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