CHAP, xii] PELL EQUATION, ax 2 +bx+c=\3. 349 



take y c and y + c as the numbers, whence 



(2y) 2 + (2yY = 2(2y* + 67/c 2 ), (2y + I) 2 = 12c 2 + 1, 

 c = 2, 28; y = 3, 48. 



For 5t/ 4 - ICOi/ 2 = c 2 ( 181, p. 249), divide by y 2 . To find ( 182) two 

 numbers whose difference is a square, and sum of squares a cube, take c 

 and c n 2 as the numbers; the sum 2c 2 2cn 2 + n 4 of their squares is 

 equated to n 6 (a restriction), whence (2c n 2 } 2 = n 4 (2n 2 1), and 2n 2 1 

 is made a square. To make ( 188, p. 253) y 2 + z 3 and y + z both squares, 

 treat the first condition by 95 with z 3 as the additive and z as the arbitrary 

 number 6; we get y = (z 2 z}/2; the second condition now becomes 

 (z 2 + 2) = p 2 , or (2z + I) 2 = 8p 2 + 1, which is a square for p = 6 or 35. 

 The sum ( 189, p. 254) of the squares of two numbers increased by their 

 product is to be a square; on adding unity to the product of their sum by 

 the root of that square, the sum shall be a square. The first condition is 

 found to be satisfied by the numbers -fc and c; then the second condition 

 (fc)(fc) + 1 = D holds if c = 6 or 180. 



E. Strachey 31 translated into English the Persian manuscript of 1634 

 of Bhascara. To solve Ax 2 -\- B = y 2 , take any square/ 2 and find a number 

 such that Af 2 + is a square, say g 2 . Then x' = 2fg, y' = A} 2 + g 

 satisfy Ax' z + 0' = y' 2 Jor /3' = /3 2 ; and 



x" = s'flr y'f, 0" = fl'flf 

 satisfy 



2 



If j8" = 5p 2 , remove the factor p from x",y"; we get a solution of the pro- 

 posed equation (if /3" = B/p 2 , multiply by p). Otherwise, we proceed as 

 before. For example, consider 8x 2 + 1 = y 2 . Take / = 1 ; then 



8/ 2 + 1 = 3 2 , 

 so that we take = 1. Then 



x' = 2-1-3 = 6, y' = 8-P + 3 2 = 17, 8-6 2 + 1 = 17 2 . 

 A new set of solutions is given by 



x" = 6-3 + 17-1 = 35, y" = 17-3 + 8-6-1 = 99. 



For the cyclic method ("operation of circulation"), choose as before 

 [relatively prime] numbers / and g such that Af 2 + /3 = g 2 . Then by an 

 earlier rule [for solving a linear Diophantine equation] choose integers X, 

 Y such that (fX + g)l& = Y. Choose an integer m so that the difference 

 between (ra/3 + X) 2 and A shall be as small as possible numerically. Now 

 (wjS + X) 2 A is divisible by 0; call the quotient 0'. Set x' = mf + Y. 

 Then Ax' 2 + 13' is a square, say y' 2 . Unless p' = Bp 2 or B/p 2 , proceed as 

 before. For example, let A = 67, B = 1. Take / = 1, = - 3; then 

 g = 8, X = 1, F = - 3, m = - 2, 



-A = 7 2 -67 = -18 = /30 / , 0' = 6> a;' =-5, 2/' = 41. 



31 Bija Ganita, or the algebra of the Hindus, London, 1813, Introduction, pp. 36-53. 



