 2.236067978...).

Clearly the largeryandxare, the better the approximation is. This can shown using the followingMapleprogramme:> n:=5:The output gives the following:

> f:=(x,y)->(2*x*y,y^{2+n*x2): >m:=0: >}x:=4

> y=:9

> while m 5 do

> m:=m+1;

> print(x,y,evalf(y/x,20), evalf(y/x-sqrt(n),50));

> a:=f(x,y);

> x:=a[1];

> y:=a[2];

> end do:

4, 9 (solutions ofxandy)

2.25000000000000000000 (y/xto 20 decimal places)

0.0139320225002103035908263312687237645593816403885 (error betweeny/xand 5)

By the 5^{th}step the following result is given:

25840354427429161536, 57780789062419261441 (5^{th}pair of solutions forxandy)

2.2360679774997896964 (y/xto 20 d.p.s)0.3348791201 10

^{-39}(error)This result is extremely accurate, and the method must be considered brilliant given it was derived by Brahmagupta in 628 AD.

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