History of Hindu Mathematics: A Source Book v2.c17.s25.12

v2.c17.s25.12

Suppose $x = 2z²(m² + n²), y = 2z²(m² - n²),$ so that x ± y are already squares. We have, therefore, only to make $xy + 1 = 4z⁴(m² + n²)(m² - n²) + 1 = a square.$ Now $4z⁴(m⁴ - n⁴) + 1 = (2z²m² - 1)² + 4z²(m² - z²n⁴).$ Hence, in order that xy + 1 may be a square, one sufficient condition is $m² = z²n⁴.$ Therefore 2z² = (2m²)/(n⁴) = ((m² + n²) + (m² - n²))/([½(m² + n²) - (m² - n²)]²).