History of Hindu Mathematics: A Source Book v2.c17.s25.76

v2.c17.s25.76

Now, we find that $x² - y² + 12 &= (w² + 2w - 3)², / x² + y² - 4 &= (w² + 2w + 1)², / ½xy + y &= (w + 1)³.$ Then the remaining condition reduces to $2w² + 7w + 3 = q².$ Completing the square on the left-hand side, we get $(4w + 7)² = 8q² + 25.$ Whence by the method of the Square-nature, we get $.q &= 5 / 4w + 7 &= 15, .q &= 175 / 4w + 7 &= 495,$ Therefore w = 2, 122, Hence (x, y) = (7, 6), (15127, 246),