ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

SOLUTION OF INDETERMINATE 259 Hence b₁ = (a₁b - 1) / a = a whole number. Now n² - N = ((a₁k - b)² - Na²) / a² = (a₁²k² - 2bka₁ + k) / a² = k(a₁²k - 2ba₁ + 1) / a² Therefore (k / a²)(a₁²k - 2ba₁ + 1) is a whole number. Since a, k have no common factor, it follows that (a₁²k - 2ba₁ + 1) / a² = (n² - N) / k = k₁ = an integer. Also k₁ = (n² - N) / k = (a₁²k - 2ba₁ + 1) / a² = (a₁²(b² - Na²) - 2ba₁ + 1) / a² = ((a₁b - 1) / a)² - Na₁². Thus having known a single solution in positive integers of the equation Nx² ± c = y², says, Brahmagupta, an infinite number of other integral solutions can be obtained by making use of the integral solutions of Nx² + 1 = y². If (p, q) be a solution of the former equation found empirically and if (α, β) be an integral solution of the latter, then by the principle of Composition x = pβ ± qα; y = qβ ± Npα will be a solution of the former. Repeating the operations, we can easily deduce as many solutions as we like. FORM Mn²x² ± c = y² : In this connection, Brahmagupta says : If the remainder is that divided by a square, the first root is that divided by its root¹. This seems to mean that if we have the equation Mn²x² ± c = y² (i) such that the multiplier (i.e. the coefficient of x²) is divisible

1. वर्गेच्छिन्ने गुणके प्रथमं तन्मूल भाजितं भवति । —BrSpSi. XVIII.70