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

SOLUTION IN POSITIVE INTEGERS 255
(ii) Let k=±2; then the auxiliary equation is
Nα²±2=β²
By Brahmagupta's Corollary, we have
N(2αβ)²+4=(β²+Nα²)²
or N (αβ)²+1=( (β²+Nα²)/2 )²
Hence the required first solution is
x=αβ, y=½(β²+Nα²)
Since Na²=β² ∓ 2,
we have ½ (β²+Nα²)=β² ∓ 1=a whole number.
(iii) Now suppose k=+4: so that
Nα²+4=β²
With an auxiliary equation like this, the first integral solution
of the equation Nx²+1=y² is
x=½αβ
y=½(β²-2);
if α is even; or
x=½