History of Hindu Mathematics: A Source Book v2.c17.s17.t2.10

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."